Symmetric functional equation $ x f ( y ) + y f ( x ) = ( x + y ) f \left( x ^ 2 + y ^ 2 \right) $ on $ \mathbb N _ 0 $: how to show $ f$ is constant?

Question

Let $ \mathbb N _ 0 $ denote the non-negative integers. Find all functions $ f : \mathbb N _ 0 \to \mathbb N _ 0 $ such that $$ x f ( y ) + y f ( x ) = ( x + y ) f \left( x ^ 2 + y ^ 2 \right) \quad \forall \, x, y \in \mathbb N _ 0 $$

I got that $ f ( x ) = f \left( 2 x ^ 2 \right) $ for non zero $ x $ by setting $ x = y $. Does this show $ f $ is always constant, since the constant case does work by plugging in?

Should add: the supposed duplicate is for functions on the natural numbers, and the answers presume that $0$ is not included. The present question allows $0$ which makes the problem a whole lot easier. — lulu, Apr 26 '21 at 01:15
@lulu That's true for the accepted answer. There is another answer which includes $ 0 $ in the domain. — Mohsen Shahriari, Apr 26 '21 at 11:29
@MohsenShahriari And, to be sure, I expect the OP was just trying to quote the same problem, and that they have no idea whether $0$ is meant to be included or not. — lulu, Apr 26 '21 at 12:33

score 1 · Answer 1 · answered Apr 26 '21 at 00:58

1

Hint: Plug in $y=0$ to find that $xf(x^2)=xf(0)$ for all $x$.

answered Apr 26 '21 at 00:58

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Symmetric functional equation $ x f ( y ) + y f ( x ) = ( x + y ) f \left( x ^ 2 + y ^ 2 \right) $ on $ \mathbb N _ 0 $: how to show $ f$ is constant?

1 Answers1