Let $p$ be an odd prime number and $\mathbb{Z}_{>0}$ be the set of positive integers. Suppose that a function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \rightarrow\{0,1\}$ satisfies the following properties:  - $f(1,1)=0$; - $f(a, b)+f(b, a)=1$ for any pair of relatively prime positive integers $(a, b)$ not both equal to 1 ; - $f(a+b, b)=f(a, b)$ for any pair of relatively prime positive integers $(a, b)$.  Prove that  $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right) \geqslant \sqrt{2 p}-2 $$

Solution: Denote by $\mathbb{A}$ the set of all pairs of coprime positive integers. Notice that for every $(a, b) \in \mathbb{A}$ there exists a pair $(u, v) \in \mathbb{Z}^{2}$ with $u a+v b=1$. Moreover, if $\left(u_{0}, v_{0}\right)$ is one such pair, then all such pairs are of the form $(u, v)=\left(u_{0}+k b, v_{0}-k a\right)$, where $k \in \mathbb{Z}$. So there exists a unique such pair $(u, v)$ with $-b / 2<u \leqslant b / 2$; we denote this pair by $(u, v)=g(a, b)$. Lemma. Let $(a, b) \in \mathbb{A}$ and $(u, v)=g(a, b)$. Then $f(a, b)=1 \Longleftrightarrow u>0$. Proof. We induct on $a+b$. The base case is $a+b=2$. In this case, we have that $a=b=1$, $g(a, b)=g(1,1)=(0,1)$ and $f(1,1)=0$, so the claim holds.  Assume now that $a+b>2$, and so $a \neq b$, since $a$ and $b$ are coprime. Two cases are possible. Case 1: $a>b$.  Notice that $g(a-b, b)=(u, v+u)$, since $u(a-b)+(v+u) b=1$ and $u \in(-b / 2, b / 2]$. Thus $f(a, b)=1 \Longleftrightarrow f(a-b, b)=1 \Longleftrightarrow u>0$ by the induction hypothesis. Case 2: $a<b$. (Then, clearly, $b \geqslant 2$.) Now we estimate $v$. Since $v b=1-u a$, we have  $$ 1+\frac{a b}{2}>v b \geqslant 1-\frac{a b}{2}, \quad \text { so } \quad \frac{1+a}{2} \geqslant \frac{1}{b}+\frac{a}{2}>v \geqslant \frac{1}{b}-\frac{a}{2}>-\frac{a}{2} . $$  Thus $1+a>2 v>-a$, so $a \geqslant 2 v>-a$, hence $a / 2 \geqslant v>-a / 2$, and thus $g(b, a)=(v, u)$. Observe that $f(a, b)=1 \Longleftrightarrow f(b, a)=0 \Longleftrightarrow f(b-a, a)=0$. We know from Case 1 that $g(b-a, a)=(v, u+v)$. We have $f(b-a, a)=0 \Longleftrightarrow v \leqslant 0$ by the inductive hypothesis. Then, since $b>a \geqslant 1$ and $u a+v b=1$, we have $v \leqslant 0 \Longleftrightarrow u>0$, and we are done.  The Lemma proves that, for all $(a, b) \in \mathbb{A}, f(a, b)=1$ if and only if the inverse of $a$ modulo $b$, taken in $\{1,2, \ldots, b-1\}$, is at most $b / 2$. Then, for any odd prime $p$ and integer $n$ such that $n \not \equiv 0(\bmod p), f\left(n^{2}, p\right)=1$ iff the inverse of $n^{2} \bmod p$ is less than $p / 2$. Since $\left\{n^{2} \bmod p: 1 \leqslant n \leqslant p-1\right\}=\left\{n^{-2} \bmod p: 1 \leqslant n \leqslant p-1\right\}$, including multiplicities (two for each quadratic residue in each set), we conclude that the desired sum is twice the number of quadratic residues that are less than $p / 2$, i.e.,  $$ \left.\sum_{n=1}^{p-1} f\left(n^{2}, p\right)=2 \left\lvert\,\left\{k: 1 \leqslant k \leqslant \frac{p-1}{2} \text { and } k^{2} \bmod p<\frac{p}{2}\right\}\right. \right\rvert\, . $$  Since the number of perfect squares in the interval $[1, p / 2)$ is $\lfloor\sqrt{p / 2}\rfloor>\sqrt{p / 2}-1$, we conclude that  $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right)>2\left(\sqrt{\frac{p}{2}}-1\right)=\sqrt{2 p}-2 $$


Solution: We provide a different proof for the Lemma. For this purpose, we use continued fractions to find $g(a, b)=(u, v)$ explicitly.  The function $f$ is completely determined on $\mathbb{A}$ by the following Claim. Represent $a / b$ as a continued fraction; that is, let $a_{0}$ be an integer and $a_{1}, \ldots, a_{k}$ be positive integers such that $a_{k} \geqslant 2$ and  $$ \frac{a}{b}=a_{0}+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{\cdots+\frac{1}{a_{k}}}}}=\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{k}\right] . $$  Then $f(a, b)=0 \Longleftrightarrow k$ is even. Proof. We induct on $b$. If $b=1$, then $a / b=[a]$ and $k=0$. Then, for $a \geqslant 1$, an easy induction shows that $f(a, 1)=f(1,1)=0$.  Now consider the case $b>1$. Perform the Euclidean division $a=q b+r$, with $0 \leqslant r<b$. We have $r \neq 0$ because $\operatorname{gcd}(a, b)=1$. Hence  $$ f(a, b)=f(r, b)=1-f(b, r), \quad \frac{a}{b}=\left[q ; a_{1}, \ldots, a_{k}\right], \quad \text { and } \quad \frac{b}{r}=\left[a_{1} ; a_{2}, \ldots, a_{k}\right] . $$  Then the number of terms in the continued fraction representations of $a / b$ and $b / r$ differ by one. Since $r<b$, the inductive hypothesis yields  $$ f(b, r)=0 \Longleftrightarrow k-1 \text { is even, } $$  and thus  $$ f(a, b)=0 \Longleftrightarrow f(b, r)=1 \Longleftrightarrow k-1 \text { is odd } \Longleftrightarrow k \text { is even. } $$  Now we use the following well-known properties of continued fractions to prove the Lemma: Let $p_{i}$ and $q_{i}$ be coprime positive integers with $\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{i}\right]=p_{i} / q_{i}$, with the notation borrowed from the Claim. In particular, $a / b=\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{k}\right]=p_{k} / q_{k}$. Assume that $k>0$ and define $q_{-1}=0$ if necessary. Then  - $q_{k}=a_{k} q_{k-1}+q_{k-2}, \quad$ and - $a q_{k-1}-b p_{k-1}=p_{k} q_{k-1}-q_{k} p_{k-1}=(-1)^{k-1}$.  Assume that $k>0$. Then $a_{k} \geqslant 2$, and  $$ b=q_{k}=a_{k} q_{k-1}+q_{k-2} \geqslant a_{k} q_{k-1} \geqslant 2 q_{k-1} \Longrightarrow q_{k-1} \leqslant \frac{b}{2} $$  with strict inequality for $k>1$, and  $$ (-1)^{k-1} q_{k-1} a+(-1)^{k} p_{k-1} b=1 $$  Now we finish the proof of the Lemma. It is immediate for $k=0$. If $k=1$, then $(-1)^{k-1}=1$, so  $$ -b / 2<0 \leqslant(-1)^{k-1} q_{k-1} \leqslant b / 2 $$  If $k>1$, we have $q_{k-1}<b / 2$, so  $$ -b / 2<(-1)^{k-1} q_{k-1}<b / 2 $$  Thus, for any $k>0$, we find that $g(a, b)=\left((-1)^{k-1} q_{k-1},(-1)^{k} p_{k-1}\right)$, and so  $$ f(a, b)=1 \Longleftrightarrow k \text { is odd } \Longleftrightarrow u=(-1)^{k-1} q_{k-1}>0 $$  Comment 1. The Lemma can also be established by observing that $f$ is uniquely defined on $\mathbb{A}$, defining $f_{1}(a, b)=1$ if $u>0$ in $g(a, b)=(u, v)$ and $f_{1}(a, b)=0$ otherwise, and verifying that $f_{1}$ satisfies all the conditions from the statement.  It seems that the main difficulty of the problem is in conjecturing the Lemma. Comment 2. The case $p \equiv 1(\bmod 4)$ is, in fact, easier than the original problem. We have, in general, for $1 \leqslant a \leqslant p-1$, $f(a, p)=1-f(p, a)=1-f(p-a, a)=f(a, p-a)=f(a+(p-a), p-a)=f(p, p-a)=1-f(p-a, p)$. If $p \equiv 1(\bmod 4)$, then $a$ is a quadratic residue modulo $p$ if and only if $p-a$ is a quadratic residue modulo $p$. Therefore, denoting by $r_{k}$ (with $1 \leqslant r_{k} \leqslant p-1$ ) the remainder of the division of $k^{2}$ by $p$, we get  $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right)=\sum_{n=1}^{p-1} f\left(r_{n}, p\right)=\frac{1}{2} \sum_{n=1}^{p-1}\left(f\left(r_{n}, p\right)+f\left(p-r_{n}, p\right)\right)=\frac{p-1}{2} $$  Comment 3. The estimate for the sum $\sum_{n=1}^{p} f\left(n^{2}, p\right)$ can be improved by refining the final argument in Solution 1. In fact, one can prove that  $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right) \geqslant \frac{p-1}{16} $$  By counting the number of perfect squares in the intervals $[k p,(k+1 / 2) p)$, we find that  $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right)=\sum_{k=0}^{p-1}\left(\left\lfloor\sqrt{\left(k+\frac{1}{2}\right) p}\right\rfloor-\lfloor\sqrt{k p}\rfloor\right) . $$  Each summand of (2) is non-negative. We now estimate the number of positive summands. Suppose that a summand is zero, i.e.,  $$ \left\lfloor\sqrt{\left(k+\frac{1}{2}\right) p}\right\rfloor=\lfloor\sqrt{k p}\rfloor=: q . $$  Then both of the numbers $k p$ and $k p+p / 2$ lie within the interval $\left[q^{2},(q+1)^{2}\right)$. Hence  $$ \frac{p}{2}<(q+1)^{2}-q^{2} $$  which implies  $$ q \geqslant \frac{p-1}{4} $$  Since $q \leqslant \sqrt{k p}$, if the $k^{\text {th }}$ summand of (2) is zero, then  $$ k \geqslant \frac{q^{2}}{p} \geqslant \frac{(p-1)^{2}}{16 p}>\frac{p-2}{16} \Longrightarrow k \geqslant \frac{p-1}{16} $$  So at least the first $\left\lceil\frac{p-1}{16}\right\rceil$ summands (from $k=0$ to $k=\left\lceil\frac{p-1}{16}\right\rceil-1$ ) are positive, and the result follows.  Comment 4. The bound can be further improved by using different methods. In fact, we prove that  $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right) \geqslant \frac{p-3}{4} $$  To that end, we use the Legendre symbol  $$ \left(\frac{a}{p}\right)= \begin{cases}0 & \text { if } p \mid a \\ 1 & \text { if } a \text { is a nonzero quadratic residue } \bmod p \\ -1 & \text { otherwise. }\end{cases} $$  We start with the following Claim, which tells us that there are not too many consecutive quadratic residues or consecutive quadratic non-residues.  Claim. $\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right)=-1$. Proof. We have $\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right)=\left(\frac{n(n+1)}{p}\right)$. For $1 \leqslant n \leqslant p-1$, we get that $n(n+1) \equiv n^{2}\left(1+n^{-1}\right)(\bmod p)$, hence $\left(\frac{n(n+1)}{p}\right)=\left(\frac{1+n^{-1}}{p}\right)$. Since $\left\{1+n^{-1} \bmod p: 1 \leqslant n \leqslant p-1\right\}=\{0,2,3, \ldots, p-1 \bmod p\}$, we find  $$ \sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right)=\sum_{n=1}^{p-1}\left(\frac{1+n^{-1}}{p}\right)=\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)-1=-1 $$  because $\sum_{n=1}^{p}\left(\frac{n}{p}\right)=0$. Observe that (1) becomes  $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right)=2|S|, \quad S=\left\{r: 1 \leqslant r \leqslant \frac{p-1}{2} \text { and }\left(\frac{r}{p}\right)=1\right\} $$  We connect $S$ with the sum from the claim by pairing quadratic residues and quadratic non-residues. To that end, define  $$ \begin{aligned} S^{\prime} & =\left\{r: 1 \leqslant r \leqslant \frac{p-1}{2} \text { and }\left(\frac{r}{p}\right)=-1\right\} \\ T & =\left\{r: \frac{p+1}{2} \leqslant r \leqslant p-1 \text { and }\left(\frac{r}{p}\right)=1\right\} \\ T^{\prime} & =\left\{r: \frac{p+1}{2} \leqslant r \leqslant p-1 \text { and }\left(\frac{r}{p}\right)=-1\right\} \end{aligned} $$  Since there are exactly $(p-1) / 2$ nonzero quadratic residues modulo $p,|S|+|T|=(p-1) / 2$. Also we obviously have $|T|+\left|T^{\prime}\right|=(p-1) / 2$. Then $|S|=\left|T^{\prime}\right|$.  For the sake of brevity, define $t=|S|=\left|T^{\prime}\right|$. If $\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right)=-1$, then exactly of one the numbers $\left(\frac{n}{p}\right)$ and $\left(\frac{n+1}{p}\right)$ is equal to 1 , so  $$ \left\lvert\,\left\{n: 1 \leqslant n \leqslant \frac{p-3}{2} \text { and }\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right)=-1\right\}|\leqslant|S|+|S-1|=2 t\right. $$  On the other hand, if $\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right)=-1$, then exactly one of $\left(\frac{n}{p}\right)$ and $\left(\frac{n+1}{p}\right)$ is equal to -1 , and  $$ \left\lvert\,\left\{n: \frac{p+1}{2} \leqslant n \leqslant p-2 \text { and }\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right)=-1\right\}\left|\leqslant\left|T^{\prime}\right|+\left|T^{\prime}-1\right|=2 t .\right.\right. $$  Thus, taking into account that the middle term $\left(\frac{(p-1) / 2}{p}\right)\left(\frac{(p+1) / 2}{p}\right)$ may happen to be -1 ,  $$ \left.\left\lvert\,\left\{n: 1 \leqslant n \leqslant p-2 \text { and }\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right)=-1\right\}\right. \right\rvert\, \leqslant 4 t+1 $$  This implies that  $$ \left.\left\lvert\,\left\{n: 1 \leqslant n \leqslant p-2 \text { and }\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right)=1\right\}\right. \right\rvert\, \geqslant(p-2)-(4 t+1)=p-4 t-3 $$  and so  $$ -1=\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\left(\frac{n+1}{p}\right) \geqslant p-4 t-3-(4 t+1)=p-8 t-4 $$  which implies $8 t \geqslant p-3$, and thus  $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right)=2 t \geqslant \frac{p-3}{4} $$  Comment 5. It is possible to prove that  $$ \sum_{n=1}^{p-1} f\left(n^{2}, p\right) \geqslant \frac{p-1}{2} $$  The case $p \equiv 1(\bmod 4)$ was already mentioned, and it is the equality case. If $p \equiv 3(\bmod 4)$, then, by a theorem of Dirichlet, we have  $$ \left.\left\lvert\,\left\{r: 1 \leqslant r \leqslant \frac{p-1}{2} \text { and }\left(\frac{r}{p}\right)=1\right\}\right. \right\rvert\,>\frac{p-1}{4} $$  which implies the result. See https://en.wikipedia.org/wiki/Quadratic_residue\#Dirichlet.27s_formulas for the full statement of the theorem. It seems that no elementary proof of it is known; a proof using complex analysis is available, for instance, in Chapter 7 of the book Quadratic Residues and Non-Residues: Selected Topics, by Steve Wright, available in https://arxiv.org/abs/1408.0235. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-96.jpg?height=138&width=204&top_left_y=2241&top_left_x=315)  BIÊNIO DA MATEMATICA BRASIL ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-96.jpg?height=249&width=683&top_left_y=2251&top_left_x=1052) ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-96.jpg?height=132&width=338&top_left_y=2607&top_left_x=1436)   [^0]:    *The name Dirichlet interval is chosen for the reason that $g$ theoretically might act similarly to the Dirichlet function on this interval.


