Solved with YaoAOPS. We claim the only solutions are functions such that $f(x) \in \{-1,1\}$ for all integers $x$ and $f(1) = -f(-1)$, and also functions such that $f(x) \in \{-2,2\}$ for all integers $x$ and $f(1)= -f(-1)$. These can be checked to work. Now we prove they are the only solutions. Let $P(m,n)$ denote the given assertion. Claim: $f$ is bounded on both sides. Proof: $P(m,0): f(m)^2 \mid f(m) - f(0)$. This implies that either $f(m) = f(0)$ or $f(m)^2 \le |f(m) - f(0)|$. However, if $f(m)$ is arbitrarily large or arbitrarily small, then $f(m)^2$ exceeds both $f(m) - f(0)$ and $f(0) - f(m)$, contradiction. $\square$ Using $P(m,0)$, we see that if $f(m) =0$, then $0\mid -f(0)$, so $f(0) = 0$. But then $P(1,-1)$ gives that $f(1) -f(-1) = 0$, contradiction. Therefore, $f(m)\ne 0$ for all integers $m$. Since $f$ is bounded on both sides, we may choose $k$ such that $|f(k)|$ is maximal. Claim: $|f(k)| \le 2$. Proof: Suppose this was not the case. $P(k,0): f(k)^2 \mid f(k) - f(0)$, so $f(k)^2 \le |f(k) - f(0)|$. If $f(k) \ne f(0)$, then $|f(k) - f(0)|$ is at most $|2f(k)|$, so we have $f(k)^2 \le |2f(k)|$, but this is not the case if $|f(k)|>2$. Thus, $f(k) = f(0)$. $P(1,-1): f(0)^2 \mid f(1) - f(-1)\implies f(k)^2 \mid f(1) - f(-1)$, so $f(k)^2 \le |f(1) - f(-1)| \le |2f(k)|$ (because $f(1) - f(-1)$ cannot be zero), which is impossible if $|f(k)|>2$. $\square$ Case 1: $|f(k)| = 1$. Then since $f(x)\ne 0$ for any integer $x$, we have $f(x)\in \{-1,1\}$ for all integers $x$. Since $f(1) \ne f(-1)$, $f(1) = -f(-1)$ must hold. This corresponds to the first solution described in the beginning. Case 2: $|f(k)| = 2$. Here we get that $f(x)\in \{-2,-1,1,2\}$ for all integers $x$. $P(k,0): 4\mid f(k) - f(0)$, so $f(0)\in \{-2,2\}$. $P(m,-m): 4\mid f(m) - f(-m)$, so either $f(m) = f(-m)$ or $f(m)$ and $f(-m)$ are even (so the parity of $f(m)$ and $f(-m)$ must be the same). This implies that $f(1)$ and $f(-1)$ are even. Now, if $f(x)$ is even for some $x$ (so $f(x)^2 = 4$), then $P(x+1, -1): 4 \mid f(x+1) - f(-1)$, so $f(x+1)$ must be even also. Now inducting from $1$ gives that $f(m)$ is even for all positive integers $m$. Now using the fact that $f(m) = f(-m)$, $f(m)$ is even for all negative integers $m$ also. Since $f(0)$ is even, we get that $f(x)$ is even for all integers $x$, so $f(x) \in \{-2,2\}$. Since $f(1)\ne f(-1)$, $f(1) = -f(-1)$. This corresponds to the second solution described in the beginning.

