problem: Let $p$ be a prime number and let $s$ be an integer with $0<s<p$. Prove that there exist integers $m$ and $n$ with $0<m<n<p$ and  $$ \left\{\frac{s m}{p}\right\}<\left\{\frac{s n}{p}\right\}<\frac{s}{p} $$  if and only if $s$ is not a divisor of $p-1$.

solution: It's equivalent to $m s \bmod p<n s \bmod p<s$, where $x \bmod p$ means the remainder when $x$ is divided by $p$, by slight abuse of notation. We will assume $s \geq 2$ for simplicity, since the case $s=1$ is clear.  For any $x \in\{1,2, \ldots, s-1\}$ we define $f(x)$ to be the unique number in $\{1, \ldots, p-1\}$ such that $s \cdot f(x) \bmod p=x$. Then, $m$ and $n$ fail to exist exactly when  $$ f(s-1)<f(s-2)<\cdots<f(1) $$  We give the following explicit description of $f$ : choose $t \equiv-s^{-1}(\bmod p), 0<t<p$. Then $f(x)=1+(s-x) \cdot t \bmod p$. So our displayed inequality is equivalent to  $$ (1+t) \bmod p<(1+2 t) \bmod p<(1+3 t) \bmod p<\cdots<(1+(s-1) t) \bmod p $$  This just means that the sequence $1+k t$ never "wraps around" modulo $p$ as we take $k=1,2, \ldots, s-1$.  Since we assumed $s \neq 1$, we have $0<1+t<p$. Now since $1+k t$ never wraps around as $k=1,2, \ldots, s-1$, and increases in increments of $t$, it follows that $1+k t<p$ for all $k=1,2, \ldots, s-1$. Finally, as $1+s t \equiv 0(\bmod p)$ we get $1+s t=p$.  In summary, $m, n$ fail to exist precisely when $1+s t=p$. That is of course equivalent to $s \mid p-1$.


problem: Let $k>0$ be a fixed integer. Compute the minimum integer $N$ (in terms of $k$ ) for which there exists a set of $2 k+1$ distinct positive integers that has sum greater than $N$, but for which every subset of size $k$ has sum at most $N / 2$.

solution: The answer is $N=k\left(2 k^{2}+3 k+3\right)$ given by  $$ S=\left\{k^{2}+1, k^{2}+2, \ldots, k^{2}+2 k+1\right\} . $$  To show this is best possible, let the set be $S=\left\{a_{0}<a_{1}<\cdots<a_{2 k}\right\}$ so that the hypothesis becomes  $$ \begin{aligned} N+1 & \leq a_{0}+a_{1}+\cdots+a_{2 k} \\ N / 2 & \geq a_{k+1}+\cdots+a_{2 k} \end{aligned} $$  Subtracting twice the latter from the former gives  $$ \begin{aligned} a_{0} & \geq 1+\left(a_{k+1}-a_{1}\right)+\left(a_{k+2}-a_{2}\right)+\cdots+\left(a_{2 k}-a_{k}\right) \\ & \geq 1+\underbrace{k+k+\cdots+k}_{k \text { terms }}=1+k^{2} . \end{aligned} $$  Now, we have  $$ \begin{aligned} N / 2 & \geq a_{k+1}+\cdots+a_{2 k} \\ & \geq\left(a_{0}+(k+1)\right)+\left(a_{0}+(k+2)\right)+\cdots+\left(a_{0}+2 k\right) \\ & =k \cdot a_{0}+((k+1)+\cdots+2 k) \\ & \geq k\left(k^{2}+1\right)+k \cdot \frac{3 k+1}{2} \end{aligned} $$  so $N \geq k\left(2 k^{2}+3 k+3\right)$. Remark. The exact value of $N$ is therefore very superficial. From playing with these concrete examples we find out we are essentially just trying to find an increasing set $S$ obeying  $$ a_{0}+a_{1}+\cdots+a_{k}>a_{k+1}+\cdots+a_{2 k} $$  and indeed given a sequence satisfying these properties one simply sets $N=2\left(a_{k+1}+\cdots+a_{2 k}\right)$. Therefore we can focus almost entirely on $a_{i}$ and not $N$.  Remark. It is relatively straightforward to figure out what is going on based on the small cases. For example, one can work out by hand that  - $\{2,3,4\}$ is optimal for $k=1$ - $\{5,6,7,8,9\}$ is optimal for $k=2$, - $\{10,11,12,13,14,15,16\}$ is optimal for $k=3$.  In all the examples, the $a_{i}$ are an arithmetic progression of difference 1 , so that $a_{j}-a_{i} \geq j-i$ is a sharp for all $i<j$, and thus this estimate may be used freely without loss of sharpness; applying it in $(\star)$ gives a lower bound on $a_{0}$ which is then good enough to get a lower bound on $N$ matching the equality cases we found empirically.


