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

Question

Find all functions $f:\mathbb{R}^{+}\to \mathbb{R}$ such that for any $x,y\in \mathbb{R}^{+}$ the following holds: $$f(x+y)=f(x^{2}+y^{2}).$$

Answer 1

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 $n \to \infty$.

Answer 2

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 $\mathbb R^+$.

Answer 3

For any positives $a,b$ we have: \begin{aligned} f(a) &=f\left(\frac{a+b}{2}+\frac{a-b}{2}\right) \\ &=f\left(\big(\frac{a+b}{2}\big)^2+\big(\frac{a-b}{2}\big)^2\right) \\ &=f\left(\big(\frac{a+b}{2}\big)^2+\big(\frac{b-a}{2}\big)^2\right) \\ &=f\left(\frac{a+b}{2}+\frac{b-a}{2}\right) \\ &=f(b)\end{aligned}

Answer 4

If

$$f(A(x,y))=f(B(x,y))$$

on a connected open set of $(x,y)$ where $A$ and $B$ are functionally independent (have nonzero Jacobian determinant), the function $f$ is constant on that set .

Locally one has both $xy$-coordinates and $AB$-coordinates. From $(A,B)$ you can get to all close enough $(A + \epsilon_1, B + \epsilon_2)$ by moving $A$-only and then $B$-only along the $AB$ coordinates and this motion does not change the value of $f$. The motion can be tracked in $xy$ coordinates (uniquely determined by $A,B$ in a small neighborhood), and such paths can reach an open set of $(x,y)$ near any given point. This shows $f$ is locally constant. On a connected set that means constant.

The translation to this problem is: $A(x,y) = x+y$, $B(x,y)=x^2+y^2$, Jacobian is $\pm(x-y)$, so that the connected open set of $(x,y)$ with $0 < x < y$ can be used. This is a subset of the allowed pairs but it covers all possible values of $A$ and $B$, and therefore of $f(A)$ and $f(B)$. So it solves the problem with slightly less than the full set of assumptions, which is still a solution.

The need for an independence hypothesis can be seen from cases like $B=A^3$ where there are continuous nonconstant solutions on some intervals.