do you have all the problems?If you have,could you post them in contedt collections?Thanks

We should avoid posting problems from ISL. (Before IMO) So, in my country, the problems are usually posted by their proposers.

Are you from Taiwan

Maybe the problems will be posted after IMO. Just wait and see if there is any volunteer willing to post Taiwan TSTs.

By request, I'm going to post a solution to the reduced version: Find all $f:\mathbb{Z}\to\mathbb{Z}$ surjective such that $f(x)f(y)=f(y+xf(y))$. One can get this reduced version by noticing that $f(0)=1$. It's easy to see that $f(0)=1$. Suppose that $f(a)=2$, then $(x,a):2f(x)=f(2x+a)$. By induction, we can see that $v_2(f(x))\geq v_2(x+a)$. Therefore $v_2(a)\leq v_2(f(0))=1$, and so $a$ is odd. By $2f(x)=f(2x+a)$, we know that $f((2^n-1)a)=2^n \quad\forall n\in\mathbb{N}_0$. Now, suppose that $f(l)=2k+1$ but $l\neq 2ka$, then take $n=v_2(l-2ka)$, we have $((2^n-1)a, l): (2k+1)2^n = f(2^n(2k+1)a-l-2ka-a)$. Taking $v_2$ on the both sides, we get that $n=v_2f(2^n(2k+1)a-l-2ka-a)\geq v_2(2^n(2k+1)a-l-2ka)\geq n+1$, which is a contradiction. Hence we have $f(2ka)=2k+1\forall k\in\mathbb{Z}$. By the identity $2f(x)=f(2x+a)$ it's not hard to get $f(ka)=k+1\quad\forall k\in\mathbb{Z}$. $(1,a^2-a): af(1)=a+1$, and so $a=\pm 1$. If $a=1$, then $f(x)=x+1$; if $a=-1$, then $f(x)=1-x$. It's easy to verify that those two are indeed solutions, and thus are the only two solutions.

Here are solutions to this problem. One for the original version and the other for the reduced version.

USJL wrote: By request, I'm going to post a solution to the reduced version: Taking $v_2$ on the both sides, we get that $n=v_2f(2^n(2k+1)a-l-2ka-a)\geq v_2(2^n(2k+1)a-l-2ka)\geq n+1$, which is a contradiction. it seems there is a mistake in your expression. shouldn’t it $n=v_2f(2^n(2k+1)a+l-2ka-a)\geq v_2(2^n(2k+1)a+l-a-2ka)\geq n+1$? but $l-a$ is an odd number and $v_2$ can’t help us.

No You are right for $-l\to+l$ but it changes nothing. He uses $v_2(f(x))\ge v_2(x+a)$ (notice the "$+a$" in RHS) And so here $v_2(f(...))\ge v_2(2^n(2k+1)a+l-2ka)$ (the "$a$" in RHS disappears) And since $n=v_2(l-2ka)$ we have $l-2ka=2^nq$ where $q$ is odd And so $v_2(2^n(2k+1)a+l-2ka)=v_2(2^n((2k+1)a+q)\ge n+1$ indeed.

USJL wrote: By induction, we can see that $v_2(f(x))\geq v_2(x+a)$. Nice solution! But pco please could you explain me why this is true? Thanks in advance!

Just write $x+a=2^nq$ for some odd $q$ Then $f(x)=f(2(2^{n-1}q-a)+a)$ $=2f(2^{n-1}q-a)$ (using $f(2x+a)=2f(x)$ But $f(2^{n-1}q-a)=f(2(2^{n-2}q-a)+a)=2f(2^{n-2}q-a)$ And so $f(x)=2^2f(2^{n-2}q-a)$ And a simple induction gives $f(x)=2^nf(q-a)$ and so $v_2(f(x))\ge n=v_2(x+a)$

