If the largest positive integer is n such that $\sqrt{n - 100} + \sqrt{n + 100}$ is a rational no. , find the value of $\sqrt{n - 1}$ . So here is the Problem :-
If the largest positive integer is n such that $\sqrt{n - 100} + \sqrt{n + 100}$ is a rational no. , find the value of $\sqrt{n - 1}$ .
What I tried :- I think that for $\sqrt{n - 100} + \sqrt{n + 100}$ to be a rational no. , both $(n - 100)$ and $(n + 100)$ have to be squares. Suppose :- $(n - 100)$  = $k^2$ and $(n + 100)$ = $m^2$ for some positive integers $k,m$  , and in the end I could only deduce that $(m + 10)(m - 10) = k^2 + 100$ , but then I couldn't proceed .
Also by guesswork, I could deduce that for $n = 125$, both nos. do become squares, although I don't know whether $n = 125$ is the highest or not.
Any hints or explanations to this problem will be greatly appreciated !

Let $\sqrt{n-100} + \sqrt{n+100} = p$, where $p$ is rational.
$$\implies 2n + 2\sqrt{n^2 - 10000} = p^2$$
But that must mean that $2\sqrt{n^2 - 10000}$ is rational.
Which must mean that $\sqrt{n^2 - 10000}$ is rational.
$$\implies n^2 - 10000 = k^2$$
$$\implies (n+k)(n-k) = 10000$$ The problem requires us to maximize $n$, notice that we'll get the maximum value of $n$ if we split $10000 = 5000 \times 2$ and set $n+k = 5000$ and $n-k = 2$ to get $n = 2501$.
Hence, $\boxed{\sqrt{n-1} = 50}$