problem: For integral $m$, let $p(m)$ be the greatest prime divisor of $m$. By convention, we set $p( \pm 1)=1$ and $p(0)=\infty$. Find all polynomials $f$ with integer coefficients such that the sequence  $$ \left\{p\left(f\left(n^{2}\right)\right)-2 n\right\}_{n \geq 0} $$  is bounded above. (In particular, this requires $f\left(n^{2}\right) \neq 0$ for $n \geq 0$.)

solution: If $f$ is the (possibly empty) product of linear factors of the form $4 n-a^{2}$, then it satisfies the condition. We will prove no other polynomials work. In what follows, assume $f$ is irreducible and nonconstant.  It suffices to show for every positive integer $c$, there exists a prime $p$ and a nonnegative integer $n$ such that $n \leq \frac{p-1}{2}-c$ and $p$ divides $f\left(n^{2}\right)$.  Firstly, recall there are infinitely many odd primes $p$, with $p>c$, such that $p$ divides some $f\left(n^{2}\right)$, by Schur's Theorem. Looking mod such a $p$ we can find $n$ between 0 and $\frac{p-1}{2}\left(\right.$ since $\left.n^{2} \equiv(-n)^{2}(\bmod p)\right)$. We claim that only finitely many $p$ from this set can fail now. For if a $p$ fails, then its $n$ must be between $\frac{p-1}{2}-c$ and $\frac{p-1}{2}$. That means for some $0 \leq k \leq c$ we have  $$ 0 \equiv f\left(\left(\frac{p-1}{2}-k\right)^{2}\right) \equiv f\left(\left(k+\frac{1}{2}\right)^{2}\right) \quad(\bmod p) $$  There are only finitely many $p$ dividing  $$ \prod_{k=1}^{c} f\left(\left(k+\frac{1}{2}\right)^{2}\right) $$  unless one of the terms in the product is zero; this means that $4 n-(2 k+1)^{2}$ divides $f(n)$. This establishes the claim and finishes the problem.


problem: Find all positive integers $n$ for which there exist an integer $k \geq 2$ and positive rational numbers $a_{1}, \ldots, a_{k}$ satisfying $a_{1}+a_{2}+\cdots+a_{k}=a_{1} a_{2} \ldots a_{k}=n$.

solution: The answer is all $n$ other than $1,2,3,5$.  We now contend that $k>2$. Indeed, if $a_{1}+a_{2}=a_{1} a_{2}=n$ then $\left(a_{1}-a_{2}\right)^{2}=$ $\left(a_{1}+a_{2}\right)^{2}-4 a_{1} a_{2}=n^{2}-4 n=(n-2)^{2}-4$ is a rational integer square, hence a perfect square. This happens only when $n=4$.  Now by AM-GM,  $$ \frac{n}{k}=\frac{a_{1}+\cdots+a_{k}}{k} \geq \sqrt[k]{a_{1} \ldots a_{k}}=n^{1 / k} $$  and so $n \geq k^{\frac{1}{1-1 / k}}=k^{\frac{k}{k-1}}$. This last quantity is always greater than 5 , since  $$ \begin{aligned} 3^{3 / 2} & =3 \sqrt{3}>5 \\ 4^{4 / 3} & =4 \sqrt[3]{4}>5 \\ k^{\frac{k}{k-1}} & >k \geq 5 \quad \forall k \geq 5 \end{aligned} $$  Now, in general:  - If $n \geq 6$ is even, we may take $\left(a_{1}, \ldots, a_{n / 2}\right)=(n / 2,2,1, \ldots, 1)$. - If $n \geq 9$ is odd, we may take $\left(a_{1}, \ldots, a_{(n-3) / 2}\right)=(n / 2,1 / 2,4,1, \ldots, 1)$. - A special case $n=7$ : one example is $(4 / 3,7 / 6,9 / 2)$. (Another is $(7 / 6,4 / 3,3 / 2,3)$.)  Remark. The main hurdle in the problem is the $n=7$ case. One good reason to believe a construction exists is that it seems quite difficult to prove that $n=7$ fails.


