How do I find the value of $f(0)$? Let $f$ be a polynomial such that $f(0)>0$ and $f(f(x))=4x+1$ for all $x\in R$, then $f(0)$ is $?$
So this is what I've tried so far,
$f(f(0))=1$ so $f(f(0))=4f(0)+1$ (*)
Also, $f(f(1))=5$,this implies $f(5)=16f(0)+5$ Now how should I find the value of $f(5)$ in terms of $f(1)$ so that I can solve (*)? Any help is appreciated .
The answer given is $1/2$

$f(x) = ax + b$ $(*)$
$f(f(x)) = a^2x + ab + b = 4x + 1$
Comparing coefficients,
$b(a+1) = 1$
$a^2 = 4$
$\implies a = 2, b = \frac{1}{3}$
($a$ is not $-2$ by the fact that $f(0) > 0$, $a = -2$ would result in $b = -\frac{1}{3} \implies f(0) = -\frac{1}{3}$)
Therefore $f(x) = 2x + \frac{1}{3}$ and $f(0) = \frac{1}{3}$
(($*$) comes from the fact that if a polynomial $f(x)$ is degree $n$, then the polynomial $f(f(x))$ is of degree $n^2$)