Let $q$ be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard:  - In the first line, Gugu writes down every number of the form $a-b$, where $a$ and $b$ are two (not necessarily distinct) numbers on his napkin. - In the second line, Gugu writes down every number of the form qab, where $a$ and $b$ are two (not necessarily distinct) numbers from the first line. - In the third line, Gugu writes down every number of the form $a^{2}+b^{2}-c^{2}-d^{2}$, where $a, b, c, d$ are four (not necessarily distinct) numbers from the first line.  Determine all values of $q$ such that, regardless of the numbers on Gugu's napkin, every number in the second line is also a number in the third line. (Austria) Answer: -2, 0,2 .

Solution: Call a number $q$ good if every number in the second line appears in the third line unconditionally. We first show that the numbers 0 and $\pm 2$ are good. The third line necessarily contains 0 , so 0 is good. For any two numbers $a, b$ in the first line, write $a=x-y$ and $b=u-v$, where $x, y, u, v$ are (not necessarily distinct) numbers on the napkin. We may now write  $$ 2 a b=2(x-y)(u-v)=(x-v)^{2}+(y-u)^{2}-(x-u)^{2}-(y-v)^{2} $$  which shows that 2 is good. By negating both sides of the above equation, we also see that -2 is good.  We now show that $-2,0$, and 2 are the only good numbers. Assume for sake of contradiction that $q$ is a good number, where $q \notin\{-2,0,2\}$. We now consider some particular choices of numbers on Gugu's napkin to arrive at a contradiction.  Assume that the napkin contains the integers $1,2, \ldots, 10$. Then, the first line contains the integers $-9,-8, \ldots, 9$. The second line then contains $q$ and $81 q$, so the third line must also contain both of them. But the third line only contains integers, so $q$ must be an integer. Furthermore, the third line contains no number greater than $162=9^{2}+9^{2}-0^{2}-0^{2}$ or less than -162 , so we must have $-162 \leqslant 81 q \leqslant 162$. This shows that the only possibilities for $q$ are $\pm 1$.  Now assume that $q= \pm 1$. Let the napkin contain $0,1,4,8,12,16,20,24,28,32$. The first line contains $\pm 1$ and $\pm 4$, so the second line contains $\pm 4$. However, for every number $a$ in the first line, $a \not \equiv 2(\bmod 4)$, so we may conclude that $a^{2} \equiv 0,1(\bmod 8)$. Consequently, every number in the third line must be congruent to $-2,-1,0,1,2(\bmod 8)$; in particular, $\pm 4$ cannot be in the third line, which is a contradiction.


Solution: Let $q$ be a good number, as defined in the first solution, and define the polynomial $P\left(x_{1}, \ldots, x_{10}\right)$ as  $$ \prod_{i<j}\left(x_{i}-x_{j}\right) \prod_{a_{i} \in S}\left(q\left(x_{1}-x_{2}\right)\left(x_{3}-x_{4}\right)-\left(a_{1}-a_{2}\right)^{2}-\left(a_{3}-a_{4}\right)^{2}+\left(a_{5}-a_{6}\right)^{2}+\left(a_{7}-a_{8}\right)^{2}\right), $$  where $S=\left\{x_{1}, \ldots, x_{10}\right\}$. We claim that $P\left(x_{1}, \ldots, x_{10}\right)=0$ for every choice of real numbers $\left(x_{1}, \ldots, x_{10}\right)$. If any two of the $x_{i}$ are equal, then $P\left(x_{1}, \ldots, x_{10}\right)=0$ trivially. If no two are equal, assume that Gugu has those ten numbers $x_{1}, \ldots, x_{10}$ on his napkin. Then, the number $q\left(x_{1}-x_{2}\right)\left(x_{3}-x_{4}\right)$ is in the second line, so we must have some $a_{1}, \ldots, a_{8}$ so that  $$ q\left(x_{1}-x_{2}\right)\left(x_{3}-x_{4}\right)-\left(a_{1}-a_{2}\right)^{2}-\left(a_{3}-a_{4}\right)^{2}+\left(a_{5}-a_{6}\right)^{2}+\left(a_{7}-a_{8}\right)^{2}=0 $$  and hence $P\left(x_{1}, \ldots, x_{10}\right)=0$. Since every polynomial that evaluates to zero everywhere is the zero polynomial, and the product of two nonzero polynomials is necessarily nonzero, we may define $F$ such that  $$ F\left(x_{1}, \ldots, x_{10}\right) \equiv q\left(x_{1}-x_{2}\right)\left(x_{3}-x_{4}\right)-\left(a_{1}-a_{2}\right)^{2}-\left(a_{3}-a_{4}\right)^{2}+\left(a_{5}-a_{6}\right)^{2}+\left(a_{7}-a_{8}\right)^{2} \equiv 0 $$  for some particular choice $a_{i} \in S$. Each of the sets $\left\{a_{1}, a_{2}\right\},\left\{a_{3}, a_{4}\right\},\left\{a_{5}, a_{6}\right\}$, and $\left\{a_{7}, a_{8}\right\}$ is equal to at most one of the four sets $\left\{x_{1}, x_{3}\right\},\left\{x_{2}, x_{3}\right\},\left\{x_{1}, x_{4}\right\}$, and $\left\{x_{2}, x_{4}\right\}$. Thus, without loss of generality, we may assume that at most one of the sets $\left\{a_{1}, a_{2}\right\},\left\{a_{3}, a_{4}\right\},\left\{a_{5}, a_{6}\right\}$, and $\left\{a_{7}, a_{8}\right\}$ is equal to $\left\{x_{1}, x_{3}\right\}$. Let $u_{1}, u_{3}, u_{5}, u_{7}$ be the indicator functions for this equality of sets: that is, $u_{i}=1$ if and only if $\left\{a_{i}, a_{i+1}\right\}=\left\{x_{1}, x_{3}\right\}$. By assumption, at least three of the $u_{i}$ are equal to 0 .  We now compute the coefficient of $x_{1} x_{3}$ in $F$. It is equal to $q+2\left(u_{1}+u_{3}-u_{5}-u_{7}\right)=0$, and since at least three of the $u_{i}$ are zero, we must have that $q \in\{-2,0,2\}$, as desired.


A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation  $$ a_{n}=-\max _{i+j=n}\left(a_{i}+a_{j}\right) \quad \text { for all } n>2017 $$  Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$. (Russia)

Solution: Set $D=2017$. Denote  $$ M_{n}=\max _{k<n} a_{k} \quad \text { and } \quad m_{n}=-\min _{k<n} a_{k}=\max _{k<n}\left(-a_{k}\right) . $$  Clearly, the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are nondecreasing. We need to prove that both are bounded.  Consider an arbitrary $n>D$; our first aim is to bound $a_{n}$ in terms of $m_{n}$ and $M_{n}$. (i) There exist indices $p$ and $q$ such that $a_{n}=-\left(a_{p}+a_{q}\right)$ and $p+q=n$. Since $a_{p}, a_{q} \leqslant M_{n}$, we have $a_{n} \geqslant-2 M_{n}$. (ii) On the other hand, choose an index $k<n$ such that $a_{k}=M_{n}$. Then, we have  $$ a_{n}=-\max _{\ell<n}\left(a_{n-\ell}+a_{\ell}\right) \leqslant-\left(a_{n-k}+a_{k}\right)=-a_{n-k}-M_{n} \leqslant m_{n}-M_{n} $$  Summarizing (i) and (ii), we get  $$ -2 M_{n} \leqslant a_{n} \leqslant m_{n}-M_{n}, $$  whence  $$ m_{n} \leqslant m_{n+1} \leqslant \max \left\{m_{n}, 2 M_{n}\right\} \quad \text { and } \quad M_{n} \leqslant M_{n+1} \leqslant \max \left\{M_{n}, m_{n}-M_{n}\right\} $$  Now, say that an index $n>D$ is lucky if $m_{n} \leqslant 2 M_{n}$. Two cases are possible. Case 1. Assume that there exists a lucky index $n$. In this case, (1) yields $m_{n+1} \leqslant 2 M_{n}$ and $M_{n} \leqslant M_{n+1} \leqslant M_{n}$. Therefore, $M_{n+1}=M_{n}$ and $m_{n+1} \leqslant 2 M_{n}=2 M_{n+1}$. So, the index $n+1$ is also lucky, and $M_{n+1}=M_{n}$. Applying the same arguments repeatedly, we obtain that all indices $k>n$ are lucky (i.e., $m_{k} \leqslant 2 M_{k}$ for all these indices), and $M_{k}=M_{n}$ for all such indices. Thus, all of the $m_{k}$ and $M_{k}$ are bounded by $2 M_{n}$. Case 2. Assume now that there is no lucky index, i.e., $2 M_{n}<m_{n}$ for all $n>D$. Then (1) shows that for all $n>D$ we have $m_{n} \leqslant m_{n+1} \leqslant m_{n}$, so $m_{n}=m_{D+1}$ for all $n>D$. Since $M_{n}<m_{n} / 2$ for all such indices, all of the $m_{n}$ and $M_{n}$ are bounded by $m_{D+1}$.  Thus, in both cases the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are bounded, as desired.


