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 !
 A: 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, $\boxed{\sqrt{n-1} = 50}$
A: Hmm, it is not stright forvard to say $n+100$ and $n-100$ are squares.
Put it this way:
$$\sqrt{n - 100} + \sqrt{n + 100}=r\in\mathbb{Q}$$
now we square it:
$$n-100+2\sqrt{n^2-100}+n+100 = r^2$$ and so $$ \sqrt{n^2-100} = {r^2\over 2}-n$$ Now we square it again and we get:
$$-100= {r^4\over 4}-r^2n$$ Now let $r={a\over b}$ where $a,b$ are relativly prime positive integers. So:
$$a^2(4nb^2-a^2)=400b^4\implies b^2(4n-400b^2)= a^4$$ and thus $$b\mid a^4\implies b=1$$
Now we have $$a^2(4n-a^2)=400\implies a\mid 400\implies a\in\{1,2,4,5,8, 10,20\}$$
Now check each possible $a$ and you are done.
