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

$\cos(\pi\sqrt{x-4})\cos(\pi\sqrt x) = 1$
implies 
$\cos(\pi\sqrt{x-4}) = \cos(\pi\sqrt x) = 1$
or 
$\cos(\pi\sqrt{x-4}) = \cos(\pi\sqrt x) = -1$
in either case 
$\sqrt{x-4}, \sqrt x$ are both integers.  In fact, they are both even, or they are both odd.  But that isn't entirely relevant.
There is only one pair of integers that are perfect squares and separated by a distance of 4.
$x = 4$ 
