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

Question

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.

Answer 1

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

$$\frac{2\pi}{\sqrt 3-1}=k\frac{2\pi}{\sqrt 3+1}\iff \frac{\sqrt 3+1}{\sqrt 3-1}=k\iff (\sqrt 3+1)^2=2k$$