Let $\mathbb{N}=\{0,1,2, \ldots\}$. Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that

$$
x f(y)+y f(x)=(x+y) f\left(x^{2}+y^{2}\right)
$$

for all $x$ and $y$ in $\mathbb{N}$.
We claim that $f$ is a constant function. Suppose, for a contradiction, that there exist $x$ and $y$ with $f(x) < f(y)$ and $f(y) - f(x) > 0$ is minimal. Then

$$
f(x)=\frac{x f(x)+y f(x)}{x+y}<\frac{x f(y)+y f(x)}{x+y}<\frac{x f(y)+y f(y)}{x+y}=f(y)
$$

so $f(x)<f\left(x^{2}+y^{2}\right)<f(y)$ and $0<f\left(x^{2}+y^{2}\right)-f(x)<f(y)-f(x)$, contradicting the choice of $x$ and $y$. Thus, $f$ is a constant function. Since $f(0)$ is in $\mathbb{N}$, the constant must be from $\mathbb{N}$.

Also, for any $c$ in $\mathbb{N}, x c+y c=(x+y) c$ for all $x$ and $y$, so $f(x)=c, c \in \mathbb{N}$ are the solutions to the equation.