![](https://cdn.mathpix.com/cropped/2024_06_05_a7c27b8cc9acec9f03deg-1.jpg?height=1131&width=785&top_left_y=228&top_left_x=630) T-1 Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$ x f(x+x y)=x f(x)+f\left(x^{2}\right) f(y) $$ for all $x, y \in \mathbb{R}$. Solution. Setting $x=y=0$ in $$ x f(x+x y)=x f(x)+f\left(x^{2}\right) f(y) $$ we get $f(0)=0$. Using this in (0) with $y=-1$ we obtain $$ x f(x)+f\left(x^{2}\right) f(-1)=0 $$ Let us distinguish the cases $f(-1)=0$ and $f(-1) \neq 0$. The case $f(-1)=0$. It follows from (1) that $f(x)=0$ for all $x \neq 0$. As we already know, $f(0)=0$. Thus we get the zero function $f(x)=0$, which is obviously a solution. The case $f(-1) \neq 0$. Setting $x=-1$ in (1) yields $f(1)=1$. Using this in (1) with $x=1$, we get $f(-1)=-1$ and hence (1) can be transformed to $$ x f(x)=f\left(x^{2}\right) $$ Put now $y=x-1$ in (0) to get $$ x f\left(x^{2}\right)=x f(x)+f\left(x^{2}\right) f(x-1) $$ Summing up (2) and (3) we obtain the equation $$ f\left(x^{2}\right)(f(x-1)-(x-1))=0 $$ Assume that $f(a)=0$ for some $a \neq 0$. Then $f\left(a^{2}\right)=0$ by (2) and hence (0) with $x=a$ implies that $a f(a+a y)=0$, i.e. $f(a+a y)=0$. Since $y$ is arbitrary here, we get $f(-1)=0$, which is not the case. Therefore, for any $x \neq 0$ we have $f(x) \neq 0$, and hence $f\left(x^{2}\right) \neq 0$ as well. Thus (4) leads to the conclusion that $f(x-1)=x-1$ for any $x \neq 0$, i.e. $f(x)=x$ for any $x \neq-1$. Since we already know that $f(-1)=-1$, we get the identity function $f(x)=x$, which is obviously a solution. T-2 Let $n \geqslant 2$ be an integer. There are $n$ positive integers written on a blackboard. In each step we choose two of the numbers on the blackboard and replace each of them by their sum. Determine all values of $n$ for which it is always possible to get $n$ identical integers in a finite number of steps. Solution. Starting from the $n$-tuple $(2,2,1,1, \ldots, 1)$ with any $n \geqslant 3$, we get always an $n$-tuple in which the number of maximal values is even. Hence no odd $n \geqslant 3$ is as required. Let us show by induction that any even $n \geqslant 2$ is satisfactory, which is obvious if $n=2$. For an even $n \geqslant 4$, by the induction hypothesis, we can transform any initial $n$-tuple to $(a, a, \ldots, a, b, b)$. If $a \neq b$, we apply repeatedly some of the following series of steps, which always lead to an $n$-tuple of type $(\underbrace{a, \ldots, a}_{k}, \underbrace{b, \ldots, b}_{n-k})$ in which the number $k$ may differ from the initial value $k=n-2$ (remaining to be even): $$ \text { series } \alpha:(\underbrace{a, \ldots, a}_{k}, \underbrace{b, \ldots, b}_{n-k}) \rightarrow(\underbrace{2 a, \ldots, 2 a}_{k}, \underbrace{b, \ldots, b}_{n-k}), $$ $$ \begin{array}{r} \text { series } \beta:(\underbrace{a, \ldots, a}_{k}, \underbrace{b, \ldots, b}_{n-k}) \rightarrow(\underbrace{a, \ldots, a}_{k}, \underbrace{2 b, \ldots, 2 b}_{n-k}), \\ \text { series } \gamma_{1}(\text { if } k \leqslant n-k):(\underbrace{a, \ldots, a}_{k}, \underbrace{b, \ldots, b}_{n-k}) \rightarrow(\underbrace{a+b, \ldots, a+b}_{2 k}, \underbrace{b, \ldots, b}_{n-2 k}), \\ \text { series } \gamma_{2}(\text { if } k \geqslant n-k):(\underbrace{a, \ldots, a}_{k}, \underbrace{b, \ldots, b}_{n-k}) \rightarrow(\underbrace{a, \ldots, a}_{2 k-n}, \underbrace{a+b, \ldots, a+b}_{2(n-k)}) . \end{array} $$ To describe our procedure, we introduce the notation $c=2^{P(c)} N(c)$ for any positive integer $c$, where $P(c) \geqslant 0$ and $N(c)$ is odd. To each $n$-tuple $(\underbrace{a, \ldots, a}_{k}, \underbrace{b, \ldots, b}_{n-k})$ with $a \neq b$, let us apply $\triangleright$ series $\alpha$ if $P(a)P(b)$, $\triangleright$ series $\gamma_{1}$ or $\gamma_{2}$ if $P(a)=P(b)$ (and hence $N(a) \neq N(b)$ ). Using the series $\alpha$ and $\beta$, the numbers $N(a), N(b)$ do not change, while the series $\gamma_{1}$ and $\gamma_{2}$ cause the changes exactly one of them, namely $$ N(b) \rightarrow \frac{N(a)+N(b)}{2^{m}}, \quad \text { or } \quad N(b) \rightarrow \frac{N(a)+N(b)}{2^{m}} \quad \text { respectively } $$ where $m=P(N(a)+N(b)) \geqslant 1$ and hence $$ \frac{N(a)+N(b)}{2^{m}} \leqslant \frac{N(a)+N(b)}{2}<\max (N(a), N(b)) $$ (recall that $N(a) \neq N(b))$. Consequently, throughout our procedure, the value of $\max (N(a), N(b))$ is a nonincreasing variable, and hence constant after a finite numbers of series. From this moment, we must still have either $N(a) \geqslant N(b)$, or $N(a) \leqslant N(b)$. This excludes either series $\gamma_{1}$, or series $\gamma_{2}$ from future applications, in which, therefore, all possible changes of the parameter $k$ are either $k \rightarrow 2 k$, or $(n-k) \rightarrow 2(n-k)$. Since this can repeat only $r$ times, where $2^{r} \leqslant n$, at the end we always we get an $n$-tuple $(a, \ldots, a, b, \ldots, b)$ for which (if $a \neq b$ ) the continuation of our procedure reduces only to the series $\alpha$ and $\beta$. Applying now either $\alpha$, or $\beta$ exactly $|P(a)-P(b)|$ times, we get an $n$-tuple $\left(a^{\prime}, \ldots, a^{\prime}, b^{\prime}, \ldots, b^{\prime}\right)$ with $P\left(a^{\prime}\right)=P\left(b^{\prime}\right)$. Since $\gamma_{1}, \gamma_{2}$ are already excluded, we have $a^{\prime}=b^{\prime}$, which completes the induction proof. Another solution (German team, adapted). We show without induction on $n$ that any even $n=2 k$ is satisfactory. At the beginning in the initial $2 k$-tuple $\left(a_{1}, \ldots, a_{2 k}\right)$ we replace every pair ( $a_{2 i-1}, a_{2 i}$ ) (for $\left.i=1, \ldots, k\right)$ by the pair $\left(a_{2 i-1}+a_{2 i}, a_{2 i-1}+a_{2 i}\right)$. From now on, we shall have always identical numbers on the $(2 i-1)$ th and $(2 i)$ th position. Hence because of brevity we shall work with $k$-tuples $(x, y, z, \ldots)$ instead of $2 k$-tuples $(x, x, y, y, z, z, \ldots)$. We are allowed to do the following transformations on the $k$-tuples: $\triangleright$ choose two of the numbers $x, y$ and replace each of them by their sum (this corresponds with two steps $(\ldots, x, x, \ldots, y, y, \ldots) \rightarrow(\ldots, x+y, x, \ldots, x+y, y, \ldots) \rightarrow$ $(\ldots, x+y, x+y, \ldots, x+y, x+y, \ldots)$ performed on the $2 k$-tuple); $\triangleright$ choose one number $x$ and multiply it by 2 (this corresponds with one step $(\ldots, x, x, \ldots) \rightarrow(\ldots, x+x, x+x, \ldots)$; $\triangleright$ divide all numbers by 2 (this obviously does not affect anything; formally we could remember how many times we have performed this dividing and multiply all the numbers by the proper power of two at the end). Our aim is to obtain $k$ identical numbers. We reach it by iterating the following algorithm: 1. While there are at least two odd numbers, find the minimum and the maximum odd number and replace each of them by their (even) sum. 2. If there is one odd number left after finishing the first step, multiply it by two. 3. Divide all numbers by 2 . Clearly, after each iteration, the maximum number among all $k$ numbers either decreases or does not change. As this maximum is permanently a positive integer, after a finite number of