$f(f(\sqrt{2}))=\sqrt{2}$ then f has a fixed point
Solution 1:
Hint: Consider $\alpha=f(\sqrt2)$. Show that $f(\alpha)=\sqrt2$. Inspect $f(x)-x$ at $\sqrt2$ and $\alpha$.
For further enjoyment: When you grok this question, try showing that if $$\overbrace{f\circ f\circ\cdots\circ f}^{n\text{ compositions}}(x_0)=x_0$$ then $f$ has a fixed point.