Determine all functions satisfying $xf(y) + yf(x) = (x+y)f(x^2+y^2)$ [duplicate]

Question

I tried to prove that the only functions $f: \mathbb{N} \to \mathbb{N}$ satisfying

$$xf(y) + yf(x) = (x+y)f(x^2+y^2)$$

for all positive integers $(x, y)$ are constant functions.

I supposed that:

$$ \exists \; x \mbox{ such that } f(x) \neq f(1) \mbox{ and } \forall y \neq x, f(y) = f(1)$$

Then we have:

$$ xf(1) + yf(x) = (x+y)f(1) $$ $$ \implies yf(x) = yf(1) \implies f(x) = f(1)$$

I would like to know if this is correct, since I am not sure that the negation of my hypothesis implies that the function is constant.

Thank you!

Answer 1

Okay, I believe I have a proof that such a function is constant. First, when $x = y$, we get that $2xf(x) = 2xf(2x^2)$. Cancelling the $2x$, we get that $f(x) = f(2x^2)$ for all $x$. In particular, for any $a$, we have $f(a) = f(2a^2) = f(2(2a^2)^2) = f(2(2(2a^2)^2)^2) = \cdots$. Let me summarize this by saying that for any $a$, there are arbitrarily large $N$ with $f(a) = f(N)$.

Now for any $x$, we have that $f(x) + xf(1) = (1+x)f(1+x^2)$, which implies that $f(x) + xf(1) \equiv 0 (\text{mod }x+1)$. Since $x \equiv -1 (\text{mod }x+1)$ are units mod $x+1$, we conclude that $f(x) \equiv f(1) (\text{mod }x+1)$.

Finally, fix an $a$. We want to show that $f(a) = f(1)$. Choose $N$ larger than $f(a) + f(1)$ such that $f(a) = f(N)$. Applying the conclusion of the previous paragraph to $x = N$, we get that \begin{equation*} f(a) = f(N) \equiv f(1) (\text{mod }N+1) \end{equation*} and thus $f(a) = f(1)$ on the nose, so $f$ is a constant function.