a_507_bc wrote: Find all functions $f:\mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(1) \neq f(-1)$ and $$f(m+n)^2 \mid f(m)-f(n)$$for all integers $m, n$. Proposed by Liam Baker, South Africa $\color{blue}\boxed{\textbf{Answer:} f \in \{ 1,-1 \} \textbf{ or }f \in\{ 2,-2 \} \textbf{ obviously with } f(1)=-f(-1)}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$f(m+n)^2|f(m)-f(n)...(\alpha)$$In $(\alpha) n=0:$ $$\Rightarrow f(m)^2|f(m)-f(0)$$$$\Rightarrow f(m)|f(m)-f(0)$$$$\Rightarrow f(m)|f(0), \forall m\in\mathbb{Z}...(I)$$If $\exists k/f(k)=0:$ In $(\alpha) n=k-m:$ $$\Rightarrow 0|f(m)-f(k-m)(\Rightarrow \Leftarrow)$$$$\Rightarrow \not\exists k/f(k)=0...(I)$$$\color{red}\boxed{\textbf{Now suppose that f takes on an infinite number of values:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow f(0) \text{ has an infinite number of divisors}$$$$\Rightarrow f(0)=0(\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow f \text{ is bounded}...(II)$$$\color{red}\boxed{\textbf{Lemma:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$f(a)|f(ak), \forall a,k\in\mathbb{Z}$$$\color{red}\rule{24cm}{0.3pt}$ We will prove this by induction: $\boxed{\text{Base case:}}$ $$f(a)|f(a)$$By $(I)$: $$f(a)|f(0)$$In $(\alpha) m=a, n=-a:$ $$\Rightarrow f(0)^2|f(a)-f(-a)$$By $(I):$ $$\Rightarrow f(a)^2|f(0)^2|f(a)-f(-a)$$$$\Rightarrow f(a)|f(a)-f(-a)$$$$\Rightarrow f(a)|f(-a)$$$$\Rightarrow f(a)|f(0), f(a)|f(a), f(a)|f(-a)$$$\boxed{\text{Inductive hypothesis:}}$ $$f(a)|f(ak), \forall a,k\in\mathbb{Z}, |k|\le t, t\in\mathbb{Z}^+$$$\boxed{\text{Inductive step:}}$ In $(\alpha) m=at, n=-a(t-1):$ $$\Rightarrow f(a)^2|f(at)-f(-a(t-1))$$$$\Rightarrow f(a)|f(at)-f(a(t-1))$$By Inductive hypothesis: $f(a)|f(a(t-1))$ $$\Rightarrow f(a)|f(at)...(1)$$In $(\alpha) m=-at, n=a:$ $$\Rightarrow f(-a(t-1))^2|f(-at)-f(-a)$$By Inductive hypothesis: $f(a)|f(a(t-1)) \text{ and } f(a)|f(-a)$ $$\Rightarrow f(a)|f(a)^2|f(-a(t-1))^2|f(-at)-f(-a)$$$$\Rightarrow f(a)|f(-at)...(2)$$By $(1)$ and $(2)$ the induction is complete $_\blacksquare$ $\color{red}\rule{24cm}{0.3pt}$ By $(I)$ we know that: $$\exists c/|f(c)| \text{ is maximum}...(\beta)$$In $(\alpha) m=1, n=-1:$ $$\Rightarrow f(c)^2|f(0)^2|f(1)-f(-1)$$$$\Rightarrow f(c)^2\le |f(1)-f(-1)|\le 2f(c)$$$$\Rightarrow f(c)\in \{ 1,-1,2,-2 \}$$$$\Rightarrow f(k)\in \{ 1,-1,2,-2 \} , \forall k\in\mathbb{Z}$$$\color{red}\boxed{|\textbf{f(c)}| \textbf{=1:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow |f(k)|\le |f(c)|=1$$By $(I):$ $$\Rightarrow f(k)\in\{ -1,1 \}, f(1)=-f(-1), \forall k\in\mathbb{Z} \text{ is a solution}$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{| \textbf{f(c)}| \textbf{=2:}}$ $\color{red}\rule{24cm}{0.3pt}$ $\color{green}\boxed{| \textbf{f(1)}| \le \textbf{1}:}$ $\color{green}\rule{24cm}{0.3pt}$ In $(\alpha) m=1, n=-1:$ $$\Rightarrow f(0)^2|f(1)-f(-1)\neq 