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

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