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 $(a-1)^2=4b$.

From $f \circ g = g \circ f$ it follows that:
$$ g^2(x) + ag(x) + b = g(x^2 + ax+b)$$
Now we plug into the above equation $x^*$ which is the unique solution of $g(x^*)=x^*$ and we get:
$$g^2(x^*) + ag(x^*) + b = g(x^{*2}+ ax^*+b)$$
and then:
$$ x^{*2} + ax^* + b = g(x^{*2} + ax^* + b) $$
But then $\tilde{x} = x^{*2} + ax^* + b$ also solves the above equation, i.e. $g(\tilde{x})=\tilde{x}$. By uniqueness of the solution we get $\tilde{x} = x^*$. Plugging in again into the definition of $\tilde{x}$ we see that:
$$ x^{*2} + (a-1)x^* + b = 0 $$
Thus the quadratic equation $x^2+(a-1)x+b=0$ has at least one solution, namely $x^*$.
EDIT: Before I claimed that in fact it only has one solution. But as Karl Kronenfeld pointed out in his answer, this is false (in fact the original question is false).
Nevertheless the above steps are still correct and because the quadratic has at least one solution, it follows that $(a-1)^2 \geq 4b$ (but equality does not necessarily hold).
