Find all functions $f$ :- $\mathbb{N}$ $\to$ $\mathbb{N}$ such that :- $xf(y) + yf(x) = (x + y)f(x^2 + y^2)$

Question

So here is the Question :-

Find all functions $f$ :- $\mathbb{N}$ $\to$ $\mathbb{N}$ such that :- $$xf(y) + yf(x) = (x + y)f(x^2 + y^2)$$ I tried substituting values for $x$ and $y$, but I couldn't reach to a possible clue to the solution. Any hints or suggestions will be greatly appreciated!

Answer 1

Suppose there exists an integer $n$ such that $f(1) <f(n)$. With $x=1$ and $y=n$, we have $$ nf(1)+f(n) = (n+1)f(n^2+1) \implies f(1)<f(g(n))<f(n),$$ where $g(n)=n^2+1$. Repeating the same argument using $f(1)<f(g(n))$, we deduce that $$f(1)<f\left(g^{f(n)-f(1)}(n)\right)<f\left(g^{f(n)-f(1)-1}(n)\right)<\ldots<f(g(n)) <f(n),\tag{1}$$ where $g^m(\cdot)$ denotes the composition of the function $g$ repeated $m$ times.

However, there cannot be more than $f(n)-f(1)-1$ integers between $f(n)$ and $f(1)$. Thus, (1) leads to contradiction because $f(x)\in\mathbb{N},\forall x\in\mathbb{N}$. Hence, $f(1)\nless f(n).$

Using similar arguments, we can show that $f(1)\ngtr f(n)$. Therefore, we conclude that $f(x)=f(1)$ for all values of $x$ which trivially satisfies the given relation.

Answer 2

Setting $x=y=1$ gives $f(1)=f(2)$. From there, try to show via induction that $f(1)=f(n)$ for all $n$ which are sums of two squares.

Remains the question if $f(n)$ can be defined arbitrarily on other $n$. For a counterexample, see comment by user Explorer (thanks!). I do not know if one can conclude further, and whether there are non-constant $f$ which satisfy the criterion.

EDIT: I realise what I had in mind here might not work. One only gets the following: If $f(x)=f(y)$, then $f(x^2+y^2)=f(x)$ as well, and starting with the above, one easily constructs infinitely many numbers $n$ with $f(n)=f(1)$. But to show that is true for all $n$, one needs more work, as in the other answer.