Composite Function Equality Let f, g be functions$: \mathbb{R} \mapsto \mathbb{R}$ with $f(x)=x^2+ax+b$ with $a,b \in \mathbb{R}$, such that $(f\circ g)=(g\circ f)$.
If the equation $g(x)=x$ has precisely one solution for all $x \in \mathbb{R}$, prove that $(a-1)^2=4b$.

This is false. Define $f$ and $g$ by 
$$f(x)=x^2+ax \qquad\text{and} \qquad g(x)=0.$$
We have, for all $x$, $f(g(x))=f(0)=0$ and $g(f(x))=0$. Thus, the condition $f\circ g=g\circ f$ holds. Moreover, $g(x)=x$ if and only if $x=0$. So, $g(x)=x$ has a unique solution.
Yet, the coefficient $a$ is arbitrary, so $(a-1)^2$ can certainly take nonzero values, contradicting the claim in the OP.