0$$$$\Rightarrow f(0)^2|2$$$$\Rightarrow f(0)=\pm 1$$By Lemma: $$f(c)|f(0)$$$$\Rightarrow |f(c)|\le 1(\Rightarrow \Leftarrow)$$$\color{green}\rule{24cm}{0.3pt}$ $\color{green}\boxed{| \textbf{f(1)} | \ge \textbf{2}:}$ $\color{green}\rule{24cm}{0.3pt}$ By Lemma: $$f(1)|f(k),\forall k \in\mathbb{Z}$$$$\Rightarrow |f(k)|=2$$$$\Rightarrow f(k)\in\{ -2,2 \}, f(1)=-f(-1), \forall k\in\mathbb{Z} \text{ is a solution}$$$\color{green}\rule{24cm}{0.3pt}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow \boxed{f \in \{ 1,-1 \} \textbf{ or }f \in\{ 2,-2 \} \textbf{ with } f(1)=-f(-1)\textbf{ are the only solutions}}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

Denote $P(m,n)$ the assertion of the given F.E. $P(m,0)$ gives $f(m) \mid f(m)^2 \mid f(m)-f(0)$ thus $f(m) \mid f(0)$ for all $m \in \mathbb Z$, also note we have that $f(m)=f(0)$ or $f(m)^2 \le |f(m)-f(0)|$ and from here and checking polynomial growth its easy to check that $f$ is bounded above and below by some constants. Also notice that if there was some $f(c)=0$ then $0 \mid f(0)$ so $f(0)=0$ and then $P(1,-1)$ would gives $f(1)=f(-1)$, a contradiction so $0 \not \in \text{Im} \; f$. Now $P(x,-x)$ gives $f(x)^2 \mid f(0)^2 \mid f(x)-f(-x)$ so $f(x) \mid f(-x)$ and since we can swap signs this means $f(x)=\pm f(-x)$ now it is clear that if $x=1$ then we now must have $f(1)=-f(-1)$, now set $P(2x,-x)$ to get $f(x)^2 \mid f(2x)-f(-x)$ but notice $f(x) \mid f(-x)$ as seen previously so $f(x) \mid f(2x)$ and thus inductively we can repeat this to get $f(x) \mid f(nx)$ for all integers $n,x$. Now suppose $f(\ell)$ is the value with maximun $| \cdot |$ then $f(\ell)^2 \mid f(0)^2 \mid f(1)-f(-1)$ so we get $f(\ell)^2 \le 2|f(\ell)|$ from triangle ineq thus $|f(\ell)| \le 2$ and this will be true for all values of $f$ as well notice now this implies since $f(x) \mid f(0)$ for all integer $x$ that if $|f(0)|=1$ then $f(x) \in \{ -1, 1 \}$ and $f(1)=-f(-1)$, which is a function that works trivially, so now if $|f(0)|=2$ we get that $f(x) \mid 2$ for all integers $x$, now if $|f(1)|=1$ then $4 \mid 2f(1)$ contradiction so $|f(1)|=2$ and thus, since $f(1) \mid f(n)$ for all integers $n$ we get that $|f(n)|=2$ for all integers $n$ and $f(1)=-f(-1)$ which works as well, thus we are done .

Small sketch: Use $P(x, 0)$ to get that $f$ is bounded; in particular, $f(x) \mid f(0) \forall x$. Now from $P(-1, 1)$, do euclidian bounding and obtain that $|f(0)| \le 2$. Neglect the case $|f(0)| = 0$ since it isn't possible. $|f(0)| = 1$ is straightforward; for $|f(0)| = 2$ (the only "hard" part of the problem in my biased opinion) induct to show $|f(x)| = 2 \forall x \ge 0$, then finish using $P(-x, x)$ to get $|f(x)| = 2 \forall x$. Of course, we do need $f(-1) = -f(1)$, and all fns satisfying any one of the above cases along with this condition can be seen to work, qed.