If the largest positive integer is n such that $\sqrt{n - 100} + \sqrt{n + 100}$ is a rational no. , find the value of $\sqrt{n - 1}$ .

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, $\boxed{\sqrt{n-1} = 50}$

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.