Solve the following equation

I wasn't able to find an easier method for this problem than this one. This is not the correct solution but will help for the time's sake.

---

Observe that $x>4$ for any real solutions to exist.

Now, on the RHS we have $1$ , which of course is a rational number.

Thus, both of $\dfrac{\sqrt{x+1}}{x-4}$ and $\dfrac{\sqrt{x-4}}{x}$ must be rational.

$\underline {\text{Case 1:}}$ $\dfrac{\sqrt{x+1}}{x-4}$ is rational.

Since the options given are integers. We know that for any $x \in \mathbb Z$ , $x-4$ is always an integer. Also, $x \neq 4$ .

So, $x+1$ must also be an integer for $\dfrac{\sqrt{x+1}}{x-4}$ to be rational.

$\therefore x \in \left \{8,15,24,35,48, \cdots \right \} \qquad \qquad \qquad \color{#D61F06}{\text{Note that} \ x>4}$

$\underline {\text{Case 2:}}$ $\dfrac{\sqrt{x-4}}{x}$ is rational.

We know that for any $x \in \mathbb Z$ , $x$ is always an integer. Also, $x \neq 0$ .

So, $x-4$ must also be an integer for $\dfrac{\sqrt{x-4}}{x}$ to be rational.

$\therefore x \in \left \{5,8,13,20,29,40, \cdots \right \} \qquad \qquad \qquad \color{#D61F06}{\text{Note that} \ x>4}$ .

At this point, I must not say that both the solution sets have only $8$ common in them because I wasn't able to prove this. But it is certain that whatever the next value of $x$ may be it will not satisfy the RHS.

But, by plugging $x=8$ we see that the equation holds true. Thus, $x = \boxed{8}$