Very fun problem. Throughout the solution, we maintain that $0$ divides $0$ and no other integer. Let $P(x,y,z)$ denote the original proposition. $P(0,0,0)$ gives $f(0)=0$, $-1$ or $1$. If $f(0)=0$, choose an $x$ satisfying $f(x)=1$, then $P(x,x,0)$ gives $f(x)=1$, contradiction! If $f(0)=-1$, choose $t$ such that $f(t)=1$. Then $P(x,y,0)$ gives $f(x-t)=-f(x)$ $$\implies \ f(x-2t)=-f(x-t)=f(x)$$Since $t \neq 0$, this gives that $f$ is periodic with period $2t$, and hence can't be surjective, contradiction! Therefore $f(0)=1$. Then $P(x,y,0)$ gives $$f(xf(y)+y)=f(x)f(y)$$Call the above $Q(x,y)$. Claim 1: For integers $a,b$, $a \equiv b \pmod{f(b)}$ $\implies$ $f(b) \mid f(a)$. Proof: If $f(b)=0$ it is obvious, else use $Q \left (\dfrac{a-b}{f(b)}, b \right )$. $\blacksquare$ Claim 2: For all integers $y$, all primes which divide $f(y)-1$ also divide $y$. Proof: If $|f(y)-1|=1$, it is vacuously true. Else let $p$ be any prime factor of $f(y)-1$. Choose a $z$ such that $f(z)=p$. Let $S$ be the set of residues $r$ modulo $p$ such that for any integer $x$, $x \equiv r \pmod{p}$ $\implies$ $p \mid f(x)$. $S$ is non-empty, since $z \in S$ by Claim 1. Choose any $t \in S$, and choose any integer $x$ satisfying $x \equiv t-y \pmod p$. Then $xf(y)+y \equiv t \pmod p$ $\implies$ $p \mid f(xf(y)+y)$. So $Q(x,y)$ gives $p \mid f(x)f(y)$ $\implies$ $p \mid f(x)$ $\implies$ $t-y \in S$ since $x$ can be chosen arbitrarily. Using this repeatedly gives $t-ny \in S$ for all positive integers $n$. If $p \nmid y$, then any residue modulo $p$ can be written as $t-ny$ for some positive integer $n$ $\implies$ $S$ contains all residues modulo $p$ $\implies$ $p \mid f(x)$ for all integers $x$, which contradicts surjectivity. Hence $p \mid y$ as required. $\blacksquare$ Claim 3: $f$ is injective over non-zero values. Proof: Let $y_1,y_2$ be integers such that $f(y_1)=f(y_2)=k \neq 0$. Then from $Q(x,y_1)$, $Q(x,y_2)$ and Claim 2 for any arbitrary $x$, all prime factors of $kf(x)-1$ divide both $xk+y_1$ and $xk+y_2$ $\implies$ they all divide $y_1-y_2$. If $y_1 \neq y_2$, then since $f$ is surjective, we can choose an $x$ such that $kf(x)-1$ is divisible by a prime greater than $|y_1-y_2|$, contradiction! Hence $y_1=y_2$ as required. $\blacksquare$ Let $a$ satisfy $f(a)=2$. Clearly $a \neq 0$. Claim 4: $f(x)=\dfrac{x}{a}+1$ if $a \mid x$, and $f(x)=0$ otherwise. Proof: First we prove that $f(x) \neq 0$ $\implies$ $f(x)=\dfrac{x}{a}+1$. Indeed, if $f(x) \neq 0$, comparing $Q(x,a)$ and $Q(a,x)$ we get $$f(2x+a)=f(af(x)+x)=2f(x) \neq 0$$$$\implies \ 2x+a=af(x)+x$$$$\implies \ f(x)=\frac{x}{a}+1$$as required. This immediately gives $f(x)=0$ if $a \nmid x$. Now assume $a \mid x$. If $x = -a$, then $Q(x,a)$ gives $f(x)=0$, consistent with the claim. Else choose $y$ such that $f(y)=\dfrac{x}{a}+1$ $\implies$ $f(y) \neq 0$ $\implies$ $f(y)=\dfrac{y}{a}+1$ $\implies$ $x=y$ as required. $\blacksquare$ If $a = \pm 1$, then by Claim 4 we get the two solutions: $$\boxed{f(x)=1+x \ \ \forall x \in \mathbb{Z}}$$$$\boxed{f(x)=1-x \ \ \forall x \in \mathbb{Z}}$$ Else $f(1)=0$ by Claim 4. Then $P(1,a^2-a)$ gives $$0=f(1)f(a^2-a)=f(f(a^2-a)+a^2-a)=f(a^2)=a+1$$which gives $a=-1$, contradiction! $\blacksquare$