$\cos(x)+\cos(x\sqrt{2})$ is not periodic Show that the function $$f(x)=\cos(x)+\cos(x\sqrt{2})$$ is not periodic. I tried $x = a$ and $a\sqrt{2}$. I am guessing that the method of contradiction would be of some help over here. What else should I try?
 A: There's only one value of $x$ for which $f(x)=2$. Proof: Since $\cos\phi\le1$ this is only possible, if $\cos x = 1$ and $\cos(x\sqrt2) = 1$ hold true at the same time. The first statement is equivalent to $\frac{x}{\pi}\in\mathbb Z$, the second is equivalent to $\frac{x}{\pi}\sqrt2\in\mathbb Z$. The value $x=0$ fulfills the requirements. If a non-zero $x$ would fulfill both $\frac{x}{\pi}\in\mathbb Z$ and $\frac{x}{\pi}\sqrt2\in\mathbb Z$, the you could devide the second integer by the first. The result would be a rational number with value $\sqrt2$. This is a contradiction. Hence, only $x=0$ fulfills both requirements.
A: Assume $ \cos(x) + \cos(\sqrt{2} x) = \cos(x+T) + \cos(\sqrt{2} x+ \sqrt{2} T) $.
Let $x=0$. Then
$$2=\cos T + \cos \sqrt{2}T$$
Since the cosine is at most one, this means that simultaneously $\cos T = 1$ and $\cos \sqrt{2} T = 1$. This is equivalent to:
$$T = 2\pi n, \ \mbox{for some } n \in \mathbb{Z}$$
$$\sqrt{2}T = 2\pi m, \ \mbox{for some } m \in \mathbb{Z}$$
Substitute $T$ from the first into the second:
$$\sqrt{2} (2\pi n)= 2\pi m$$
$$\sqrt{2}=\frac{m}{n}$$
So $\sqrt{2} \in \mathbb{Q}$ which is a contradiction, or $T=0$. Either way, you're done.
