# 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 !

• This looks like a good start. Note that if $n\pm100$ are squares as you say then $$200=m^2-k^2=(m+k)(m-k)$$ – Peter Foreman Jul 25 at 15:50
• Oh yeah!, @PeterForeman , I totally forgot about that . I actually had got before that $$m^2 - 100 = k^2 + 100$$ . – Anonymous Jul 25 at 15:52
• It's generally a good idea to wait some time before choosing the best answer, just to give other people enough time to provide their answer (unless this is some kind of speed competition). – Sil Jul 25 at 16:08

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

• Excellent and the quickest solution . Thank you! – Anonymous Jul 25 at 15:54
• $k$ is rational or integer? – Aqua Jul 25 at 15:55
• @Aqua If the square root of an integer is a rational then this square root is also an integer. – mathcounterexamples.net Jul 25 at 16:03
• "we'll get the maximum value of $n$ if we split $10000 = 5000 \times 2$" - you might want to elaborate more on that, why do you get maximum value this way? Did you try them all? Not saying it's wrong, but it might improve the answer. – Sil Jul 25 at 16:12
• @Aqua This perfectly makes sense and would had avoided this list of comments! – mathcounterexamples.net Jul 25 at 16:14

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.

• I got your point . But @Nikunj's solution was the best . – Anonymous Jul 25 at 15:55
• But his solution lack of steps. If someone would have to grade it it should not give him all points. – Aqua Jul 25 at 16:18
• His solution do lacks some steps, but anybody reading his solution can understand it easily . – Anonymous Jul 25 at 16:41