I want to find integer soutions to p=(n^2-1)/m^2 where p is a prime number

Question

I want to find all integer solutions for m and n such that $p=\frac{n^2-1}{m^2}$, where p is a prime number

as an example, I plugged in p=3 in wolfram and I got that the following values of m and n would work

$m=\pm \frac{(2-\sqrt3)^k-(2+\sqrt3)^k}{2 \sqrt3}$

$n=\pm \frac{1}{2} (-(2-\sqrt3)^k-(2+\sqrt3)^k), k\in Z, k\ge1$

but, how does wolfram get this? how could I do it for a given prime p other than 3?

Why do those solutions matter? Don't you want integer solutions for $n$ and $m$? — John Lou, May 17 '17 at 00:59
sorry, forgot to add the k exponents on the value for m, that way these are integer solutions for any k — Alosapien, May 17 '17 at 01:16

score 0 · Answer 1 · answered May 17 '17 at 09:31

The number of primes that exist is infinite, you will get infinite solutions. This statement will never stop giving solutions but if you want to start taking them out, I suggest you think that if you make $m$ be $1$, you are going to draw a large number of them. I hope it has been helpful for you :))

score 0 · Answer 2 · answered May 17 '17 at 09:33

0

It's just Pell's equation. It's hard to solve the prpblem if you don't already know this equation

answered May 17 '17 at 09:33

Why not you write out Pell's equation so the OP doesn't have to go searching? An answer should provide a full answer, with reference links if necessary – lioness99a May 17 '17 at 09:55

I want to find integer soutions to p=(n^2-1)/m^2 where p is a prime number

2 Answers2