Solution: As in the previous solution, let $D=2017$. If the sequence is bounded above, say, by $Q$, then we have that $a_{n} \geqslant \min \left\{a_{1}, \ldots, a_{D},-2 Q\right\}$ for all $n$, so the sequence is bounded. Assume for sake of contradiction that the sequence is not bounded above. Let $\ell=\min \left\{a_{1}, \ldots, a_{D}\right\}$, and $L=\max \left\{a_{1}, \ldots, a_{D}\right\}$. Call an index $n$ good if the following criteria hold:  $$ a_{n}>a_{i} \text { for each } i<n, \quad a_{n}>-2 \ell, \quad \text { and } \quad n>D $$  We first show that there must be some good index $n$. By assumption, we may take an index $N$ such that $a_{N}>\max \{L,-2 \ell\}$. Choose $n$ minimally such that $a_{n}=\max \left\{a_{1}, a_{2}, \ldots, a_{N}\right\}$. Now, the first condition in (2) is satisfied because of the minimality of $n$, and the second and third conditions are satisfied because $a_{n} \geqslant a_{N}>L,-2 \ell$, and $L \geqslant a_{i}$ for every $i$ such that $1 \leqslant i \leqslant D$.  Let $n$ be a good index. We derive a contradiction. We have that  $$ a_{n}+a_{u}+a_{v} \leqslant 0 $$  whenever $u+v=n$. We define the index $u$ to maximize $a_{u}$ over $1 \leqslant u \leqslant n-1$, and let $v=n-u$. Then, we note that $a_{u} \geqslant a_{v}$ by the maximality of $a_{u}$.  Assume first that $v \leqslant D$. Then, we have that  $$ a_{N}+2 \ell \leqslant 0, $$  because $a_{u} \geqslant a_{v} \geqslant \ell$. But this contradicts our assumption that $a_{n}>-2 \ell$ in the second criteria of (2).  Now assume that $v>D$. Then, there exist some indices $w_{1}, w_{2}$ summing up to $v$ such that  $$ a_{v}+a_{w_{1}}+a_{w_{2}}=0 $$  But combining this with (3), we have  $$ a_{n}+a_{u} \leqslant a_{w_{1}}+a_{w_{2}} $$  Because $a_{n}>a_{u}$, we have that $\max \left\{a_{w_{1}}, a_{w_{2}}\right\}>a_{u}$. But since each of the $w_{i}$ is less than $v$, this contradicts the maximality of $a_{u}$.  Comment 1. We present two harder versions of this problem below. Version 1. Let $a_{1}, a_{2}, \ldots$ be a sequence of numbers that satisfies the relation  $$ a_{n}=-\max _{i+j+k=n}\left(a_{i}+a_{j}+a_{k}\right) \quad \text { for all } n>2017 $$  Then, this sequence is bounded. Proof. Set $D=2017$. Denote  $$ M_{n}=\max _{k<n} a_{k} \quad \text { and } \quad m_{n}=-\min _{k<n} a_{k}=\max _{k<n}\left(-a_{k}\right) . $$  Clearly, the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are nondecreasing. We need to prove that both are bounded. Consider an arbitrary $n>2 D$; our first aim is to bound $a_{n}$ in terms of $m_{i}$ and $M_{i}$. Set $k=\lfloor n / 2\rfloor$. (i) Choose indices $p, q$, and $r$ such that $a_{n}=-\left(a_{p}+a_{q}+a_{r}\right)$ and $p+q+r=n$. Without loss of generality, $p \geqslant q \geqslant r$.  Assume that $p \geqslant k+1(>D)$; then $p>q+r$. Hence  $$ -a_{p}=\max _{i_{1}+i_{2}+i_{3}=p}\left(a_{i_{1}}+a_{i_{2}}+a_{i_{3}}\right) \geqslant a_{q}+a_{r}+a_{p-q-r} $$  and therefore $a_{n}=-\left(a_{p}+a_{q}+a_{r}\right) \geqslant\left(a_{q}+a_{r}+a_{p-q-r}\right)-a_{q}-a_{r}=a_{p-q-r} \geqslant-m_{n}$. Otherwise, we have $k \geqslant p \geqslant q \geqslant r$. Since $n<3 k$, we have $r<k$. Then $a_{p}, a_{q} \leqslant M_{k+1}$ and $a_{r} \leqslant M_{k}$, whence $a_{n} \geqslant-2 M_{k+1}-M_{k}$.  Thus, in any case $a_{n} \geqslant-\max \left\{m_{n}, 2 M_{k+1}+M_{k}\right\}$. (ii) On the other hand, choose $p \leqslant k$ and $q \leqslant k-1$ such that $a_{p}=M_{k+1}$ and $a_{q}=M_{k}$. Then $p+q<n$, so $a_{n} \leqslant-\left(a_{p}+a_{q}+a_{n-p-q}\right)=-a_{n-p-q}-M_{k+1}-M_{k} \leqslant m_{n}-M_{k+1}-M_{k}$.  To summarize,  $$ -\max \left\{m_{n}, 2 M_{k+1}+M_{k}\right\} \leqslant a_{n} \leqslant m_{n}-M_{k+1}-M_{k} $$  whence  $$ m_{n} \leqslant m_{n+1} \leqslant \max \left\{m_{n}, 2 M_{k+1}+M_{k}\right\} \quad \text { and } \quad M_{n} \leqslant M_{n+1} \leqslant \max \left\{M_{n}, m_{n}-M_{k+1}-M_{k}\right\} . $$  Now, say that an index $n>2 D$ is lucky if $m_{n} \leqslant 2 M_{\lfloor n / 2\rfloor+1}+M_{\lfloor n / 2]}$. Two cases are possible. Case 1. Assume that there exists a lucky index $n$; set $k=\lfloor n / 2\rfloor$. In this case, (4) yields $m_{n+1} \leqslant$ $2 M_{k+1}+M_{k}$ and $M_{n} \leqslant M_{n+1} \leqslant M_{n}$ (the last relation holds, since $m_{n}-M_{k+1}-M_{k} \leqslant\left(2 M_{k+1}+\right.$ $\left.M_{k}\right)-M_{k+1}-M_{k}=M_{k+1} \leqslant M_{n}$ ). Therefore, $M_{n+1}=M_{n}$ and $m_{n+1} \leqslant 2 M_{k+1}+M_{k}$; the last relation shows that the index $n+1$ is also lucky.  Thus, all indices $N>n$ are lucky, and $M_{N}=M_{n} \geqslant m_{N} / 3$, whence all the $m_{N}$ and $M_{N}$ are bounded by $3 M_{n}$. Case 2. Conversely, assume that there is no lucky index, i.e., $2 M_{\lfloor n / 2\rfloor+1}+M_{\lfloor n / 2\rfloor}<m_{n}$ for all $n>2 D$. Then (4) shows that for all $n>2 D$ we have $m_{n} \leqslant m_{n+1} \leqslant m_{n}$, i.e., $m_{N}=m_{2 D+1}$ for all $N>2 D$. Since $M_{N}<m_{2 N+1} / 3$ for all such indices, all the $m_{N}$ and $M_{N}$ are bounded by $m_{2 D+1}$.  Thus, in both cases the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are bounded, as desired. Version 2. Let $a_{1}, a_{2}, \ldots$ be a sequence of numbers that satisfies the relation  $$ a_{n}=-\max _{i_{1}+\cdots+i_{k}=n}\left(a_{i_{1}}+\cdots+a_{i_{k}}\right) \quad \text { for all } n>2017 $$  Then, this sequence is bounded. Proof. As in the solutions above, let $D=2017$. If the sequence is bounded above, say, by $Q$, then we have that $a_{n} \geqslant \min \left\{a_{1}, \ldots, a_{D},-k Q\right\}$ for all $n$, so the sequence is bounded. Assume for sake of contradiction that the sequence is not bounded above. Let $\ell=\min \left\{a_{1}, \ldots, a_{D}\right\}$, and $L=\max \left\{a_{1}, \ldots, a_{D}\right\}$. Call an index $n$ good if the following criteria hold:  $$ a_{n}>a_{i} \text { for each } i<n, \quad a_{n}>-k \ell, \quad \text { and } \quad n>D $$  We first show that there must be some good index $n$. By assumption, we may take an index $N$ such that $a_{N}>\max \{L,-k \ell\}$. Choose $n$ minimally such that $a_{n}=\max \left\{a_{1}, a_{2}, \ldots, a_{N}\right\}$. Now, the first condition is satisfied because of the minimality of $n$, and the second and third conditions are satisfied because $a_{n} \geqslant a_{N}>L,-k \ell$, and $L \geqslant a_{i}$ for every $i$ such that $1 \leqslant i \leqslant D$.  Let $n$ be a good index. We derive a contradiction. We have that  $$ a_{n}+a_{v_{1}}+\cdots+a_{v_{k}} \leqslant 0 $$  whenever $v_{1}+\cdots+v_{k}=n$. We define the sequence of indices $v_{1}, \ldots, v_{k-1}$ to greedily maximize $a_{v_{1}}$, then $a_{v_{2}}$, and so forth, selecting only from indices such that the equation $v_{1}+\cdots+v_{k}=n$ can be satisfied by positive integers $v_{1}, \ldots, v_{k}$. More formally, we define them inductively so that the following criteria are satisfied by the $v_{i}$ :  1. $1 \leqslant v_{i} \leqslant n-(k-i)-\left(v_{1}+\cdots+v_{i-1}\right)$. 2. $a_{v_{i}}$ is maximal among all choices of $v_{i}$ from the first criteria.  First of all, we note that for each $i$, the first criteria is always satisfiable by some $v_{i}$, because we are guaranteed that  $$ v_{i-1} \leqslant n-(k-(i-1))-\left(v_{1}+\cdots+v_{i-2}\right), $$  which implies  $$ 1 \leqslant n-(k-i)-\left(v_{1}+\cdots+v_{i-1}\right) . $$  Secondly, the sum $v_{1}+\cdots+v_{k-1}$ is at most $n-1$. Define $v_{k}=n-\left(v_{1}+\cdots+v_{k-1}\right)$. Then, (6) is satisfied by the $v_{i}$. We also note that $a_{v_{i}} \geqslant a_{v_{j}}$ for all $i<j$; otherwise, in the definition of $v_{i}$, we could have selected $v_{j}$ instead.  Assume first that $v_{k} \leqslant D$. Then, from (6), we have that  $$ a_{n}+k \ell \leqslant 0, $$  by using that $a_{v_{1}} \geqslant \cdots \geqslant a_{v_{k}} \geqslant \ell$. But this contradicts our assumption that $a_{n}>-k \ell$ in the second criteria of (5).  Now assume that $v_{k}>D$, and then we must have some indices $w_{1}, \ldots, w_{k}$ summing up to $v_{k}$ such that  $$ a_{v_{k}}+a_{w_{1}}+\cdots+a_{w_{k}}=0 $$  But combining this with (6), we have  $$ a_{n}+a_{v_{1}}+\cdots+a_{v_{k-1}} \leqslant a_{w_{1}}+\cdots+a_{w_{k}} . $$  Because $a_{n}>a_{v_{1}} \geqslant \cdots \geqslant a_{v_{k-1}}$, we have that $\max \left\{a_{w_{1}}, \ldots, a_{w_{k}}\right\}>a_{v_{k-1}}$. But since each of the $w_{i}$ is less than $v_{k}$, in the definition of the $v_{k-1}$ we could have chosen one of the $w_{i}$ instead, which is a contradiction.  Comment 2. It seems that each sequence satisfying the condition in Version 2 is eventually periodic, at least when its terms are integers.  However, up to this moment, the Problem Selection Committee is not aware of a proof for this fact (even in the case $k=2$ ).


An integer $n \geqslant 3$ is given. We call an $n$-tuple of real numbers $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ Shiny if for each permutation $y_{1}, y_{2}, \ldots, y_{n}$ of these numbers we have  $$ \sum_{i=1}^{n-1} y_{i} y_{i+1}=y_{1} y_{2}+y_{2} y_{3}+y_{3} y_{4}+\cdots+y_{n-1} y_{n} \geqslant-1 $$  Find the largest constant $K=K(n)$ such that  $$ \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \geqslant K $$  holds for every Shiny $n$-tuple $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$. (Serbia) Answer: $K=-(n-1) / 2$.

Solution: First of all, we show that we may not take a larger constant $K$. Let $t$ be a positive number, and take $x_{2}=x_{3}=\cdots=t$ and $x_{1}=-1 /(2 t)$. Then, every product $x_{i} x_{j}(i \neq j)$ is equal to either $t^{2}$ or $-1 / 2$. Hence, for every permutation $y_{i}$ of the $x_{i}$, we have  $$ y_{1} y_{2}+\cdots+y_{n-1} y_{n} \geqslant(n-3) t^{2}-1 \geqslant-1 . $$  This justifies that the $n$-tuple $\left(x_{1}, \ldots, x_{n}\right)$ is Shiny. Now, we have  $$ \sum_{i<j} x_{i} x_{j}=-\frac{n-1}{2}+\frac{(n-1)(n-2)}{2} t^{2} $$  Thus, as $t$ approaches 0 from above, $\sum_{i<j} x_{i} x_{j}$ gets arbitrarily close to $-(n-1) / 2$. This shows that we may not take $K$ any larger than $-(n-1) / 2$. It remains to show that $\sum_{i<j} x_{i} x_{j} \geqslant$ $-(n-1) / 2$ for any Shiny choice of the $x_{i}$.  From now onward, assume that $\left(x_{1}, \ldots, x_{n}\right)$ is a Shiny $n$-tuple. Let the $z_{i}(1 \leqslant i \leqslant n)$ be some permutation of the $x_{i}$ to be chosen later. The indices for $z_{i}$ will always be taken modulo $n$. We will first split up the sum $\sum_{i<j} x_{i} x_{j}=\sum_{i<j} z_{i} z_{j}$ into $\lfloor(n-1) / 2\rfloor$ expressions, each of the form $y_{1} y_{2}+\cdots+y_{n-1} y_{n}$ for some permutation $y_{i}$ of the $z_{i}$, and some leftover terms. More specifically, write where $L=z_{1} z_{-1}+z_{2} z_{-2}+\cdots+z_{(n-1) / 2} z_{-(n-1) / 2}$ if $n$ is odd, and $L=z_{1} z_{-1}+z_{1} z_{-2}+z_{2} z_{-2}+$ $\cdots+z_{(n-2) / 2} z_{-n / 2}$ if $n$ is even. We note that for each $p=1,2, \ldots,\lfloor(n-1) / 2\rfloor$, there is some permutation $y_{i}$ of the $z_{i}$ such that  $$ \sum_{\substack{i+j \equiv 2 p-1,2 p \\ i \neq j}(\bmod n)} z_{i} z_{j}=\sum_{k=1}^{n-1} y_{k} y_{k+1} $$  because we may choose $y_{2 i-1}=z_{i+p-1}$ for $1 \leqslant i \leqslant(n+1) / 2$ and $y_{2 i}=z_{p-i}$ for $1 \leqslant i \leqslant n / 2$. We show (1) graphically for $n=6,7$ in the diagrams below. The edges of the graphs each represent a product $z_{i} z_{j}$, and the dashed and dotted series of lines represents the sum of the edges, which is of the form $y_{1} y_{2}+\cdots+y_{n-1} y_{n}$ for some permutation $y_{i}$ of the $z_{i}$ precisely when the series of lines is a Hamiltonian path. The filled edges represent the summands of $L$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-24.jpg?height=791&width=1656&top_left_y=185&top_left_x=195)  Now, because the $z_{i}$ are Shiny, we have that (1) yields the following bound:  $$ \sum_{i<j} z_{i} z_{j} \geqslant-\left\lfloor\frac{n-1}{2}\right\rfloor+L $$  It remains to show that, for each $n$, there exists some permutation $z_{i}$ of the $x_{i}$ such that $L \geqslant 0$ when $n$ is odd, and $L \geqslant-1 / 2$ when $n$ is even. We now split into cases based on the parity of $n$ and provide constructions of the permutations $z_{i}$.  Since we have not made any assumptions yet about the $x_{i}$, we may now assume without loss of generality that  $$ x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{k} \leqslant 0 \leqslant x_{k+1} \leqslant \cdots \leqslant x_{n} $$  Case 1: $n$ is odd. Without loss of generality, assume that $k$ (from (2)) is even, because we may negate all the $x_{i}$ if $k$ is odd. We then have $x_{1} x_{2}, x_{3} x_{4}, \ldots, x_{n-2} x_{n-1} \geqslant 0$ because the factors are of the same sign. Let $L=x_{1} x_{2}+x_{3} x_{4}+\cdots+x_{n-2} x_{n-1} \geqslant 0$. We choose our $z_{i}$ so that this definition of $L$ agrees with the sum of the leftover terms in (1). Relabel the $x_{i}$ as $z_{i}$ such that  $$ \left\{z_{1}, z_{n-1}\right\},\left\{z_{2}, z_{n-2}\right\}, \ldots,\left\{z_{(n-1) / 2}, z_{(n+1) / 2}\right\} $$  are some permutation of  $$ \left\{x_{1}, x_{2}\right\},\left\{x_{3}, x_{4}\right\}, \ldots,\left\{x_{n-2}, x_{n-1}\right\} $$  and $z_{n}=x_{n}$. Then, we have $L=z_{1} z_{n-1}+\cdots+z_{(n-1) / 2} z_{(n+1) / 2}$, as desired. Case 2: $n$ is even. Let $L=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n-1} x_{n}$. Assume without loss of generality $k \neq 1$. Now, we have  $$ \begin{gathered} 2 L=\left(x_{1} x_{2}+\cdots+x_{n-1} x_{n}\right)+\left(x_{1} x_{2}+\cdots+x_{n-1} x_{n}\right) \geqslant\left(x_{2} x_{3}+\cdots+x_{n-1} x_{n}\right)+x_{k} x_{k+1} \\ \geqslant x_{2} x_{3}+\cdots+x_{n-1} x_{n}+x_{n} x_{1} \geqslant-1 \end{gathered} $$  where the first inequality holds because the only negative term in $L$ is $x_{k} x_{k+1}$, the second inequality holds because $x_{1} \leqslant x_{k} \leqslant 0 \leqslant x_{k+1} \leqslant x_{n}$, and the third inequality holds because the $x_{i}$ are assumed to be Shiny. We thus have that $L \geqslant-1 / 2$. We now choose a suitable $z_{i}$ such that the definition of $L$ matches the leftover terms in (1).  Relabel the $x_{i}$ with $z_{i}$ in the following manner: $x_{2 i-1}=z_{-i}, x_{2 i}=z_{i}$ (again taking indices modulo $n$ ). We have that  $$ L=\sum_{\substack{i+j \equiv 0,-1(\bmod n) \\ i \neq j \\(\bmod n)}} z_{i} z_{j}, $$  as desired.


