Check the proof that $f(x) = \cos({x})\cdot\cos({\sqrt{3}x})$ is not periodic 
I have to prove that $f(x) = \cos({x})\cdot\cos({\sqrt{3}x})$ is not periodic

If the function is periodic then:
$$
f(x) = f(x+T)\\
\cos(x)\cdot\cos(\sqrt{3}x) = \cos(x+T)\cdot\cos(\sqrt{3}(x+T))
$$
Consider the function at $0$:
$$
\cos(T)\cdot\cos(\sqrt{3}T) = 1
$$
But this equation has only one solution at $T=0$ which contradicts the initial assumption that there exists a positive period.
Or the other way:
$$
\cos(T)\cdot\cos(\sqrt{3}T) = 1
$$
Let $\cos(T) = 1$ and $\cos(\sqrt{3}T) = 1$, hence
$$
T = 2 \pi m \\
\sqrt{3}T=2\pi n
$$
Substituting $T$ in the second equation gives:
$$
2\pi m\sqrt{3} = 2\pi n \\
\sqrt{3} = \frac{n}{m}
$$
But $m,n \in \mathbb N$ and $\sqrt{3} \in \{\mathbb R \setminus \mathbb Q\}$ which gives a contradiction.

Is the prove above valid?

Update:
For $\cos{T} = -1$ and $\cos\sqrt{3}T = -1$:
$$
T = \pi+ 2 \pi m \\
\sqrt{3}T=\pi + 2\pi n \\
\sqrt{3} = \frac{2n + 1}{2m+1}
$$
Which is a contradiction.

As an alternative, we have that
$$\cos({x})\cdot \cos({\sqrt{3}x})=\frac12 \cos(x(\sqrt3-1))+\frac12\cos(x(\sqrt 3+1))$$
and $\cos(x(\sqrt3-1))$ has period $\frac{2\pi}{\sqrt 3-1} $ while $\cos(x(\sqrt3+1)+$ has period $\frac{2\pi}{\sqrt 3+1}$
and $\not\exists k\in\mathbb{Z}$ such that
$$\frac{2\pi}{\sqrt 3-1}=k\frac{2\pi}{\sqrt 3+1}\iff \frac{\sqrt 3+1}{\sqrt 3-1}=k\iff (\sqrt 3+1)^2=2k$$
