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

Question

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

Answer 1

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.

Answer 2

$\cos(\pi\sqrt{x-4})\cos(\pi\sqrt x) = 1$

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.

$x = 4$

Answer 3

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.)

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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 $x-4<0$.)