Problem

Show that the function $f(x) = \cos (x) + \cos(\sqrt{2}x)$ is not periodic.

(Note: there was a typo in the email. I mistakenly wrote $f(x) = \cos(x) + \cos(\sqrt{2}+x)$, which, as several people pointed out to me, is periodic. Apologies for the mistake here!.)

Solution (by Marc Caelles)

Assume that there exists some $T\in \mathbb{R}$ such that $f(x)=f(x+T)$ for all $x\in \mathbb{R}$. In particular, $f(0)=f(T)$ and

$$ 2 = \cos T + \cos(\sqrt{2} T). $$

Since $|\cos x|\leq 1$, then $\cos T = \cos(\sqrt{2} T) = 1$. Therefore, there exist $n, m \in \mathbb{Z}$ such that

$$ T = 2\pi n, \quad \sqrt{2}T = 2\pi m. $$

But then, $\sqrt{2} n = m$, which is only possible if $n=m=0$. Thus, $T=0$ and the function is not periodic.

Thank you!

Special thanks to Damian Albrecht (Warsaw), Enrico M. (Italy), Amulya Srivastava (Pune), Nathaniel Kingsbury (US), Jaron Bailey (Australia), and Marc Caelles (Barcelona, Spain) for submitting a solution for this challenge problem.