Solution: We present another proof that $\sum_{i<j} x_{i} x_{j} \geqslant-(n-1) / 2$ for any Shiny $n$-tuple $\left(x_{1}, \ldots, x_{n}\right)$. Assume an ordering of the $x_{i}$ as in (2), and let $\ell=n-k$. Assume without loss of generality that $k \geqslant \ell$. Also assume $k \neq n$, (as otherwise, all of the $x_{i}$ are nonpositive, and so the inequality is trivial). Define the sets of indices $S=\{1,2, \ldots, k\}$ and $T=\{k+1, \ldots, n\}$. Define the following sums:  $$ K=\sum_{\substack{i<j \\ i, j \in S}} x_{i} x_{j}, \quad M=\sum_{\substack{i \in S \\ j \in T}} x_{i} x_{j}, \quad \text { and } \quad L=\sum_{\substack{i<j \\ i, j \in T}} x_{i} x_{j} $$  By definition, $K, L \geqslant 0$ and $M \leqslant 0$. We aim to show that $K+L+M \geqslant-(n-1) / 2$. We split into cases based on whether $k=\ell$ or $k>\ell$. Case 1: $k>\ell$. Consider all permutations $\phi:\{1,2, \ldots, n\} \rightarrow\{1,2, \ldots, n\}$ such that $\phi^{-1}(T)=\{2,4, \ldots, 2 \ell\}$. Note that there are $k!!$ ! such permutations $\phi$. Define  $$ f(\phi)=\sum_{i=1}^{n-1} x_{\phi(i)} x_{\phi(i+1)} $$  We know that $f(\phi) \geqslant-1$ for every permutation $\phi$ with the above property. Averaging $f(\phi)$ over all $\phi$ gives  $$ -1 \leqslant \frac{1}{k!\ell!} \sum_{\phi} f(\phi)=\frac{2 \ell}{k \ell} M+\frac{2(k-\ell-1)}{k(k-1)} K $$  where the equality holds because there are $k \ell$ products in $M$, of which $2 \ell$ are selected for each $\phi$, and there are $k(k-1) / 2$ products in $K$, of which $k-\ell-1$ are selected for each $\phi$. We now have  $$ K+L+M \geqslant K+L+\left(-\frac{k}{2}-\frac{k-\ell-1}{k-1} K\right)=-\frac{k}{2}+\frac{\ell}{k-1} K+L $$  Since $k \leqslant n-1$ and $K, L \geqslant 0$, we get the desired inequality. Case 2: $k=\ell=n / 2$. We do a similar approach, considering all $\phi:\{1,2, \ldots, n\} \rightarrow\{1,2, \ldots, n\}$ such that $\phi^{-1}(T)=$ $\{2,4, \ldots, 2 \ell\}$, and defining $f$ the same way. Analogously to Case 1 , we have  $$ -1 \leqslant \frac{1}{k!\ell!} \sum_{\phi} f(\phi)=\frac{2 \ell-1}{k \ell} M $$  because there are $k \ell$ products in $M$, of which $2 \ell-1$ are selected for each $\phi$. Now, we have that  $$ K+L+M \geqslant M \geqslant-\frac{n^{2}}{4(n-1)} \geqslant-\frac{n-1}{2} $$  where the last inequality holds because $n \geqslant 4$.


Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that  $$ f(f(x) f(y))+f(x+y)=f(x y) $$  for all $x, y \in \mathbb{R}$. (Albania) Answer: There are 3 solutions:  $$ x \mapsto 0 \quad \text { or } \quad x \mapsto x-1 \quad \text { or } \quad x \mapsto 1-x \quad(x \in \mathbb{R}) . $$

Solution: An easy check shows that all the 3 above mentioned functions indeed satisfy the original equation (*).  In order to show that these are the only solutions, first observe that if $f(x)$ is a solution then $-f(x)$ is also a solution. Hence, without loss of generality we may (and will) assume that $f(0) \leqslant 0$ from now on. We have to show that either $f$ is identically zero or $f(x)=x-1$ $(\forall x \in \mathbb{R})$.  Observe that, for a fixed $x \neq 1$, we may choose $y \in \mathbb{R}$ so that $x+y=x y \Longleftrightarrow y=\frac{x}{x-1}$, and therefore from the original equation (*) we have  $$ f\left(f(x) \cdot f\left(\frac{x}{x-1}\right)\right)=0 \quad(x \neq 1) $$  In particular, plugging in $x=0$ in (1), we conclude that $f$ has at least one zero, namely $(f(0))^{2}$ :  $$ f\left((f(0))^{2}\right)=0 $$  We analyze two cases (recall that $f(0) \leqslant 0$ ): Case 1: $f(0)=0$. Setting $y=0$ in the original equation we get the identically zero solution:  $$ f(f(x) f(0))+f(x)=f(0) \Longrightarrow f(x)=0 \text { for all } x \in \mathbb{R} $$  From now on, we work on the main Case 2: $f(0)<0$. We begin with the following


Sir Alex plays the following game on a row of 9 cells. Initially, all cells are empty. In each move, Sir Alex is allowed to perform exactly one of the following two operations: (1) Choose any number of the form $2^{j}$, where $j$ is a non-negative integer, and put it into an empty cell. (2) Choose two (not necessarily adjacent) cells with the same number in them; denote that number by $2^{j}$. Replace the number in one of the cells with $2^{j+1}$ and erase the number in the other cell.  At the end of the game, one cell contains the number $2^{n}$, where $n$ is a given positive integer, while the other cells are empty. Determine the maximum number of moves that Sir Alex could have made, in terms of $n$. (Thailand) Answer: $2 \sum_{j=0}^{8}\binom{n}{j}-1$.

Solution: We will solve a more general problem, replacing the row of 9 cells with a row of $k$ cells, where $k$ is a positive integer. Denote by $m(n, k)$ the maximum possible number of moves Sir Alex can make starting with a row of $k$ empty cells, and ending with one cell containing the number $2^{n}$ and all the other $k-1$ cells empty. Call an operation of type (1) an insertion, and an operation of type (2) a merge.  Only one move is possible when $k=1$, so we have $m(n, 1)=1$. From now on we consider $k \geqslant 2$, and we may assume Sir Alex's last move was a merge. Then, just before the last move, there were exactly two cells with the number $2^{n-1}$, and the other $k-2$ cells were empty.  Paint one of those numbers $2^{n-1}$ blue, and the other one red. Now trace back Sir Alex's moves, always painting the numbers blue or red following this rule: if $a$ and $b$ merge into $c$, paint $a$ and $b$ with the same color as $c$. Notice that in this backward process new numbers are produced only by reversing merges, since reversing an insertion simply means deleting one of the numbers. Therefore, all numbers appearing in the whole process will receive one of the two colors.  Sir Alex's first move is an insertion. Without loss of generality, assume this first number inserted is blue. Then, from this point on, until the last move, there is always at least one cell with a blue number.  Besides the last move, there is no move involving a blue and a red number, since all merges involves numbers with the same color, and insertions involve only one number. Call an insertion of a blue number or merge of two blue numbers a blue move, and define a red move analogously.  The whole sequence of blue moves could be repeated on another row of $k$ cells to produce one cell with the number $2^{n-1}$ and all the others empty, so there are at most $m(n-1, k)$ blue moves.  Now we look at the red moves. Since every time we perform a red move there is at least one cell occupied with a blue number, the whole sequence of red moves could be repeated on a row of $k-1$ cells to produce one cell with the number $2^{n-1}$ and all the others empty, so there are at most $m(n-1, k-1)$ red moves. This proves that  $$ m(n, k) \leqslant m(n-1, k)+m(n-1, k-1)+1 . $$  On the other hand, we can start with an empty row of $k$ cells and perform $m(n-1, k)$ moves to produce one cell with the number $2^{n-1}$ and all the others empty, and after that perform $m(n-1, k-1)$ moves on those $k-1$ empty cells to produce the number $2^{n-1}$ in one of them, leaving $k-2$ empty. With one more merge we get one cell with $2^{n}$ and the others empty, proving that  $$ m(n, k) \geqslant m(n-1, k)+m(n-1, k-1)+1 . $$  It follows that  $$ m(n, k)=m(n-1, k)+m(n-1, k-1)+1, $$  for $n \geqslant 1$ and $k \geqslant 2$. If $k=1$ or $n=0$, we must insert $2^{n}$ on our first move and immediately get the final configuration, so $m(0, k)=1$ and $m(n, 1)=1$, for $n \geqslant 0$ and $k \geqslant 1$. These initial values, together with the recurrence relation (1), determine $m(n, k)$ uniquely.  Finally, we show that  $$ m(n, k)=2 \sum_{j=0}^{k-1}\binom{n}{j}-1 $$  for all integers $n \geqslant 0$ and $k \geqslant 1$. We use induction on $n$. Since $m(0, k)=1$ for $k \geqslant 1,(2)$ is true for the base case. We make the induction hypothesis that (2) is true for some fixed positive integer $n$ and all $k \geqslant 1$. We have $m(n+1,1)=1=2\binom{n+1}{0}-1$, and for $k \geqslant 2$ the recurrence relation (1) and the induction hypothesis give us  $$ \begin{aligned} & m(n+1, k)=m(n, k)+m(n, k-1)+1=2 \sum_{j=0}^{k-1}\binom{n}{j}-1+2 \sum_{j=0}^{k-2}\binom{n}{j}-1+1 \\ & \quad=2 \sum_{j=0}^{k-1}\binom{n}{j}+2 \sum_{j=0}^{k-1}\binom{n}{j-1}-1=2 \sum_{j=0}^{k-1}\left(\binom{n}{j}+\binom{n}{j-1}\right)-1=2 \sum_{j=0}^{k-1}\binom{n+1}{j}-1, \end{aligned} $$  which completes the proof.  Comment 1. After deducing the recurrence relation (1), it may be convenient to homogenize the recurrence relation by defining $h(n, k)=m(n, k)+1$. We get the new relation  $$ h(n, k)=h(n-1, k)+h(n-1, k), $$  for $n \geqslant 1$ and $k \geqslant 2$, with initial values $h(0, k)=h(n, 1)=2$, for $n \geqslant 0$ and $k \geqslant 1$. This may help one to guess the answer, and also with other approaches like the one we develop next.  Comment 2. We can use a generating function to find the answer without guessing. We work with the homogenized recurrence relation (3). Define $h(n, 0)=0$ so that (3) is valid for $k=1$ as well. Now we set up the generating function $f(x, y)=\sum_{n, k \geqslant 0} h(n, k) x^{n} y^{k}$. Multiplying the recurrence relation (3) by $x^{n} y^{k}$ and summing over $n, k \geqslant 1$, we get  $$ \sum_{n, k \geqslant 1} h(n, k) x^{n} y^{k}=x \sum_{n, k \geqslant 1} h(n-1, k) x^{n-1} y^{k}+x y \sum_{n, k \geqslant 1} h(n-1, k-1) x^{n-1} y^{k-1} . $$  Completing the missing terms leads to the following equation on $f(x, y)$ :  $$ f(x, y)-\sum_{n \geqslant 0} h(n, 0) x^{n}-\sum_{k \geqslant 1} h(0, k) y^{k}=x f(x, y)-x \sum_{n \geqslant 0} h(n, 0) x^{n}+x y f(x, y) . $$  Substituting the initial values, we obtain  $$ f(x, y)=\frac{2 y}{1-y} \cdot \frac{1}{1-x(1+y)} . $$  Developing as a power series, we get  $$ f(x, y)=2 \sum_{j \geqslant 1} y^{j} \cdot \sum_{n \geqslant 0}(1+y)^{n} x^{n} . $$  The coefficient of $x^{n}$ in this power series is  $$ 2 \sum_{j \geqslant 1} y^{j} \cdot(1+y)^{n}=2 \sum_{j \geqslant 1} y^{j} \cdot \sum_{i \geqslant 0}\binom{n}{i} y^{i} $$  and extracting the coefficient of $y^{k}$ in this last expression we finally obtain the value for $h(n, k)$,  $$ h(n, k)=2 \sum_{j=0}^{k-1}\binom{n}{j} $$  This proves that  $$ m(n, k)=2 \sum_{j=0}^{k-1}\binom{n}{j}-1 $$  The generating function approach also works if applied to the non-homogeneous recurrence relation (1), but the computations are less straightforward.


