Number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$ 
Find the number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$

\begin{align}
2\cos(\pi\sqrt{x-4})&.\cos(\pi\sqrt{x})=2\\\implies\cos\Big[\pi(\sqrt{x-4}+\sqrt{x})\Big]&+\cos\Big[\pi(\sqrt{x-4}-\sqrt{x})\Big]=2\\
\implies\cos\Big[\pi(\sqrt{x-4}+\sqrt{x})\Big]=1\quad&\&\quad\cos\Big[\pi(\sqrt{x-4}-\sqrt{x})\Big]=1\\
\pi(\sqrt{x-4}+\sqrt{x})=2n\pi\quad&\&\quad\pi(\sqrt{x-4}-\sqrt{x})=2m\pi\\
\sqrt{x-4}+\sqrt{x}=2n\quad&\&\quad\sqrt{x-4}-\sqrt{x}=2m\\
2\sqrt{x}=2(n-m)\quad&\&\quad2\sqrt{x-4}=2(n+m)\\
\sqrt{x}=n-m\quad&\&\quad\sqrt{x-4}=n+m\quad\&\quad x\geq4
\end{align}
How do I properly find the solutions ?
Or can I simply say
$$
x=(n-m)^2=(n+m)^2-4nm=x-4-4nm\implies nm=-1\\
\implies x=\bigg[n+\frac{1}{n}\bigg]^2\in\mathbb{Z}\implies n,\frac{1}{n}\in\mathbb{Z}\\
\implies n\neq0\implies n=1,x=4\text{ is the only solution}
$$

More simply we need as a necessary condition


*

*$\pi\sqrt{x-4}=k_1\pi$

*$\pi\sqrt{x}=k_2\pi$
that is


*

*$\sqrt{x-4}=k_1 \implies x=k_1^2+4$

*$\sqrt{x}=k_2\implies x=k_2^2$
that is
$$k_1^2+4=k_2^2 \iff k_2^2-k_1^2=4$$
and the only possible solutions is $k_1=0 \implies x=4$ which works by inspection.
