$f(f(x)) = 1 + x^2$, then what is f(1)?

Question

I get $f(f(a)) = a^2 + 1 = f(f(-a))$, and so $f(a)^2 + 1 = f(a^2 + 1) = f(-a)^2 + 1$, so $f(a) = f(-a)$ or $f(a) = -f(-a)$, but then I donot know what to do next. Thanks for any help.

Answer 1

A piece of information is still required

$f(f(x)) = 1+x^2$

Let's say $y = f(x)$ is the function, then $y = g(x)$ is the inverse function

$f(x) = g(1+x^2)$, $f(1) = g(2)$

$f(f(x)) = 1+x^2 = f(f(-x))$ $f(f(x)) = f(f(-x))$ Therefore $f(x) = f(-x)$ $f(x)$ is a even function

$f(f(x)) = 1+x^2$, say $x → g(x)$

$f(x) = 1+g(x)^2$, $f(1) = 1+g(1)^2$

$1+g(x)^2 = g(1+x^2)$

Say $x → f(x)$ $1+x^2 = g(1+f(x)^2)$, $f(1+x^2) = 1+f(x)^2$

$f(x) = $

$1+f(x)^2 = f(1+x^2)$

I can't find $f(1)$ unless I knew $f(n)$

$f(1+x^2) = 1+f(x)^2$, Remember $f(x) = 1+g(x)^2$

$f(1+x^2) = 1+(g(x)^2+1)^2$, $f(1+x^2) = g(x)^4+2*g(x)^2+2$

Recall $f(x) = g(1+x^2)$, say $1+x^2→x$ $g(x) = f(\sqrt(x-1))$

$f(1+x^2) = f(\sqrt(x-1))^4+2*f(\sqrt(x-1))^2+2$