iterations, it fixes at the value $M$ and does not change anymore. From now on, look at the number $N$ of $M$ 's in the $k$-tuple. Obviously $M$ is odd (otherwise it would decrease in the third step in the next iteration). If $N\angle A C^{\prime} B=\angle A B C^{\prime}=\angle A B E^{\prime}>\angle A B E$ (because $E$ lies between $A$ and $E^{\prime}$ ), which contradicts to $\angle A C B=\angle E B A$. Similarly, if $C$ does not lie on the segment $C^{\prime} D$, then $C$ lies outside $\mathcal{K}$ (on the same side of the line $A B$ as $C^{\prime}$ ), therefore $\angle A C B<\angle A C^{\prime} B=\angle A B C^{\prime}=\angle A B E^{\prime}<\angle A B E$ (because $E^{\prime}$ lies between $A$ and $E$ ), which again contradicts to $\angle A C B=\angle E B A$. ![](https://cdn.mathpix.com/cropped/2024_06_05_a7c27b8cc9acec9f03deg-5.jpg?height=702&width=526&top_left_y=660&top_left_x=411) Fig. 4 ![](https://cdn.mathpix.com/cropped/2024_06_05_a7c27b8cc9acec9f03deg-5.jpg?height=654&width=525&top_left_y=707&top_left_x=1094) Fig. 5 Another solution (Karel Horák, Czech Republic). From the given equalities of angles it follows that triangles $A B D$ and $C E D$ are similar (Fig. 5). From that similarity we immediately get that triangles $A C D$ and $B E D$ are similar (by sas, same angles at common vertex $D$, and proportional sides). From the equal angles $B E D$ and $A C D$ if follows that the sum of three angles $B C A, A C D$ and $D C E$ is equal to the sum of angles in the triangle $A B E$, so $E, C$, and $B$ are collinear. T-4 Let $n$ be a positive integer. Prove that if the sum of all positive divisors of $n$ is a perfect power of 2 , then the number of these divisors is also a perfect power of 2 . Solution. Suppose that $n=p_{1}^{s_{1}} p_{2}^{s_{2}} \ldots p_{k}^{s_{k}}$, where $p_{1}, \ldots, p_{k}$ are distinct primes and $s_{i} \geqslant 1$ for each $i$, and that the sum of all positive divisors of $n$, which is given by $$ \left(1+p_{1}+p_{1}^{2}+\cdots+p_{1}^{s_{1}}\right)\left(1+p_{2}+p_{2}^{2}+\cdots+p_{2}^{s_{2}}\right) \ldots\left(1+p_{k}+p_{k}^{2}+\cdots+p_{k}^{s_{k}}\right) $$ is a perfect power of 2 . Then each of the factors $$ f_{i}=1+p_{i}+p_{i}^{2}+\cdots+p_{i}^{s_{i}} $$ is also a perfect power of 2 greater than 1 and hence both $p_{i}$ and $s_{i}$ are odd. Suppose that $s_{i}>1$. In this case we have $$ f_{i}=\left(1+p_{i}\right)\left(1+p_{i}^{2}+p_{i}^{4}+\cdots+p_{i}^{s_{i}-1}\right) $$ Since $f_{i}$ has no odd divisor greater than 1 , the even integer $s_{i}-1$ (which is supposed to be positive) must be of the form $4 k+2$ and thus we can make another factorization $$ f_{i}=\left(1+p_{i}\right)\left(1+p_{i}^{2}\right)\left(1+p_{i}^{4}+p_{i}^{8}+\cdots+p_{i}^{s_{i}-3}\right) $$ Consequently, both $1+p_{i}$ and $1+p_{i}^{2}$ are powers of 2 , hence $1+p_{i} \mid 1+p_{i}^{2}$, which contradicts to $1+p_{i}^{2}=\left(1+p_{i}\right)\left(p_{i}-1\right)+2$ (as $1+p_{i} \mid 2$ is impossible). This means that $s_{i}=1$ for each $i$ and thus the number of divisors of $n$ equals $2^{k}$. Note that the above solution can be finished without observing the fact that $1+p_{i}$ and $1+p_{i}^{2}$ cannot be powers of 2 at the same time. Indeed, repeating the procedure of factorization we get finally $$ f_{i}=\left(1+p_{i}\right)\left(1+p_{i}^{2}\right)\left(1+p_{i}^{4}\right) \ldots\left(1+p_{i}^{2_{i}}\right) $$ hence $s_{i}=2^{t_{i}+1}-1$ with some $t_{i} \geqslant 0$ for each $i$ and thus the number of divisors of $n$ equals $2^{k+t_{1}+t_{2}+\cdots+t_{k}}$. (As we know from the original solution, $t_{i}=0$ for each $i$.)