m^2 = n^2 * 2 has no solution for integers m and n because root two is irrational.

 

But m^2 = n^2 * 2 -1 does have solutions, the first of these being 7&5, 41&29, 239&169, 1393&985, 8119&5741.

I believe that there are an infinite number of solutions, in other words for all N, there exists a solution with m and n both greater than N.

 

Can anyone give me a proof ?

Why don't you try substituting m=n+d where d is some constant and solve the resulting equation.

I don't have a full solution, but try the substitution

 

[math]n = \sqrt 2 p[/math]
Which leads to the condition

 

[math]p = \sqrt {\frac{{1 + {m^2}}}{4}} [/math]

Which should be easier to handle.

Edited February 17, 201411 yr by studiot

But m^2 = n^2 * 2 -1 does have solutions, the first of these being 7&5, 41&29, 239&169, 1393&985, 8119&5741.

I believe that there are an infinite number of solutions, in other words for all N, there exists a solution with m and n both greater than N.

 

Can anyone give me a proof ?

This is an example of Pell’s equation.

Edited March 4, 201411 yr by Olinguito