MarizOzawa wrote: Find all functions $f:\mathbb{N} \rightarrow \mathbb{N}$ such that: $xf(y)+yf(x)=(x+y)f(x^2+y^2), \forall x,y \in \mathbb{N}$ Let $P(x,y)$ be the assertion $xf(y)+yf(x)=(x+y)f(x^2+y^2)$ Note that $P(x,y)$ implies assertion $Q(x,y)$ : $x+y|y(f(x)-f(y))$ Let $a=f(1)$ $P(x,x)$ $\implies$ $f(2x^2)=f(x)$ and simple induction gives $f(2^{2^n-1})=a$ $\forall$ non negative integer $n$ $Q(2n+1,2^{2^m-1})$ $\implies$ $2n+1+2^{2^m-1}|2^{2^m-1}(f(2n+1)-a)$ and so $2n+1+2^{2^m-1}|f(2n+1)-a$ Setting there $m\to+\infty$, we get $f(2n+1)=a$ Then $P(2n,1)$ $\implies$ $f(2n)=a$ Hence rthe answer : $\boxed{f(x)=a\text{ }\forall x\in\mathbb N}$ which indeed is a solution, whatever is $a\in\mathbb N$

the another way(it is familar but nice) we consider S={ f(x)-f(y)|x,y s.t f(x)>f(y)} because f(x) on N then S have min ,assume a,b from F.E f(a)>f(a^2+b^2)>f(b) then x=a^2+b^2 , y=b we have contract ( we can chane x^2+y^2 is f(x,y) s.t f(x,y)>x,y)

See France JBMO TST 2019