I'm studying the equation $(2m^2-1)^2=2n^2-1$ ($\ast$), where $m$, $n$ are positive integers. It is known that $m$ can only be 1 or 2 by using Wolfram Alpha. Now I want to prove that result.
That is my attempt: If a prime $p$ | gcd($m$, $n$), ($\ast$) is equivalent to $(2m^2-1)(m^2-1)=(n-m)(n+m)$, $p$ divides RHS but not LHS, which is impossible. So gcd($m$, $n$)=1. Module 8 to ($\ast$) and we find $n$ must be odd. This is as far as I got. Can someone help me?
COMMENT (just for fun).- Your equation has solution modulo all integer, because if $f(m,n)=2m^4-2m^2-n^2+1$ then $f(2,5)=f(1,1)=0$, so all "local" method is not useful (unless restrictions but for the first primes till $19$ there are other solutions and for greater primes brute force becomes hard). Because of squares we can just to try with positive integers. The problem is difficult.
We have $(2m^2-1)^2=2n^2-1\iff 2m^4-2m^2-n^2+1=0$ solving the quadratic in $m^2$ we get $$m^2=\frac{1+\sqrt{2n^2-1}}{2}\hspace1cm(1)$$ which imposes for some integer $x$, $$2n^2-1=x^2\hspace1cm(2)$$ Equation $(2)$ has infinitely many solutions given by $$x=-\frac12[(1+\sqrt2)(3-2\sqrt2)^k+(1-\sqrt2)(3+2\sqrt2)^k]; \space k\ge0$$ for example, for $k=1,2,3,4,5$, we have respectively $x=1,7,41,239,1393$ which in equation $(1)$ gives $m^2=1,4,21,120,697$ respectively so only the two first values given a square. We get the two solutions $(m,n)=(1,1),(2,5)$ given by the O.P.
Consequently a way of proving these are the only solutions, can be to prove that $$4m^2-2=-[(-7+5\sqrt2)(3-2\sqrt2)^{k-2}-(7+5\sqrt2)(3+2\sqrt2)^{k-2}]\hspace1cm(3)$$ has no solution for all integer $k\gt2\iff h=k-2\gt0$.
Maybe someone can finish using equation $(3)$.
There is already an elementary proof in $\textit{Number Theory: Conceptions and Problems}$ written by Titu Andreescu et al. , which is example 3.56.