Find all functions $f(x+y)=f(x^{2}+y^{2})$ for positive $x,y$

Solution 1:

If $x+y=a$ then $x^2+y^2$ is onto the range $\left[\frac{a^2}{2},a^2\right)$. So $f$ must be constant on any interval $\left[b,2b\right)$ with $b\in\mathbb R^+$.

Now define $b_n=2^{n/2}$ for $n\in\mathbb Z$. Then show that $\bigcup_n [b_n,2b_n)=\mathbb R^+$ and that $b_{n+1}\in(b_n,2b_n)$. This means that $f$ must be constant on all of $\mathbb R^+$.

Solution 2:

Take $y=x$. Then $f(2x)=f(2x^2)$, whence, taking $t=2x^2$, $f(t)=f(\sqrt{2t})$. Using this, you see that $f(t)=f(\sqrt{2t})=f(\sqrt{2\sqrt{2t}})=f(2^{3/4}t^{1/4})=...$. You should be able to show that for any positive integer $n$, $f(t)=f(2^{1/2+1/2^n}t^{1/2^n})$. If you assume continuity, you should now be able to observe something interesting by letting $n \to \infty$.