if $f(f(x+y))=f(x^2)+f(y^2)$ then $f(x)=?$

Question

if $f(f(x+y))=f(x^2)+f(y^2)$ then $f(x)=?$ for all integers ($f: \mathbb Z \rightarrow \mathbb Z$)

I know how to solve the following problem though: if $f(f(x+y))=f(2x)+2f(y)$ then $f(x)=?$ We can easily analyze that $f(x)$ here (in the second problem) is a linear function. And hence solve it by using the linear equation.

But, As for the main problem (the one I mentioned first) I don't know how to proceed. Should I use the quadratic equation? (my calculation says its a quadratic function)

Would be greatful if anyone could help me with this...

Thank you

Answer 1

Substituting $y=-x$ grants $f(f(0))=2f(x^2)$. Thus $f(x^2)$ is constant for all $x$, i.e. $f$ is constant for nonnegative numbers, the range of values for $x^2$. (If your $f$ has domain $\mathbb R$ or something else, you'll need to specify to proceed further.)

Answer 2

in this answer, $x$ and $y$ are integers.
let $y=-x$.
then $2f(x^2)=f(f(0))$
let $x=0$
then $2f(0)=f(f(0))$
Let $a=f(0)$
$f(a)=2a$
$f(f(x+y))=f(x^2)+f(y^2)=2a$
$f(f(x))=2a$
$f(2a)=f(f(a))=2a$
$f$ has only one fixed point $2f(0)$.
also, the fixed point cannot be a square or twice a square.

so the solutions $f$ must satisfy 2 conditions:

• $f(f(x))=2f(0)$
• $f(x^2)=f(0)$

$f$ cannot be a polynomial, because if it were, its second iterate is a polynomial and a constant at the same time, contradiction. in particular, $f$ cannot be linear or quadratic. also the qn mentioned another functional eqn, but the solution of one won't help with solving the other.