A bit confusing. If $f(k)=f(l)$ then the denominator doesn’t make sense so are we going to assume $f$ is injective?

If $f$ works so does $f+c$ for $c \in \mathbb Z$ so wlog $f(0)=0$. Also $f$ is injective else LHS makes no sense. Let $P(k,l)$ the assertion of the given F.E. $P(p,0)$ where $p$ is a prime number $$f(p) \mid p^2 \implies f(p) \in \{-p^2, -p, -1, 1, p, p^2 \}$$Also note that if $f$ works so does $-f$ so wlog $f(1) \ge 0$, now $P(1,0)$ gives $f(1)=1$ $P(p,1)$ $$f(p)-1 \mid p^2-1 \implies f(p) \in \{-p, -1,p,p^2 \}$$Clearly there is at most one prime such that $f(p)=-1$, now we go with these cases. Case 1: There exists infinitely many primes with $f(p)=p^2$ If so then we do $P(k,p)$ and we set a arbitrarily large prime $p$ $$f(k)-p^2 \mid k^2-p^2 \implies f(k)-p^2 \mid k^2-f(k) \implies f(x)=x^2 \; \forall x \in \mathbb N_0$$Case 2: There exists infinitely many primes with $f(p)=p$ If so then we do $P(k,p)$ setting a arbitrarily large prime $p$ $$f(k)-p \mid k^2-p^2 \implies f(k)-p \mid k^2-f(k)^2 \implies f(x)=(-1)^{g(x)} x \; \forall x \in \mathbb N_0 \; \; \text{where} \; g(x) \in (0,1)$$Case 3: There exists infinitely many primes with $f(p)=-p$ If so then we do $P(p,k)$ for $p$ arbitrarily large prime $$p+f(k) \mid p^2-k^2 \implies p+f(k) \mid f(k)^2-k^2 \implies f(x)=(-1)^{g(x)} x \; \forall x \in \mathbb N_0 \; \; \text{where} \; g(x) \in (0,1)$$Case 4: There is only finitely many primes $p$ with $f(p)=p$ and only finitely many primes $q$ with $f(q)=q^2$ and only finitely many primes $s$ with $f(s)=-s$ Let $\mathcal P$ the set of such finite primes $p$, let $\mathcal Q$ the set of such finite primes $q$ and let $\mathcal S$ the set of such finite primes $s$, call $\mathbb P$ the set of all prime numbers, now note that for any prime $r$ in the set $\mathbb P$ but doesnt lie in the sets $\mathcal P$, $\mathcal Q$ or $\mathcal S$ has to satisfy that $f(r)=-1$, but there is infinitly many of those $r$ while $f(r)=-1$ can be held by at most 1 prime so contradiction!!. Hence $f(x)=(-1)^{g(x)} x+c, x^2+c, -x^2+c$ for all $x \in \mathbb N_0$ with $g(x) \in (0,1)$ and a constant $c \in \mathbb Z$ works thus we are done . Edit: Thx to @RevolveWithMe101 for pointing out my previous solution that i forgot that the function $f$ also covered integers (becuase i said $f(x)^2=x^2$ mean $f(x)=x$, oops...), now it is corrected .

MathLuis wrote: If $f$ works so does $f+c$ for $c \in \mathbb Z$ so wlog $f(0)=0$. Also $f$ is injective else LHS makes no sense. Let $P(k,l)$ the assertion of the given F.E. $P(p,0)$ where $p$ is a prime number $$f(p) \mid p^2 \implies f(p) \in \{-p^2, -p, -1, 1, p, p^2 \}$$Also note that if $f$ works so does $-f$ so wlog $f(1) \ge 0$, now $P(1,0)$ gives $f(1)=1$ $P(p,1)$ $$f(p)-1 \mid p^2-1 \implies f(p) \in \{-1,p,p^2 \}$$Clearly there is at most one prime such that $f(p)=-1$, now we go with these cases. Case 1: There exists infinitly many primes with $f(p)=p^2$ If so then we do $P(k,p)$ and we set a arbitrarily large prime $p$ $$f(k)-p^2 \mid k^2-p^2 \implies f(k)-p^2 \mid k^2-f(k) \implies f(x)=x^2 \; \forall x \in \mathbb N_0$$Case 2: There exists infinity many primes with $f(p)=p$ If so then we do $P(k,p)$ setting a arbitrarily large prime $p$ $$f(k)-p \mid k^2-p^2 \implies f(k)-p \mid k^2-f(k)^2 \implies f(x)=x \; \forall x \in \mathbb N_0$$Case 3: There is only finitely many primes $p$ with $f(p)=p$ and only finitely many primes $q$ with $f(q)=q^2$ Let $\mathcal P$ the set of such finite primes $p$ and let $\mathcal Q$ the set of such finite primes $q$, call $\mathbb P$ the set of all prime numbers, now note that for any prime $r$ in the set $\mathbb P$ but doesnt lie in the sets $\mathcal P$ or $\mathcal Q$ has to satisfy that $f(r)=-1$, but there is infinitly many of those $r$ while $f(r)=-1$ can be held by at most 1 prime so contradiction!!. Hence $f(x)=x+c, -x+c, x^2+c, -x^2+c$ for all $x \in \mathbb N_0$ and a constant $c \in \mathbb Z$ works thus we are done . Doesn't $f(x) = (-1)^{g(x)} x$ works for any random $g : \mathbb{N}_0 \to \{ 0, 1 \}$?

official English version Find all sequences of integers $a_0, a_1, a_2, \ldots$ such that for any integers $k, \ell \geq 0$, we have $$ a_k-a_{\ell} \mid k^2-\ell^2, $$that is, for any integers $k, \ell \geq 0$, there exists some integer $z$ such that $\left(a_k-a_{\ell}\right) z=$ $k^2-\ell^2$.

Same idea in IMO Shortlist 2011 N3 works which is the fact that there are infinitely many prime $f(p)=+-p^t$ for fixed $t$. Then put $l=p$ and take $p$ large enough.