Find all integral solutions to the equation $$\left(\frac{1}{n}\right)^{-1/2}=\sqrt{a+\sqrt{15}}-\sqrt{a-\sqrt{15}}.$$
Clearly, $\displaystyle \left(\frac{1}{n}\right)^{-1/2}=\sqrt{n}$, so $n\geq 0$, and the equation becomes $$\sqrt{n}=\sqrt{a+\sqrt{15}}-\sqrt{a-\sqrt{15}}.$$ Squaring both sides (which I am aware may introduce additional solutions) gives $$n=2a-2\sqrt{a^2-15}.$$ We require $a+\sqrt{15}\geq 0$ and $a-\sqrt{15}\geq 0$, so $a\geq 4$. Is it possible to find an upper bound for $a$, or bounds on $n$?
Rearranging and squaring again (which may also introduce superfluous solutions) gives $$4a^2-4an+n^2=4a^2-60,$$ or $$n^2-4an+60=0.$$ Perhaps now writing $$n(4a-n)=60$$ helps, since we know $n,4a-n\in\mathbb{Z}$, and the integral factors of $60$ are $1,2,3,4,5,6,10,12,15,20,30,60,-1,-2,-3,-4,-5,-6,-10,-12,-15,-20,-30,-60$. Since I have already shown $a$ and $n$ are non-negative, the negative factors can be excluded.
Moreover, $n$ must be even, since if $n$ is odd, both $n$ and $4a-n$ are odd, so their product cannot be $60$.
From here it is tedious (but easy) to find the solutions and then double check that they satisfy the original equation.
Questions: 1) Does this solution appear to be correct?
2) Is there an easier solution? I would welcome any suggestions for alternative solutions.