Largest possible solution for equation

Question

I have the following equation $$x^2=n^2+2016n, n>0$$ The questions wants to find the largest possible integer value for n. I have tried solving using difference of squares but it doesn't lead anywhere (substituting into '2016') $$x^2-n^2 = 2016n$$ $$(x-n)(x+n)=2016n$$ Alternatively, I have tried isolating x but that doesn't work either $$x^2 + 1008^2=n^2+2016n+1008^2$$ $$x^2 + 1016064=(n+1008)^2$$ $$x= \sqrt{(n+1008)^2-1016064}$$ And since that's derived from the initial, subbing it back into the initial doesn't help. Is there any other way/continuation to the problem or to construct another equation (thus solving with simultaneous eq.)? Thank you for your help!

Answer 1

$x= \sqrt{(n+1008)^2-1016064}$ works wonders, you just have to think a bit about what it says.

It says that you can take the square number $(n+1008)^2$, subtract the number $1016064$, then you get a new perfect square.

The distance between square numbers increase as the numbers themselves increase, and they increase quite predictably. The distance between consecutive squares actually follow the odd numbers.

So, back to your problem, we want $n$, and by extension, $(n+1008)^2$ to be as large as possible. The limitation is that there has to be another square number $1016064$ below it, and it has to be as "adjacent" as possible. Now, it can't be the previous square, because then the difference would be odd. But it can be the one before that, namely $(n+1008 - 2)^2 = (n+1006)^2$. So, we want $$ (n+1008)^2 - 1016064 = (n+1006)^2\\ n^2 + 2016n = n^2 + 2012n + 1012036\\ 4n = 1012036\\ n = 253009 $$