Solution: Define merges and insertions as in   We will need the following lemma. Lemma. If the binary representation of a positive integer $A$ has $d$ nonzero digits, then $A$ cannot be represented as a sum of fewer than $d$ powers of 2 . Moreover, any representation of $A$ as a sum of $d$ powers of 2 must coincide with its binary representation. Proof. Let $s$ be the minimum number of summands in all possible representations of $A$ as sum of powers of 2 . Suppose there is such a representation with $s$ summands, where two of the summands are equal to each other. Then, replacing those two summands with the result of their sum, we obtain a representation with fewer than $s$ summands, which is a contradiction. We deduce that in any representation with $s$ summands, the summands are all distinct, so any such representation must coincide with the unique binary representation of $A$, and $s=d$.  Now we split the solution into a sequence of claims. Claim 1. After every move, the number $S$ is the sum of at most $k-1$ distinct powers of 2 . Proof. If $S$ is the sum of $k$ (or more) distinct powers of 2 , the Lemma implies that the $k$ cells are filled with these numbers. This is a contradiction since no more merges or insertions can be made.  Let $A(n, k-1)$ denote the set of all positive integers not exceeding $2^{n}$ with at most $k-1$ nonzero digits in its base 2 representation. Since every insertion increases the value of $S$, by Claim 1, the total number of insertions is at most $|A(n, k-1)|$. We proceed to prove that it is possible to achieve this number of insertions. Claim 2. Let $A(n, k-1)=\left\{a_{1}, a_{2}, \ldots, a_{m}\right\}$, with $a_{1}<a_{2}<\cdots<a_{m}$. If after some of Sir Alex's moves the value of $S$ is $a_{j}$, with $j \in\{1,2, \ldots, m-1\}$, then there is a sequence of moves after which the value of $S$ is exactly $a_{j+1}$. Proof. Suppose $S=a_{j}$. Performing all possible merges, we eventually get different powers of 2 in all nonempty cells. After that, by Claim 1 there will be at least one empty cell, in which we want to insert $a_{j+1}-a_{j}$. It remains to show that $a_{j+1}-a_{j}$ is a power of 2 .  For this purpose, we notice that if $a_{j}$ has less than $k-1$ nonzero digits in base 2 then $a_{j+1}=a_{j}+1$. Otherwise, we have $a_{j}=2^{b_{k-1}}+\cdots+2^{b_{2}}+2^{b_{1}}$ with $b_{1}<b_{2}<\cdots<b_{k-1}$. Then, adding any number less than $2^{b_{1}}$ to $a_{j}$ will result in a number with more than $k-1$ nonzero binary digits. On the other hand, $a_{j}+2^{b_{1}}$ is a sum of $k$ powers of 2 , not all distinct, so by the Lemma it will be a sum of less then $k$ distinct powers of 2 . This means that $a_{j+1}-a_{j}=2^{b_{1}}$, completing the proof.  Claims 1 and 2 prove that the maximum number of insertions is $|A(n, k-1)|$. We now compute this number. Claim 3. $|A(n, k-1)|=\sum_{j=0}^{k-1}\binom{n}{j}$. Proof. The number $2^{n}$ is the only element of $A(n, k-1)$ with $n+1$ binary digits. Any other element has at most $n$ binary digits, at least one and at most $k-1$ of them are nonzero (so they are ones). For each $j \in\{1,2, \ldots, k-1\}$, there are $\binom{n}{j}$ such elements with exactly $j$ binary digits equal to one. We conclude that $|A(n, k-1)|=1+\sum_{j=1}^{k-1}\binom{n}{j}=\sum_{j=0}^{k-1}\binom{n}{j}$.  Recalling that the number of insertions exceeds that of merges by exactly 1 , we deduce that the maximum number of moves is $2 \sum_{j=0}^{k-1}\binom{n}{j}-1$.


A hunter and an invisible rabbit play a game in the Euclidean plane. The hunter's starting point $H_{0}$ coincides with the rabbit's starting point $R_{0}$. In the $n^{\text {th }}$ round of the game ( $n \geqslant 1$ ), the following happens. (1) First the invisible rabbit moves secretly and unobserved from its current point $R_{n-1}$ to some new point $R_{n}$ with $R_{n-1} R_{n}=1$. (2) The hunter has a tracking device (e.g. dog) that returns an approximate position $R_{n}^{\prime}$ of the rabbit, so that $R_{n} R_{n}^{\prime} \leqslant 1$. (3) The hunter then visibly moves from point $H_{n-1}$ to a new point $H_{n}$ with $H_{n-1} H_{n}=1$.  Is there a strategy for the hunter that guarantees that after $10^{9}$ such rounds the distance between the hunter and the rabbit is below 100 ? (Austria) Answer: There is no such strategy for the hunter. The rabbit "wins".

Solution: If the answer were "yes", the hunter would have a strategy that would "work", no matter how the rabbit moved or where the radar pings $R_{n}^{\prime}$ appeared. We will show the opposite: with bad luck from the radar pings, there is no strategy for the hunter that guarantees that the distance stays below 100 in $10^{9}$ rounds.  So, let $d_{n}$ be the distance between the hunter and the rabbit after $n$ rounds. Of course, if $d_{n} \geqslant 100$ for any $n<10^{9}$, the rabbit has won - it just needs to move straight away from the hunter, and the distance will be kept at or above 100 thereon.  We will now show that, while $d_{n}<100$, whatever given strategy the hunter follows, the rabbit has a way of increasing $d_{n}^{2}$ by at least $\frac{1}{2}$ every 200 rounds (as long as the radar pings are lucky enough for the rabbit). This way, $d_{n}^{2}$ will reach $10^{4}$ in less than $2 \cdot 10^{4} \cdot 200=4 \cdot 10^{6}<10^{9}$ rounds, and the rabbit wins.  Suppose the hunter is at $H_{n}$ and the rabbit is at $R_{n}$. Suppose even that the rabbit reveals its position at this moment to the hunter (this allows us to ignore all information from previous radar pings). Let $r$ be the line $H_{n} R_{n}$, and $Y_{1}$ and $Y_{2}$ be points which are 1 unit away from $r$ and 200 units away from $R_{n}$, as in the figure below. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-46.jpg?height=446&width=1140&top_left_y=1636&top_left_x=464)  The rabbit's plan is simply to choose one of the points $Y_{1}$ or $Y_{2}$ and hop 200 rounds straight towards it. Since all hops stay within 1 distance unit from $r$, it is possible that all radar pings stay on $r$. In particular, in this case, the hunter has no way of knowing whether the rabbit chose $Y_{1}$ or $Y_{2}$.  Looking at such pings, what is the hunter going to do? If the hunter's strategy tells him to go 200 rounds straight to the right, he ends up at point $H^{\prime}$ in the figure. Note that the hunter does not have a better alternative! Indeed, after these 200 rounds he will always end up at a point to the left of $H^{\prime}$. If his strategy took him to a point above $r$, he would end up even further from $Y_{2}$; and if his strategy took him below $r$, he would end up even further from $Y_{1}$. In other words, no matter what strategy the hunter follows, he can never be sure his distance to the rabbit will be less than $y \xlongequal{\text { def }} H^{\prime} Y_{1}=H^{\prime} Y_{2}$ after these 200 rounds.  To estimate $y^{2}$, we take $Z$ as the midpoint of segment $Y_{1} Y_{2}$, we take $R^{\prime}$ as a point 200 units to the right of $R_{n}$ and we define $\varepsilon=Z R^{\prime}$ (note that $H^{\prime} R^{\prime}=d_{n}$ ). Then  $$ y^{2}=1+\left(H^{\prime} Z\right)^{2}=1+\left(d_{n}-\varepsilon\right)^{2} $$  where  $$ \varepsilon=200-R_{n} Z=200-\sqrt{200^{2}-1}=\frac{1}{200+\sqrt{200^{2}-1}}>\frac{1}{400} $$  In particular, $\varepsilon^{2}+1=400 \varepsilon$, so  $$ y^{2}=d_{n}^{2}-2 \varepsilon d_{n}+\varepsilon^{2}+1=d_{n}^{2}+\varepsilon\left(400-2 d_{n}\right) $$  Since $\varepsilon>\frac{1}{400}$ and we assumed $d_{n}<100$, this shows that $y^{2}>d_{n}^{2}+\frac{1}{2}$. So, as we claimed, with this list of radar pings, no matter what the hunter does, the rabbit might achieve $d_{n+200}^{2}>d_{n}^{2}+\frac{1}{2}$. The wabbit wins.  Comment 1. Many different versions of the solution above can be found by replacing 200 with some other number $N$ for the number of hops the rabbit takes between reveals. If this is done, we have:  $$ \varepsilon=N-\sqrt{N^{2}-1}>\frac{1}{N+\sqrt{N^{2}-1}}>\frac{1}{2 N} $$  and  $$ \varepsilon^{2}+1=2 N \varepsilon $$  so, as long as $N>d_{n}$, we would find  $$ y^{2}=d_{n}^{2}+\varepsilon\left(2 N-2 d_{n}\right)>d_{n}^{2}+\frac{N-d_{n}}{N} $$  For example, taking $N=101$ is already enough-the squared distance increases by at least $\frac{1}{101}$ every 101 rounds, and $101^{2} \cdot 10^{4}=1.0201 \cdot 10^{8}<10^{9}$ rounds are enough for the rabbit. If the statement is made sharper, some such versions might not work any longer.  Comment 2. The original statement asked whether the distance could be kept under $10^{10}$ in $10^{100}$ rounds.


Let $n>1$ be an integer. An $n \times n \times n$ cube is composed of $n^{3}$ unit cubes. Each unit cube is painted with one color. For each $n \times n \times 1$ box consisting of $n^{2}$ unit cubes (of any of the three possible orientations), we consider the set of the colors present in that box (each color is listed only once). This way, we get $3 n$ sets of colors, split into three groups according to the orientation. It happens that for every set in any group, the same set appears in both of the other groups. Determine, in terms of $n$, the maximal possible number of colors that are present. (Russia) Answer: The maximal number is $\frac{n(n+1)(2 n+1)}{6}$.

Solution: Call a $n \times n \times 1$ box an $x$-box, a $y$-box, or a $z$-box, according to the direction of its short side. Let $C$ be the number of colors in a valid configuration. We start with the upper bound for $C$.  Let $\mathcal{C}_{1}, \mathcal{C}_{2}$, and $\mathcal{C}_{3}$ be the sets of colors which appear in the big cube exactly once, exactly twice, and at least thrice, respectively. Let $M_{i}$ be the set of unit cubes whose colors are in $\mathcal{C}_{i}$, and denote $n_{i}=\left|M_{i}\right|$.  Consider any $x$-box $X$, and let $Y$ and $Z$ be a $y$ - and a $z$-box containing the same set of colors as $X$ does. Claim. $4\left|X \cap M_{1}\right|+\left|X \cap M_{2}\right| \leqslant 3 n+1$. Proof. We distinguish two cases. Case 1: $X \cap M_{1} \neq \varnothing$. A cube from $X \cap M_{1}$ should appear in all three boxes $X, Y$, and $Z$, so it should lie in $X \cap Y \cap Z$. Thus $X \cap M_{1}=X \cap Y \cap Z$ and $\left|X \cap M_{1}\right|=1$.  Consider now the cubes in $X \cap M_{2}$. There are at most $2(n-1)$ of them lying in $X \cap Y$ or $X \cap Z$ (because the cube from $X \cap Y \cap Z$ is in $M_{1}$ ). Let $a$ be some other cube from $X \cap M_{2}$. Recall that there is just one other cube $a^{\prime}$ sharing a color with $a$. But both $Y$ and $Z$ should contain such cube, so $a^{\prime} \in Y \cap Z$ (but $a^{\prime} \notin X \cap Y \cap Z$ ). The map $a \mapsto a^{\prime}$ is clearly injective, so the number of cubes $a$ we are interested in does not exceed $|(Y \cap Z) \backslash X|=n-1$. Thus $\left|X \cap M_{2}\right| \leqslant 2(n-1)+(n-1)=3(n-1)$, and hence $4\left|X \cap M_{1}\right|+\left|X \cap M_{2}\right| \leqslant 4+3(n-1)=3 n+1$. Case 2: $X \cap M_{1}=\varnothing$.  In this case, the same argument applies with several changes. Indeed, $X \cap M_{2}$ contains at most $2 n-1$ cubes from $X \cap Y$ or $X \cap Z$. Any other cube $a$ in $X \cap M_{2}$ corresponds to some $a^{\prime} \in Y \cap Z$ (possibly with $a^{\prime} \in X$ ), so there are at most $n$ of them. All this results in $\left|X \cap M_{2}\right| \leqslant(2 n-1)+n=3 n-1$, which is even better than we need (by the assumptions of our case).  Summing up the inequalities from the Claim over all $x$-boxes $X$, we obtain  $$ 4 n_{1}+n_{2} \leqslant n(3 n+1) $$  Obviously, we also have $n_{1}+n_{2}+n_{3}=n^{3}$. Now we are prepared to estimate $C$. Due to the definition of the $M_{i}$, we have $n_{i} \geqslant i\left|\mathcal{C}_{i}\right|$, so  $$ C \leqslant n_{1}+\frac{n_{2}}{2}+\frac{n_{3}}{3}=\frac{n_{1}+n_{2}+n_{3}}{3}+\frac{4 n_{1}+n_{2}}{6} \leqslant \frac{n^{3}}{3}+\frac{3 n^{2}+n}{6}=\frac{n(n+1)(2 n+1)}{6} $$  It remains to present an example of an appropriate coloring in the above-mentioned number of colors. For each color, we present the set of all cubes of this color. These sets are:  1. $n$ singletons of the form $S_{i}=\{(i, i, i)\}$ (with $1 \leqslant i \leqslant n$ ); 2. $3\binom{n}{2}$ doubletons of the forms $D_{i, j}^{1}=\{(i, j, j),(j, i, i)\}, D_{i, j}^{2}=\{(j, i, j),(i, j, i)\}$, and $D_{i, j}^{3}=$ $\{(j, j, i),(i, i, j)\}$ (with $1 \leqslant i<j \leqslant n)$; 3. $2\binom{n}{3}$ triplets of the form $T_{i, j, k}=\{(i, j, k),(j, k, i),(k, i, j)\}$ (with $1 \leqslant i<j<k \leqslant n$ or $1 \leqslant i<k<j \leqslant n)$.  One may easily see that the $i^{\text {th }}$ boxes of each orientation contain the same set of colors, and that  $$ n+\frac{3 n(n-1)}{2}+\frac{n(n-1)(n-2)}{3}=\frac{n(n+1)(2 n+1)}{6} $$  colors are used, as required.


