functional equation for $x^2$ $f(f(x))=x^4$

Question

If $f(f(x))=x^4$ for all real $x$ and $f(1)=1$ find $f(0)$. It seems that $f(x)=x^2$ but can we solve without this explicit form of $f$?

Answer 1

Assume that $f(0) = a$. Then the functional equation for $x = a$ gives $$ f(f(a)) = a^4 $$ but $f(a) = 0$ (since $0 = f(f(0)) = f(a)$), so we insert that and get $f(0) = a^4$. We therefore have $a = a^4$, which drastically limits our options for what $f(0)$ can be (it's either $1$ or $0$).

Assume (for contradiction) that $f(0) = 1$. We then have $$ 0 = f(f(0)) = f(1) = 1 $$ which cannot be the case. However, $f(0) = 0$ is consistent, as the example $f(x) = x^2$ shows.

Note that we haven't come close to any explicit solution of $f$, and I'm not even going to try. There are many solutions, most of them quite complicated, although it's possible only one of them ($f(x) = x^2$) is continuous.

Answer 2

Apply $f$ to both parts of the equation:

$$(f(x))^4=f(f(f(x)))=f(x^4).$$ You get $(f(0))^4 = f(0).$ Can you conclude?

Answer 3

You can have $f(a^n)=a^{-2n}$, and $f(b^n)=b^{2n}$ so long as the sequences $a^{2^n}$ and $b^{2^n}$ don't coincide. That gives uncountably many different functions $f$, although it seems all have the same value for $f(0)$