problem: A mathematical frog jumps along the number line. The frog starts at 1 , and jumps according to the following rule: if the frog is at integer $n$, then it can jump either to $n+1$ or to $n+2^{m_{n}+1}$ where $2^{m_{n}}$ is the largest power of 2 that is a factor of $n$. Show that if $k \geq 2$ is a positive integer and $i$ is a nonnegative integer, then the minimum number of jumps needed to reach $2^{i} k$ is greater than the minimum number of jumps needed to reach $2^{i}$.

solution: We will think about the problem in terms of finite sequences of jumps $\left(s_{1}, s_{2}, \ldots, s_{\ell}\right)$, which we draw as  $$ 1=x_{0} \xrightarrow{s_{1}} x_{1} \xrightarrow{s_{2}} x_{2} \xrightarrow{s_{3}} \ldots \xrightarrow{s_{\ell}} x_{\ell} $$  where $s_{k}=x_{k}-x_{k-1}$ is the length of some hop. We say the sequence is valid if it has the property required by the problem: for each $k$, either $s_{k}=1$ or $s_{k}=2^{m_{x_{k-1}}+1}$.  An example is shown below.  ## Lemma  Let $\left(s_{1}, \ldots, s_{\ell}\right)$ be a sequence of jumps. Suppose we delete pick an index $k$ and exponent $e>0$, and delete any jumps after the $k$ th one which are divisible by $2^{e}$. The resulting sequence is still valid.   But since $s<2^{e}$, we have $\nu_{2}\left(x^{\prime}\right)<e$ and hence $\nu_{2}(x)=\nu_{2}\left(x^{\prime}\right)$ so the jump is valid. ![](https://cdn.mathpix.com/cropped/2024_11_19_e43d520d3d3bb47ac469g-08.jpg?height=375&width=1420&top_left_y=1743&top_left_x=318)  Now let's consider a valid path to $2^{i} k$ with $\ell$ steps, say  $$ 1=x_{0} \xrightarrow{s_{1}} x_{1} \xrightarrow{s_{2}} x_{2} \xrightarrow{s_{3}} \ldots \xrightarrow{s_{\ell}} x_{\ell}=2^{i} \cdot k $$  where $s_{i}=x_{i}-x_{i-1}$ is the distance jumped. We delete jumps in the following way: starting from the largest $e$ and going downwards until $e=0$, we delete all the jumps of length $2^{e}$ which end at a point exceeding the target $2^{i}$.  By the lemma, at each stage, the path remains valid. We claim more:  Claim - Let $e \geq 0$. After the jumps of length greater than $2^{e}$ are deleted, the resulting end-point is at least $2^{i}$, and divisible by $2^{\min (i, e)}$.   It is also divisible by $2^{\min (i, e)}$ by induction hypothesis, since we are changing the end-point by multiples of $2^{e}$. And the smallest multiple of $2^{\min (i, e)}$ exceeding $x$ is $2^{i}$.  On the other hand by construction when the process ends the reduced path ends at a point at most $2^{i}$, so it is $2^{i}$ as desired.  Therefore we have taken a path to $2^{i} k$ and reduced it to one to $2^{i}$ by deleting some jumps. This proves the result.


problem: Let $A B C D$ be a quadrilateral, and let $E$ and $F$ be points on sides $A D$ and $B C$, respectively, such that $\frac{A E}{E D}=\frac{B F}{F C}$. Ray $F E$ meets rays $B A$ and $C D$ at $S$ and $T$, respectively. Prove that the circumcircles of triangles $S A E, S B F, T C F$, and $T D E$ pass through a common point.

solution: ![](https://cdn.mathpix.com/cropped/2024_11_19_e43d520d3d3bb47ac469g-10.jpg?height=518&width=712&top_left_y=815&top_left_x=643)  Let $M$ be the Miquel point of $A B C D$. Then $M$ is the center of a spiral similarity taking $A D$ to $B C$. The condition guarantees that it also takes $E$ to $F$. Hence, we see that $M$ is the center of a spiral similarity taking $\overline{A B}$ to $\overline{E F}$, and consequently the circumcircles of $Q A B, Q E F, S A E, S B F$ concur at point $M$.  In other words, the Miquel point of $A B C D$ is also the Miquel point of $A B F E$. Similarly, $M$ is also the Miquel point of $E D C F$, so all four circles concur at $M$.