Solution: We will approach a new version of the original problem. In this new version, each cube may have a color, or be invisible (not both). Now we make sets of colors for each $n \times n \times 1$ box as before (where "invisible" is not considered a color) and group them by orientation, also as before. Finally, we require that, for every non-empty set in any group, the same set must appear in the other 2 groups. What is the maximum number of colors present with these new requirements?  Let us call strange a big $n \times n \times n$ cube whose painting scheme satisfies the new requirements, and let $D$ be the number of colors in a strange cube. Note that any cube that satisfies the original requirements is also strange, so $\max (D)$ is an upper bound for the original answer. Claim. $D \leqslant \frac{n(n+1)(2 n+1)}{6}$. Proof. The proof is by induction on $n$. If $n=1$, we must paint the cube with at most 1 color. Now, pick a $n \times n \times n$ strange cube $A$, where $n \geqslant 2$. If $A$ is completely invisible, $D=0$ and we are done. Otherwise, pick a non-empty set of colors $\mathcal{S}$ which corresponds to, say, the boxes $X, Y$ and $Z$ of different orientations.  Now find all cubes in $A$ whose colors are in $\mathcal{S}$ and make them invisible. Since $X, Y$ and $Z$ are now completely invisible, we can throw them away and focus on the remaining $(n-1) \times(n-1) \times(n-1)$ cube $B$. The sets of colors in all the groups for $B$ are the same as the sets for $A$, removing exactly the colors in $\mathcal{S}$, and no others! Therefore, every nonempty set that appears in one group for $B$ still shows up in all possible orientations (it is possible that an empty set of colors in $B$ only matched $X, Y$ or $Z$ before these were thrown away, but remember we do not require empty sets to match anyway). In summary, $B$ is also strange.  By the induction hypothesis, we may assume that $B$ has at most $\frac{(n-1) n(2 n-1)}{6}$ colors. Since there were at most $n^{2}$ different colors in $\mathcal{S}$, we have that $A$ has at most $\frac{(n-1) n(2 n-1)}{6}+n^{2}=$ $\frac{n(n+1)(2 n+1)}{6}$ colors.  Finally, the construction in the previous solution shows a painting scheme (with no invisible cubes) that reaches this maximum, so we are done.


For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\mathrm{th}}$ smallest positive integer not in $X$, and let  $$ X * Y=X \cup\left\{f_{X}(y): y \in Y\right\} $$  Let $A$ be a set of $a>0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, then  $$ \underbrace{A *(A * \cdots *(A *(A * A)) \ldots)}_{A \text { appears } b \text { times }}=\underbrace{B *(B * \cdots *(B *(B * B)) \ldots)}_{B \text { appears } a \text { times }} . $$

Solution: For any function $g: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ and any subset $X \subset \mathbb{Z}_{>0}$, we define $g(X)=$ $\{g(x): x \in X\}$. We have that the image of $f_{X}$ is $f_{X}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash X$. We now show a general lemma about the operation *, with the goal of showing that $*$ is associative. Lemma 1. Let $X$ and $Y$ be finite sets of positive integers. The functions $f_{X * Y}$ and $f_{X} \circ f_{Y}$ are equal. Proof. We have $f_{X * Y}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash(X * Y)=\left(\mathbb{Z}_{>0} \backslash X\right) \backslash f_{X}(Y)=f_{X}\left(\mathbb{Z}_{>0}\right) \backslash f_{X}(Y)=f_{X}\left(\mathbb{Z}_{>0} \backslash Y\right)=f_{X}\left(f_{Y}\left(\mathbb{Z}_{>0}\right)\right)$. Thus, the functions $f_{X * Y}$ and $f_{X} \circ f_{Y}$ are strictly increasing functions with the same range. Because a strictly function is uniquely defined by its range, we have $f_{X * Y}=f_{X} \circ f_{Y}$.  Lemma 1 implies that * is associative, in the sense that $(A * B) * C=A *(B * C)$ for any finite sets $A, B$, and $C$ of positive integers. We prove the associativity by noting  $$ \begin{gathered} \mathbb{Z}_{>0} \backslash((A * B) * C)=f_{(A * B) * C}\left(\mathbb{Z}_{>0}\right)=f_{A * B}\left(f_{C}\left(\mathbb{Z}_{>0}\right)\right)=f_{A}\left(f_{B}\left(f_{C}\left(\mathbb{Z}_{>0}\right)\right)\right) \\ =f_{A}\left(f_{B * C}\left(\mathbb{Z}_{>0}\right)=f_{A *(B * C)}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash(A *(B * C))\right. \end{gathered} $$  In light of the associativity of *, we may drop the parentheses when we write expressions like $A *(B * C)$. We also introduce the notation  $$ X^{* k}=\underbrace{X *(X * \cdots *(X *(X * X)) \ldots)}_{X \text { appears } k \text { times }} $$  Our goal is then to show that $A * B=B * A$ implies $A^{* b}=B^{* a}$. We will do so via the following general lemma. Lemma 2. Suppose that $X$ and $Y$ are finite sets of positive integers satisfying $X * Y=Y * X$ and $|X|=|Y|$. Then, we must have $X=Y$. Proof. Assume that $X$ and $Y$ are not equal. Let $s$ be the largest number in exactly one of $X$ and $Y$. Without loss of generality, say that $s \in X \backslash Y$. The number $f_{X}(s)$ counts the $s^{t h}$ number not in $X$, which implies that  $$ f_{X}(s)=s+\left|X \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right| $$  Since $f_{X}(s) \geqslant s$, we have that  $$ \left\{f_{X}(s)+1, f_{X}(s)+2, \ldots\right\} \cap X=\left\{f_{X}(s)+1, f_{X}(s)+2, \ldots\right\} \cap Y $$  which, together with the assumption that $|X|=|Y|$, gives  $$ \left|X \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right|=\left|Y \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right| $$  Now consider the equation  $$ t-|Y \cap\{1,2, \ldots, t\}|=s $$  This equation is satisfied only when $t \in\left[f_{Y}(s), f_{Y}(s+1)\right)$, because the left hand side counts the number of elements up to $t$ that are not in $Y$. We have that the value $t=f_{X}(s)$ satisfies the above equation because of (1) and (2). Furthermore, since $f_{X}(s) \notin X$ and $f_{X}(s) \geqslant s$, we have that $f_{X}(s) \notin Y$ due to the maximality of $s$. Thus, by the above discussion, we must have $f_{X}(s)=f_{Y}(s)$.  Finally, we arrive at a contradiction. The value $f_{X}(s)$ is neither in $X$ nor in $f_{X}(Y)$, because $s$ is not in $Y$ by assumption. Thus, $f_{X}(s) \notin X * Y$. However, since $s \in X$, we have $f_{Y}(s) \in Y * X$, a contradiction.  We are now ready to finish the proof. Note first of all that $\left|A^{* b}\right|=a b=\left|B^{* a}\right|$. Moreover, since $A * B=B * A$, and $*$ is associative, it follows that $A^{* b} * B^{* a}=B^{* a} * A^{* b}$. Thus, by Lemma 2, we have $A^{* b}=B^{* a}$, as desired.  Comment 1. Taking $A=X^{* k}$ and $B=X^{* l}$ generates many non-trivial examples where $A * B=B * A$. There are also other examples not of this form. For example, if $A=\{1,2,4\}$ and $B=\{1,3\}$, then $A * B=\{1,2,3,4,6\}=B * A$.


Solution: We will use Lemma 1 from   $$ f_{X}=f_{Y} \Longleftrightarrow f_{X}\left(\mathbb{Z}_{>0}\right)=f_{Y}\left(\mathbb{Z}_{>0}\right) \Longleftrightarrow\left(\mathbb{Z}_{>0} \backslash X\right)=\left(\mathbb{Z}_{>0} \backslash Y\right) \Longleftrightarrow X=Y, $$  where the first equivalence is because $f_{X}$ and $f_{Y}$ are strictly increasing functions, and the second equivalence is because $f_{X}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash X$ and $f_{Y}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash Y$.  Denote $g=f_{A}$ and $h=f_{B}$. The given relation $A * B=B * A$ is equivalent to $f_{A * B}=f_{B * A}$ because of (3), and by Lemma 1 of the first solution, this is equivalent to $g \circ h=h \circ g$. Similarly, the required relation $A^{* b}=B^{* a}$ is equivalent to $g^{b}=h^{a}$. We will show that  $$ g^{b}(n)=h^{a}(n) $$  for all $n \in \mathbb{Z}_{>0}$, which suffices to solve the problem. To start, we claim that (4) holds for all sufficiently large $n$. Indeed, let $p$ and $q$ be the maximal elements of $A$ and $B$, respectively; we may assume that $p \geqslant q$. Then, for every $n \geqslant p$ we have $g(n)=n+a$ and $h(n)=n+b$, whence $g^{b}(n)=n+a b=h^{a}(n)$, as was claimed.  In view of this claim, if (4) is not identically true, then there exists a maximal $s$ with $g^{b}(s) \neq$ $h^{a}(s)$. Without loss of generality, we may assume that $g(s) \neq s$, for if we had $g(s)=h(s)=s$, then $s$ would satisfy (4). As $g$ is increasing, we then have $g(s)>s$, so (4) holds for $n=g(s)$. But then we have  $$ g\left(g^{b}(s)\right)=g^{b+1}(s)=g^{b}(n)=h^{a}(n)=h^{a}(g(s))=g\left(h^{a}(s)\right), $$  where the last equality holds in view of $g \circ h=h \circ g$. By the injectivity of $g$, the above equality yields $g^{b}(s)=h^{a}(s)$, which contradicts the choice of $s$. Thus, we have proved that (4) is identically true on $\mathbb{Z}_{>0}$, as desired.  Comment 2. We present another proof of Lemma 2 of the first solution. Let $x=|X|=|Y|$. Say that $u$ is the smallest number in $X$ and $v$ is the smallest number in $Y$; assume without loss of generality that $u \leqslant v$.  Let $T$ be any finite set of positive integers, and define $t=|T|$. Enumerate the elements of $X$ as $x_{1}<x_{2}<\cdots<x_{n}$. Define $S_{m}=f_{\left(T * X^{*(m-1)}\right)}(X)$, and enumerate its elements $s_{m, 1}<s_{m, 2}<\cdots<$ $s_{m, n}$. Note that the $S_{m}$ are pairwise disjoint; indeed, if we have $m<m^{\prime}$, then  $$ S_{m} \subset T * X^{* m} \subset T * X^{*\left(m^{\prime}-1\right)} \quad \text { and } \quad S_{m^{\prime}}=\left(T * X^{* m^{\prime}}\right) \backslash\left(T * X^{*\left(m^{\prime}-1\right)}\right) $$  We claim the following statement, which essentially says that the $S_{m}$ are eventually linear translates of each other:  Claim. For every $i$, there exists some $m_{i}$ and $c_{i}$ such that for all $m>m_{i}$, we have that $s_{m, i}=t+m n-c_{i}$. Furthermore, the $c_{i}$ do not depend on the choice of $T$.  First, we show that this claim implies Lemma 2. We may choose $T=X$ and $T=Y$. Then, there is some $m^{\prime}$ such that for all $m \geqslant m^{\prime}$, we have  $$ f_{X * m}(X)=f_{(Y * X *(m-1))}(X) $$  Because $u$ is the minimum element of $X, v$ is the minimum element of $Y$, and $u \leqslant v$, we have that  $$ \left(\bigcup_{m=m^{\prime}}^{\infty} f_{X * m}(X)\right) \cup X^{* m^{\prime}}=\left(\bigcup_{m=m^{\prime}}^{\infty} f_{\left(Y * X^{*(m-1)}\right)}(X)\right) \cup\left(Y * X^{*\left(m^{\prime}-1\right)}\right)=\{u, u+1, \ldots\}, $$  and in both the first and second expressions, the unions are of pairwise distinct sets. By (5), we obtain $X^{* m^{\prime}}=Y * X^{*\left(m^{\prime}-1\right)}$. Now, because $X$ and $Y$ commute, we get $X^{* m^{\prime}}=X^{*\left(m^{\prime}-1\right)} * Y$, and so $X=Y$.  We now prove the claim. Proof of the claim. We induct downwards on $i$, first proving the statement for $i=n$, and so on. Assume that $m$ is chosen so that all elements of $S_{m}$ are greater than all elements of $T$ (which is possible because $T$ is finite). For $i=n$, we have that $s_{m, n}>s_{k, n}$ for every $k<m$. Thus, all ( $m-1$ ) $n$ numbers of the form $s_{k, u}$ for $k<m$ and $1 \leqslant u \leqslant n$ are less than $s_{m, n}$. We then have that $s_{m, n}$ is the $\left((m-1) n+x_{n}\right)^{t h}$ number not in $T$, which is equal to $t+(m-1) n+x_{n}$. So we may choose $c_{n}=x_{n}-n$, which does not depend on $T$, which proves the base case for the induction.  For $i<n$, we have again that all elements $s_{m, j}$ for $j<i$ and $s_{p, i}$ for $p<m$ are less than $s_{m, i}$, so $s_{m, i}$ is the $\left((m-1) i+x_{i}\right)^{t h}$ element not in $T$ or of the form $s_{p, j}$ for $j>i$ and $p<m$. But by the inductive hypothesis, each of the sequences $s_{p, j}$ is eventually periodic with period $n$, and thus the sequence $s_{m, i}$ such must be as well. Since each of the sequences $s_{p, j}-t$ with $j>i$ eventually do not depend on $T$, the sequence $s_{m, i}-t$ eventually does not depend on $T$ either, so the inductive step is complete. This proves the claim and thus Lemma 2.


