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
$$
x=(n-m)^2=(n+m)^2-4nm=x-4-4nm\implies nm=-1\\
\implies x=\bigg[n+\frac{1}{n}\bigg]^2\in\mathbb{Z}\implies n,\frac{1}{n}\in\mathbb{Z}\\
\implies n\neq0\implies n=1,x=4\text{ is the only solution}
$$

I suppose $x$ is real in the following and that $\cos$ is the function $\cos:\Bbb R\to\Bbb R$. (There is an other function $\cos:\Bbb C\to\Bbb C$, to use it i have to ask for the branch of the square root(s) first.) 

The two $\cos$ functions in the product (evaluated at those two places) must have (in a correlated way) the value $\pm 1$. This makes things easier maybe to decide that $\sqrt x$ and $\sqrt {x-4}$ are two integers of same parity. In particular $x\ge 4$. Starting with the perfect square $2^2=4$ the distance between two perfect squares is $\ge 3^2-2^2=9-4=5$, so it is at least $5$. So the bigger perfect square $x$ (among $x,x-4$) is at most $4$. We get the solution $x=2^2=4$. (There is no $x=1^2$ or $x=0^2$ as solution, because $x-4<0$.) 
