Indian Statistical Institute B.Math & B.Stat : Continuity

Let $P: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $P(x)=x$ has no real solution. Prove that $P(P(x))=x$ has no real solution. If possible let, $P(P(x))=x$ has a real solution for $x=x_0$. Then $P(P(x_0))=x_0\dots(1)$ Now let, $P(x_0)=y_0$ $\implies P(y_0)=x_0$ using $(1)$ Note that $x_0 \neq y_0$, otherwise we will have a solution to the equation $P(x)=x$! A contradiction to the hypothesis. Without loss of generality assume that $x_0 < y_0$ Construct a function $Q:[x_0,y_0]\rightarrow \mathbb{R}$ where $Q(x)=P(x)-x$,since $P$ is given to be continuous on $\mathbb{R}$, $Q$ is continuous on $[x_0,y_0].$ Observe that, $Q(x_0)=P(x_0)-x_0=y_0-x_0 > 0$ and $Q(y_0)=P(y_0)-y_0=x_0-y_0 < 0$ $\implies Q(x_0)Q(y_0) < 0 \implies$ there exists a point $c\in$ $[x_0,y_0]$ such that $Q(c)=0$ using $Intermediate-Value-Theorem$ $\implies P(c)-c=0 \implies P(c)=c$ for a real value, which contradicts the hypothesis, thus the assumption $P(P(x))=x$ has a real solution is not tenable.