Show that $f(x) = x$ if $f(f(f(x))) = x$.

Solution 1:

Suppose that $f(x_0)>x_0$ for some $x_0$. Then $$ f(f(f(x_0)))>f(f(x_0))>f(x_0)>x_0, $$ contradicting the hypothesis that $f(f(f(x_0)))=x_0$. We get a similar contradiction if $f(x_0)<x_0$. So the only option is that $f(x)=x$ for all $x$.

Solution 2:

If f is strictly increasing, then if x>y, f(x)>f(y). Suppose f(f(f(x)))=f(x).

Case 1: Suppose $f(x) < x$. So, $x>f(x)$. Thus, $f(x)>f(f(x))$ since $f$ is strictly increasing. It then would follow that $f(f(f(x)))>f(f(x))$, since $f(f(f(x)))=f(x)$. But, since f is strictly increasing we can also infer that $f(f(x))>f(f(f(x)))$. Thus we have a contradiction and it is not the case that $f(x) < x$.

Case 2: Suppose $f(x)>x$. The reasoning of this case comes as similar to that of case 1. So, it does not hold that $f(x)>x$.

By trichotomy it follows that $f(x)=x$