Let $n$ be a given positive integer. In the Cartesian plane, each lattice point with nonnegative coordinates initially contains a butterfly, and there are no other butterflies. The neighborhood of a lattice point $c$ consists of all lattice points within the axis-aligned $(2 n+1) \times$ $(2 n+1)$ square centered at $c$, apart from $c$ itself. We call a butterfly lonely, crowded, or comfortable, depending on whether the number of butterflies in its neighborhood $N$ is respectively less than, greater than, or equal to half of the number of lattice points in $N$.  Every minute, all lonely butterflies fly away simultaneously. This process goes on for as long as there are any lonely butterflies. Assuming that the process eventually stops, determine the number of comfortable butterflies at the final state. (Bulgaria) Answer: $n^{2}+1$.

Solution: We always identify a butterfly with the lattice point it is situated at. For two points $p$ and $q$, we write $p \geqslant q$ if each coordinate of $p$ is at least the corresponding coordinate of $q$. Let $O$ be the origin, and let $\mathcal{Q}$ be the set of initially occupied points, i.e., of all lattice points with nonnegative coordinates. Let $\mathcal{R}_{\mathrm{H}}=\{(x, 0): x \geqslant 0\}$ and $\mathcal{R}_{\mathrm{V}}=\{(0, y): y \geqslant 0\}$ be the sets of the lattice points lying on the horizontal and vertical boundary rays of $\mathcal{Q}$. Denote by $N(a)$ the neighborhood of a lattice point $a$.  1. Initial observations. We call a set of lattice points up-right closed if its points stay in the set after being shifted by any lattice vector $(i, j)$ with $i, j \geqslant 0$. Whenever the butterflies form a up-right closed set $\mathcal{S}$, we have $|N(p) \cap \mathcal{S}| \geqslant|N(q) \cap \mathcal{S}|$ for any two points $p, q \in \mathcal{S}$ with $p \geqslant q$. So, since $\mathcal{Q}$ is up-right closed, the set of butterflies at any moment also preserves this property. We assume all forthcoming sets of lattice points to be up-right closed.  When speaking of some set $\mathcal{S}$ of lattice points, we call its points lonely, comfortable, or crowded with respect to this set (i.e., as if the butterflies were exactly at all points of $\mathcal{S}$ ). We call a set $\mathcal{S} \subset \mathcal{Q}$ stable if it contains no lonely points. In what follows, we are interested only in those stable sets whose complements in $\mathcal{Q}$ are finite, because one can easily see that only a finite number of butterflies can fly away on each minute.  If the initial set $\mathcal{Q}$ of butterflies contains some stable set $\mathcal{S}$, then, clearly no butterfly of this set will fly away. On the other hand, the set $\mathcal{F}$ of all butterflies in the end of the process is stable. This means that $\mathcal{F}$ is the largest (with respect to inclusion) stable set within $\mathcal{Q}$, and we are about to describe this set. 2. A description of a final set. The following notion will be useful. Let $\mathcal{U}=\left\{\vec{u}_{1}, \vec{u}_{2}, \ldots, \vec{u}_{d}\right\}$ be a set of $d$ pairwise non-parallel lattice vectors, each having a positive $x$ - and a negative $y$-coordinate. Assume that they are numbered in increasing order according to slope. We now define a $\mathcal{U}$-curve to be the broken line $p_{0} p_{1} \ldots p_{d}$ such that $p_{0} \in \mathcal{R}_{\mathrm{V}}, p_{d} \in \mathcal{R}_{\mathrm{H}}$, and $\overrightarrow{p_{i-1} p_{i}}=\vec{u}_{i}$ for all $i=1,2, \ldots, m$ (see the Figure below to the left). ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-53.jpg?height=546&width=766&top_left_y=2194&top_left_x=191)  Construction of $\mathcal{U}$-curve ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-53.jpg?height=415&width=803&top_left_y=2322&top_left_x=1066)  Construction of $\mathcal{D}$  Now, let $\mathcal{K}_{n}=\{(i, j): 1 \leqslant i \leqslant n,-n \leqslant j \leqslant-1\}$. Consider all the rays emerging at $O$ and passing through a point from $\mathcal{K}_{n}$; number them as $r_{1}, \ldots, r_{m}$ in increasing order according to slope. Let $A_{i}$ be the farthest from $O$ lattice point in $r_{i} \cap \mathcal{K}_{n}$, set $k_{i}=\left|r_{i} \cap \mathcal{K}_{n}\right|$, let $\vec{v}_{i}=\overrightarrow{O A_{i}}$, and finally denote $\mathcal{V}=\left\{\vec{v}_{i}: 1 \leqslant i \leqslant m\right\}$; see the Figure above to the right. We will concentrate on the $\mathcal{V}$-curve $d_{0} d_{1} \ldots d_{m}$; let $\mathcal{D}$ be the set of all lattice points $p$ such that $p \geqslant p^{\prime}$ for some (not necessarily lattice) point $p^{\prime}$ on the $\mathcal{V}$-curve. In fact, we will show that $\mathcal{D}=\mathcal{F}$.  Clearly, the $\mathcal{V}$-curve is symmetric in the line $y=x$. Denote by $D$ the convex hull of $\mathcal{D}$. 3. We prove that the set $\mathcal{D}$ contains all stable sets. Let $\mathcal{S} \subset \mathcal{Q}$ be a stable set (recall that it is assumed to be up-right closed and to have a finite complement in $\mathcal{Q}$ ). Denote by $S$ its convex hull; clearly, the vertices of $S$ are lattice points. The boundary of $S$ consists of two rays (horizontal and vertical ones) along with some $\mathcal{V}_{*}$-curve for some set of lattice vectors $\mathcal{V}_{*}$. Claim 1. For every $\vec{v}_{i} \in \mathcal{V}$, there is a $\vec{v}_{i}^{*} \in \mathcal{V}_{*}$ co-directed with $\vec{v}$ with $\left|\vec{v}_{i}^{*}\right| \geqslant|\vec{v}|$. Proof. Let $\ell$ be the supporting line of $S$ parallel to $\vec{v}_{i}$ (i.e., $\ell$ contains some point of $S$, and the set $S$ lies on one side of $\ell$ ). Take any point $b \in \ell \cap \mathcal{S}$ and consider $N(b)$. The line $\ell$ splits the set $N(b) \backslash \ell$ into two congruent parts, one having an empty intersection with $\mathcal{S}$. Hence, in order for $b$ not to be lonely, at least half of the set $\ell \cap N(b)$ (which contains $2 k_{i}$ points) should lie in $S$. Thus, the boundary of $S$ contains a segment $\ell \cap S$ with at least $k_{i}+1$ lattice points (including $b$ ) on it; this segment corresponds to the required vector $\vec{v}_{i}^{*} \in \mathcal{V}_{*}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-54.jpg?height=600&width=1626&top_left_y=1199&top_left_x=221)  Claim 2. Each stable set $\mathcal{S} \subseteq \mathcal{Q}$ lies in $\mathcal{D}$. Proof. To show this, it suffices to prove that the $\mathcal{V}_{*}$-curve lies in $D$, i.e., that all its vertices do so. Let $p^{\prime}$ be an arbitrary vertex of the $\mathcal{V}_{*^{-}}$-curve; $p^{\prime}$ partitions this curve into two parts, $\mathcal{X}$ (being down-right of $p$ ) and $\mathcal{Y}$ (being up-left of $p$ ). The set $\mathcal{V}$ is split now into two parts: $\mathcal{V}_{\mathcal{X}}$ consisting of those $\vec{v}_{i} \in \mathcal{V}$ for which $\vec{v}_{i}^{*}$ corresponds to a segment in $\mathcal{X}$, and a similar part $\mathcal{V}_{\mathcal{Y}}$. Notice that the $\mathcal{V}$-curve consists of several segments corresponding to $\mathcal{V}_{\mathcal{X}}$, followed by those corresponding to $\mathcal{V}_{\mathcal{Y}}$. Hence there is a vertex $p$ of the $\mathcal{V}$-curve separating $\mathcal{V}_{\mathcal{X}}$ from $\mathcal{V}_{\mathcal{Y}}$. Claim 1 now yields that $p^{\prime} \geqslant p$, so $p^{\prime} \in \mathcal{D}$, as required.  Claim 2 implies that the final set $\mathcal{F}$ is contained in $\mathcal{D}$. 4. $\mathcal{D}$ is stable, and its comfortable points are known. Recall the definitions of $r_{i}$; let $r_{i}^{\prime}$ be the ray complementary to $r_{i}$. By our definitions, the set $N(O)$ contains no points between the rays $r_{i}$ and $r_{i+1}$, as well as between $r_{i}^{\prime}$ and $r_{i+1}^{\prime}$. Claim 3. In the set $\mathcal{D}$, all lattice points of the $\mathcal{V}$-curve are comfortable. Proof. Let $p$ be any lattice point of the $\mathcal{V}$-curve, belonging to some segment $d_{i} d_{i+1}$. Draw the line $\ell$ containing this segment. Then $\ell \cap \mathcal{D}$ contains exactly $k_{i}+1$ lattice points, all of which lie in $N(p)$ except for $p$. Thus, exactly half of the points in $N(p) \cap \ell$ lie in $\mathcal{D}$. It remains to show that all points of $N(p)$ above $\ell$ lie in $\mathcal{D}$ (recall that all the points below $\ell$ lack this property).  Notice that each vector in $\mathcal{V}$ has one coordinate greater than $n / 2$; thus the neighborhood of $p$ contains parts of at most two segments of the $\mathcal{V}$-curve succeeding $d_{i} d_{i+1}$, as well as at most two of those preceding it.  The angles formed by these consecutive segments are obtained from those formed by $r_{j}$ and $r_{j-1}^{\prime}$ (with $i-1 \leqslant j \leqslant i+2$ ) by shifts; see the Figure below. All the points in $N(p)$ above $\ell$ which could lie outside $\mathcal{D}$ lie in shifted angles between $r_{j}, r_{j+1}$ or $r_{j}^{\prime}, r_{j-1}^{\prime}$. But those angles, restricted to $N(p)$, have no lattice points due to the above remark. The claim is proved. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-55.jpg?height=404&width=1052&top_left_y=592&top_left_x=503)  Proof of Claim 3 Claim 4. All the points of $\mathcal{D}$ which are not on the boundary of $D$ are crowded. Proof. Let $p \in \mathcal{D}$ be such a point. If it is to the up-right of some point $p^{\prime}$ on the curve, then the claim is easy: the shift of $N\left(p^{\prime}\right) \cap \mathcal{D}$ by $\overrightarrow{p^{\prime} p}$ is still in $\mathcal{D}$, and $N(p)$ contains at least one more point of $\mathcal{D}$ - either below or to the left of $p$. So, we may assume that $p$ lies in a right triangle constructed on some hypothenuse $d_{i} d_{i+1}$. Notice here that $d_{i}, d_{i+1} \in N(p)$.  Draw a line $\ell \| d_{i} d_{i+1}$ through $p$, and draw a vertical line $h$ through $d_{i}$; see Figure below. Let $\mathcal{D}_{\mathrm{L}}$ and $\mathcal{D}_{\mathrm{R}}$ be the parts of $\mathcal{D}$ lying to the left and to the right of $h$, respectively (points of $\mathcal{D} \cap h$ lie in both parts). ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-55.jpg?height=538&width=1120&top_left_y=1556&top_left_x=468)  Notice that the vectors $\overrightarrow{d_{i} p}, \overrightarrow{d_{i+1} d_{i+2}}, \overrightarrow{d_{i} d_{i+1}}, \overrightarrow{d_{i-1} d_{i}}$, and $\overrightarrow{p d_{i+1}}$ are arranged in non-increasing order by slope. This means that $\mathcal{D}_{\mathrm{L}}$ shifted by $\overrightarrow{d_{i} p}$ still lies in $\mathcal{D}$, as well as $\mathcal{D}_{\mathrm{R}}$ shifted by $\overrightarrow{d_{i+1} p}$. As we have seen in the proof of Claim 3, these two shifts cover all points of $N(p)$ above $\ell$, along with those on $\ell$ to the left of $p$. Since $N(p)$ contains also $d_{i}$ and $d_{i+1}$, the point $p$ is crowded.  Thus, we have proved that $\mathcal{D}=\mathcal{F}$, and have shown that the lattice points on the $\mathcal{V}$-curve are exactly the comfortable points of $\mathcal{D}$. It remains to find their number.  Recall the definition of $\mathcal{K}_{n}$ (see Figure on the first page of the solution). Each segment $d_{i} d_{i+1}$ contains $k_{i}$ lattice points different from $d_{i}$. Taken over all $i$, these points exhaust all the lattice points in the $\mathcal{V}$-curve, except for $d_{1}$, and thus the number of lattice points on the $\mathcal{V}$-curve is $1+\sum_{i=1}^{m} k_{i}$. On the other hand, $\sum_{i=1}^{m} k_{i}$ is just the number of points in $\mathcal{K}_{n}$, so it equals $n^{2}$. Hence the answer to the problem is $n^{2}+1$.  Comment 1. The assumption that the process eventually stops is unnecessary for the problem, as one can see that, in fact, the process stops for every $n \geqslant 1$. Indeed, the proof of Claims 3 and 4 do not rely essentially on this assumption, and they together yield that the set $\mathcal{D}$ is stable. So, only butterflies that are not in $\mathcal{D}$ may fly away, and this takes only a finite time.  This assumption has been inserted into the problem statement in order to avoid several technical details regarding finiteness issues. It may also simplify several other arguments.  Comment 2. The description of the final set $\mathcal{F}(=\mathcal{D})$ seems to be crucial for the solution; the Problem Selection Committee is not aware of any solution that completely avoids such a description.  On the other hand, after the set $\mathcal{D}$ has been defined, the further steps may be performed in several ways. For example, in order to prove that all butterflies outside $\mathcal{D}$ will fly away, one may argue as follows. (Here we will also make use of the assumption that the process eventually stops.)  First of all, notice that the process can be modified in the following manner: Each minute, exactly one of the lonely butterflies flies away, until there are no more lonely butterflies. The modified process necessarily stops at the same state as the initial one. Indeed, one may observe, as in solution above, that the (unique) largest stable set is still the final set for the modified process.  Thus, in order to prove our claim, it suffices to indicate an order in which the butterflies should fly away in the new process; if we are able to exhaust the whole set $\mathcal{Q} \backslash \mathcal{D}$, we are done.  Let $\mathcal{C}_{0}=d_{0} d_{1} \ldots d_{m}$ be the $\mathcal{V}$-curve. Take its copy $\mathcal{C}$ and shift it downwards so that $d_{0}$ comes to some point below the origin $O$. Now we start moving $\mathcal{C}$ upwards continuously, until it comes back to its initial position $\mathcal{C}_{0}$. At each moment when $\mathcal{C}$ meets some lattice points, we convince all the butterflies at those points to fly away in a certain order. We will now show that we always have enough arguments for butterflies to do so, which will finish our argument for the claim..  Let $\mathcal{C}^{\prime}=d_{0}^{\prime} d_{1}^{\prime} \ldots d_{m}^{\prime}$ be a position of $\mathcal{C}$ when it meets some butterflies. We assume that all butterflies under this current position of $\mathcal{C}$ were already convinced enough and flied away. Consider the lowest butterfly $b$ on $\mathcal{C}^{\prime}$. Let $d_{i}^{\prime} d_{i+1}^{\prime}$ be the segment it lies on; we choose $i$ so that $b \neq d_{i+1}^{\prime}$ (this is possible because $\mathcal{C}$ as not yet reached $\mathcal{C}_{0}$ ).  Draw a line $\ell$ containing the segment $d_{i}^{\prime} d_{i+1}^{\prime}$. Then all the butterflies in $N(b)$ are situated on or above $\ell$; moreover, those on $\ell$ all lie on the segment $d_{i} d_{i+1}$. But this segment now contains at most $k_{i}$ butterflies (including $b$ ), since otherwise some butterfly had to occupy $d_{i+1}^{\prime}$ which is impossible by the choice of $b$. Thus, $b$ is lonely and hence may be convinced to fly away.  After $b$ has flied away, we switch to the lowest of the remaining butterflies on $\mathcal{C}^{\prime}$, and so on. Claims 3 and 4 also allow some different proofs which are not presented here.  This page is intentionally left blank


