Composite Function Equality

Question

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

Answer 1

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).

Answer 2

From the equallity, we have

$g^2(x) + ag(x) + b = g(x^2 + ax + b)$

If we have a solution $c$ for $g(x) = x$, so:

$c^2 + ac + b = g(c^2 + ac + b)$

But from the unicity of the solution of $g(x) = x$, we have

$c^2 + ac + b = c \implies c^2 + (a-1)c + b = 0$

If $\Delta < 0$, then there no exist such $c$. If $\Delta > 0$, then we would have other solution different from $c$ to $g(x) = x$. So, $\Delta = (a-1)^2 - 4b = 0$

Answer 3

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.