One of my favorite problem in 2024 imoc : D Let $\square$ denotes the set of perfect squares Let $P(x,y)$ denotes the assertion of $|xf(y)-yf(x)| \in \square$ Choose a prime $p$ with form $4k+1$ $P(p,x)$ $\implies$ $|pf(x)-xf(p)| \in \square$, $\therefore$ $(\frac{xf(p)}{p})=0$ $\vee$ $1$(Note that $(\frac{-1}{p})=1$ since $4|p-1$, so the plus–minus sign doesn't matter at all) if $p \nmid f(p)$ then we can choose $x$ such that $(\frac{x}{p})=-(\frac{f(p)}{p})$, then $1=(\frac{xf(p)}{p})=(\frac{x}{p})(\frac{f(p)}{p})=-1$, contradiction!! So we must have $p|f(p)$ $\forall p \in \mathbb{P}_{4k+1}$ Define another function $g:\mathbb{N} \to \mathbb{Q}^+$ (from the previous section we know $f(x) \in \mathbb{N}$ $\forall p \in \mathbb{P}_{4k+1}$) Let $Q(x,y)$ denote the assertion of $xy|g(x)-g(y)| \in \square$. Choose three distinct primes with form $4k+1$: $p$, $x$, $y$ $Q(p,x)$ $\implies$ $px|g(p)-g(x)| \in \square$ $\overset{p, x \in \mathbb{P}}{\implies}$ $px|g(p)-g(x)$ $\implies$ $p|g(p)-g(x) \quad (1)$ $Q(p,y)$ $\implies$ $py|g(p)-g(y)| \in \square$ $\overset{p, y \in \mathbb{P}}{\implies}$ $py|g(p)-g(y)$ $\implies$ $p|g(p)-g(y) \quad (2)$ $(1)-(2)$ $\implies$ $p|g(y)-g(x)$ since there exist infinity primes with form $4k+1$, $g(y)-g(x)$ has infinity prime divisors, which implies $g(y)-g(x)=0$ $\therefore$ $f(x)$ is a constant $c$ $\forall x \in \mathbb{P}_{4k+1}$ Finally, choose $p \in \mathbb{P}_{4k+1}$ and $x \in \mathbb{N}$ ($p \nmid x$). $Q(x,p)$ $\implies$ $px|g(x)-c| \in square$ $\implies$ $p|g(x)-c$ Again, since there exist infinity primes with form $4k+1$, $g(x)-c$ has infinity prime divisors, which implies $g(x)-c=0$ so $g(x)=c$ $\forall x \in \mathbb{N}$. Namely, $\boxed{f(x)=cx\quad \forall x}$ is a solution, where $c$ is an arbtitary positive integer (which indeed fit since $|xf(y)-yf(x)|=|cxy-cxy|=0 \in \square$ $\forall x,y$)

Another weird solution: $(x,y)=(1,3),(1,5),(3,5)$ can find out that $$\frac{f(1)}{1}=\frac{f(3)}{3}=\frac{f(5)}{5}$$so $(x,y)=(x,1),(x,3)$ shows that $f(x)=cx$ for all $x$

I claim $f(n)=cn$ for all $n$, where $c\in\mathbb{N}$ is arbitrary. Let $P(x,y)$ denotes the assertion that $|xf(y)-yf(x)|$ is a perfect square for every $x,y\in\mathbb{N}$. Consider $P(p,y)$ where $p\equiv 1\pmod{4}$ is an arbitrary prime, we find $yf(p)$ is a quadratic residue modulo $p$---as $-1$ is a quadratic residue modulo any such prime. So, unless $p\mid f(p)$, we arrive at a contradiction by taking $y$ to be a quadratic non-residue. Thus, $p\mid f(p)$ for every $p\equiv 1\pmod{4}$, a prime. Next, let $f(n)=n\phi(n)$, where $\phi(p)\in\mathbb{N}$ for every prime $p\equiv 1\pmod{4}$ and $f(1)=\phi(1)\in\mathbb{N}$. Considering $P(p,1)$, we obtain $|pf(1)-f(p)| = |p\phi(1) - p\phi(p)|=p|\phi(p)-\phi(1)|$ is a perfect square. Thus, $\phi(p)\equiv \phi(1)\pmod{p}$, for every prime $p\equiv 1\pmod{4}$. Next, consider $P(p,q)$ where $p\equiv q\equiv 1\pmod{4}$ are distinct primes. We have $|pf(q)-qf(p)| = pq|\phi(p)-\phi(q)|$ a perfect square. Using modulo $p$ and the fact $q\ne p$, we must have $p\mid \phi(p)-\phi(q)$. As $\phi(p)\equiv \phi(1)\pmod{p}$, we find $p\mid \bigl|\phi(1)-\phi(q)\bigr|$. Unless $\phi(q)=\phi(1)$, we arrive at a contradiction by taking $p>|\phi(p)-\phi(1)|>0$. So, for $c:=\phi(1)$, we find $f(p) = cp$ for every prime $p\equiv 1\pmod{4}$. We now prove $f(n)=n$ for all $n$. Consider $P(p,n)$ where $p\equiv 1\pmod{4}$ is a prime to get $|pf(n) -nf(p)| = p|f(n)-cn|$ is a square. If $f(n)\ne cn$ for some $n$, then taking $p$ large we find $p\nmid |f(n)-cn|$, a contradiction.

grupyorum wrote: Now if $p\nmid f(p)$ then by taking $y$ to be a quadratic residue and quadratic non-residue, we arrive at a contradiction. So, $p\mid f(p)$ for all primes $p$. I think there is a little mistake since $|pf(y)-yf(p)|$ is equal to $pf(y)-y(p)$ or $yf(p)-pf(y)$, so both $|pf(a)-af(p)|$ and $|pf(b)-af(b)|$ may be a quadratic residue (suppose that $p\nmid f(p)$ and $(\frac{a}{p})=1$ and $(\frac{b}{p})=1$)