How to prove that the following two sets are equal?

Question

Let $M=\{x|f(x)=x\},~N=\{x|f(f(x))=x\}$, and $f(x)=x^2+ax+b$, where $a,b\in R$ are such that $4b=(a-1)^2$. Show that $M=N$.

Answer 1

Proceed to show the desired result by showing $M\subseteq N$ and $N\subseteq M$. The first part follows easily since \begin{align} &x\in M \Rightarrow f(x)=x \\ &\Rightarrow f(f(x))=f(x)=x\\ &\Rightarrow x\in N. \end{align}

Thus, $M\subseteq N$. In order to show that $N\subseteq M$, consider an $x$ such that $f(f(x))=x$. Then, we have \begin{align} x-f(x)=f(f(x))-f(x)&=(f(x))^{2}+a(f(x))+b-(x^{2}+ax+b)\\ &=(f(x)-x)(f(x)+x+a). \end{align}

This implies that \begin{eqnarray} (f(x)-x)(f(x)+x+a+1)=0. \end{eqnarray}

From here, we conclude that the only possibility is $f(x)=x$, since if $f(x)=-x-a-1$, then $f(f(x))\neq x$. Thus, we have shown that if $x\in N$, then $x\in M$, which implies $N\subseteq M$.

Hence, $M=N$.