There are 2017 mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard, one at a time, so that no tangent segment intersects any other circles or previously drawn tangent segments. Luciano keeps drawing tangent segments until no more can be drawn. Find all possible numbers of tangent segments when he stops drawing. (Australia) Answer: If there were $n$ circles, there would always be exactly $3(n-1)$ segments; so the only possible answer is $3 \cdot 2017-3=6048$.

Solution: First, consider a particular arrangement of circles $C_{1}, C_{2}, \ldots, C_{n}$ where all the centers are aligned and each $C_{i}$ is eclipsed from the other circles by its neighbors - for example, taking $C_{i}$ with center $\left(i^{2}, 0\right)$ and radius $i / 2$ works. Then the only tangent segments that can be drawn are between adjacent circles $C_{i}$ and $C_{i+1}$, and exactly three segments can be drawn for each pair. So Luciano will draw exactly $3(n-1)$ segments in this case. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-72.jpg?height=258&width=1335&top_left_y=1059&top_left_x=355)  For the general case, start from a final configuration (that is, an arrangement of circles and segments in which no further segments can be drawn). The idea of the solution is to continuously resize and move the circles around the plane, one by one (in particular, making sure we never have 4 circles with a common tangent line), and show that the number of segments drawn remains constant as the picture changes. This way, we can reduce any circle/segment configuration to the particular one mentioned above, and the final number of segments must remain at $3 n-3$.  Some preliminary considerations: look at all possible tangent segments joining any two circles. A segment that is tangent to a circle $A$ can do so in two possible orientations - it may come out of $A$ in clockwise or counterclockwise orientation. Two segments touching the same circle with the same orientation will never intersect each other. Each pair $(A, B)$ of circles has 4 choices of tangent segments, which can be identified by their orientations - for example, $(A+, B-)$ would be the segment which comes out of $A$ in clockwise orientation and comes out of $B$ in counterclockwise orientation. In total, we have $2 n(n-1)$ possible segments, disregarding intersections.  Now we pick a circle $C$ and start to continuously move and resize it, maintaining all existing tangent segments according to their identifications, including those involving $C$. We can keep our choice of tangent segments until the configuration reaches a transition. We lose nothing if we assume that $C$ is kept at least $\varepsilon$ units away from any other circle, where $\varepsilon$ is a positive, fixed constant; therefore at a transition either: (1) a currently drawn tangent segment $t$ suddenly becomes obstructed; or (2) a currently absent tangent segment $t$ suddenly becomes unobstructed and available. Claim. A transition can only occur when three circles $C_{1}, C_{2}, C_{3}$ are tangent to a common line $\ell$ containing $t$, in a way such that the three tangent segments lying on $\ell$ (joining the three circles pairwise) are not obstructed by any other circles or tangent segments (other than $C_{1}, C_{2}, C_{3}$ ). Proof. Since (2) is effectively the reverse of (1), it suffices to prove the claim for (1). Suppose $t$ has suddenly become obstructed, and let us consider two cases.  Case 1: $t$ becomes obstructed by a circle ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-73.jpg?height=318&width=1437&top_left_y=292&top_left_x=315)  Then the new circle becomes the third circle tangent to $\ell$, and no other circles or tangent segments are obstructing $t$.


Solution: First note that all tangent segments lying on the boundary of the convex hull of the circles are always drawn since they do not intersect anything else. Now in the final picture, aside from the $n$ circles, the blackboard is divided into regions. We can consider the picture as a plane (multi-)graph $G$ in which the circles are the vertices and the tangent segments are the edges. The idea of this solution is to find a relation between the number of edges and the number of regions in $G$; then, once we prove that $G$ is connected, we can use Euler's formula to finish the problem.  The boundary of each region consists of 1 or more (for now) simple closed curves, each made of arcs and tangent segments. The segment and the arc might meet smoothly (as in $S_{i}$, $i=1,2, \ldots, 6$ in the figure below) or not (as in $P_{1}, P_{2}, P_{3}, P_{4}$; call such points sharp corners of the boundary). In other words, if a person walks along the border, her direction would suddenly turn an angle of $\pi$ at a sharp corner. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-75.jpg?height=626&width=577&top_left_y=178&top_left_x=748)  Claim 1. The outer boundary $B_{1}$ of any internal region has at least 3 sharp corners. Proof. Let a person walk one lap along $B_{1}$ in the counterclockwise orientation. As she does so, she will turn clockwise as she moves along the circle arcs, and not turn at all when moving along the lines. On the other hand, her total rotation after one lap is $2 \pi$ in the counterclockwise direction! Where could she be turning counterclockwise? She can only do so at sharp corners, and, even then, she turns only an angle of $\pi$ there. But two sharp corners are not enough, since at least one arc must be present-so she must have gone through at least 3 sharp corners.  Claim 2. Each internal region is simply connected, that is, has only one boundary curve. Proof. Suppose, by contradiction, that some region has an outer boundary $B_{1}$ and inner boundaries $B_{2}, B_{3}, \ldots, B_{m}(m \geqslant 2)$. Let $P_{1}$ be one of the sharp corners of $B_{1}$.  Now consider a car starting at $P_{1}$ and traveling counterclockwise along $B_{1}$. It starts in reverse, i.e., it is initially facing the corner $P_{1}$. Due to the tangent conditions, the car may travel in a way so that its orientation only changes when it is moving along an arc. In particular, this means the car will sometimes travel forward. For example, if the car approaches a sharp corner when driving in reverse, it would continue travel forward after the corner, instead of making an immediate half-turn. This way, the orientation of the car only changes in a clockwise direction since the car always travels clockwise around each arc.  Now imagine there is a laser pointer at the front of the car, pointing directly ahead. Initially, the laser endpoint hits $P_{1}$, but, as soon as the car hits an arc, the endpoint moves clockwise around $B_{1}$. In fact, the laser endpoint must move continuously along $B_{1}$ ! Indeed, if the endpoint ever jumped (within $B_{1}$, or from $B_{1}$ to one of the inner boundaries), at the moment of the jump the interrupted laser would be a drawable tangent segment that Luciano missed (see figure below for an example). ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-75.jpg?height=637&width=587&top_left_y=2140&top_left_x=740)  Now, let $P_{2}$ and $P_{3}$ be the next two sharp corners the car goes through, after $P_{1}$ (the previous lemma assures their existence). At $P_{2}$ the car starts moving forward, and at $P_{3}$ it will start to move in reverse again. So, at $P_{3}$, the laser endpoint is at $P_{3}$ itself. So while the car moved counterclockwise between $P_{1}$ and $P_{3}$, the laser endpoint moved clockwise between $P_{1}$ and $P_{3}$. That means the laser beam itself scanned the whole region within $B_{1}$, and it should have crossed some of the inner boundaries.  Claim 3. Each region has exactly 3 sharp corners. Proof. Consider again the car of the previous claim, with its laser still firmly attached to its front, traveling the same way as before and going through the same consecutive sharp corners $P_{1}, P_{2}$ and $P_{3}$. As we have seen, as the car goes counterclockwise from $P_{1}$ to $P_{3}$, the laser endpoint goes clockwise from $P_{1}$ to $P_{3}$, so together they cover the whole boundary. If there were a fourth sharp corner $P_{4}$, at some moment the laser endpoint would pass through it. But, since $P_{4}$ is a sharp corner, this means the car must be on the extension of a tangent segment going through $P_{4}$. Since the car is not on that segment itself (the car never goes through $P_{4}$ ), we would have 3 circles with a common tangent line, which is not allowed. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-76.jpg?height=538&width=792&top_left_y=1007&top_left_x=643)  We are now ready to finish the solution. Let $r$ be the number of internal regions, and $s$ be the number of tangent segments. Since each tangent segment contributes exactly 2 sharp corners to the diagram, and each region has exactly 3 sharp corners, we must have $2 s=3 r$. Since the graph corresponding to the diagram is connected, we can use Euler's formula $n-s+r=1$ and find $s=3 n-3$ and $r=2 n-2$.


The sequence $a_{0}, a_{1}, a_{2}, \ldots$ of positive integers satisfies  $$ a_{n+1}=\left\{\begin{array}{ll} \sqrt{a_{n}}, & \text { if } \sqrt{a_{n}} \text { is an integer } \\ a_{n}+3, & \text { otherwise } \end{array} \quad \text { for every } n \geqslant 0\right. $$  Determine all values of $a_{0}>1$ for which there is at least one number $a$ such that $a_{n}=a$ for infinitely many values of $n$. (South Africa) Answer: All positive multiples of 3.

Solution: Since the value of $a_{n+1}$ only depends on the value of $a_{n}$, if $a_{n}=a_{m}$ for two different indices $n$ and $m$, then the sequence is eventually periodic. So we look for the values of $a_{0}$ for which the sequence is eventually periodic. Claim 1. If $a_{n} \equiv-1(\bmod 3)$, then, for all $m>n, a_{m}$ is not a perfect square. It follows that the sequence is eventually strictly increasing, so it is not eventually periodic. Proof. A square cannot be congruent to -1 modulo 3 , so $a_{n} \equiv-1(\bmod 3)$ implies that $a_{n}$ is not a square, therefore $a_{n+1}=a_{n}+3>a_{n}$. As a consequence, $a_{n+1} \equiv a_{n} \equiv-1(\bmod 3)$, so $a_{n+1}$ is not a square either. By repeating the argument, we prove that, from $a_{n}$ on, all terms of the sequence are not perfect squares and are greater than their predecessors, which completes the proof. Claim 2. If $a_{n} \not \equiv-1(\bmod 3)$ and $a_{n}>9$ then there is an index $m>n$ such that $a_{m}<a_{n}$. Proof. Let $t^{2}$ be the largest perfect square which is less than $a_{n}$. Since $a_{n}>9, t$ is at least 3. The first square in the sequence $a_{n}, a_{n}+3, a_{n}+6, \ldots$ will be $(t+1)^{2},(t+2)^{2}$ or $(t+3)^{2}$, therefore there is an index $m>n$ such that $a_{m} \leqslant t+3<t^{2}<a_{n}$, as claimed. Claim 3. If $a_{n} \equiv 0(\bmod 3)$, then there is an index $m>n$ such that $a_{m}=3$. Proof. First we notice that, by the definition of the sequence, a multiple of 3 is always followed by another multiple of 3 . If $a_{n} \in\{3,6,9\}$ the sequence will eventually follow the periodic pattern $3,6,9,3,6,9, \ldots$ If $a_{n}>9$, let $j$ be an index such that $a_{j}$ is equal to the minimum value of the set $\left\{a_{n+1}, a_{n+2}, \ldots\right\}$. We must have $a_{j} \leqslant 9$, otherwise we could apply Claim 2 to $a_{j}$ and get a contradiction on the minimality hypothesis. It follows that $a_{j} \in\{3,6,9\}$, and the proof is complete. Claim 4. If $a_{n} \equiv 1(\bmod 3)$, then there is an index $m>n$ such that $a_{m} \equiv-1(\bmod 3)$. Proof. In the sequence, 4 is always followed by $2 \equiv-1(\bmod 3)$, so the claim is true for $a_{n}=4$. If $a_{n}=7$, the next terms will be $10,13,16,4,2, \ldots$ and the claim is also true. For $a_{n} \geqslant 10$, we again take an index $j>n$ such that $a_{j}$ is equal to the minimum value of the set $\left\{a_{n+1}, a_{n+2}, \ldots\right\}$, which by the definition of the sequence consists of non-multiples of 3 . Suppose $a_{j} \equiv 1(\bmod 3)$. Then we must have $a_{j} \leqslant 9$ by Claim 2 and the minimality of $a_{j}$. It follows that $a_{j} \in\{4,7\}$, so $a_{m}=2<a_{j}$ for some $m>j$, contradicting the minimality of $a_{j}$. Therefore, we must have $a_{j} \equiv-1(\bmod 3)$.  It follows from the previous claims that if $a_{0}$ is a multiple of 3 the sequence will eventually reach the periodic pattern $3,6,9,3,6,9, \ldots$; if $a_{0} \equiv-1(\bmod 3)$ the sequence will be strictly increasing; and if $a_{0} \equiv 1(\bmod 3)$ the sequence will be eventually strictly increasing.  So the sequence will be eventually periodic if, and only if, $a_{0}$ is a multiple of 3 .