Find big perfect squares in the form 2x^2 - 1

Question

I am trying to find integer solutions for $x$ $$x = \sqrt{2a^2 - 1}$$ Where $a$ is an odd integer bigger than $\mathrm{1\times10^{12}}$.

I have no idea where to start with this. Is there any general solution, except running through possible values of $a$ bigger than $\mathrm{1\times10^{12}}$?

Are you familiar with Pell's equation? You want integer solutions to $x^2 - 2a^2 = - 1$. — Calvin Lin, Dec 27 '23 at 16:04
You should write "1e12" here as $10^{12}$. In any case, why is it important that $a > 10^{12}$? Study this problem with small $a$ to see patterns. Requiring $a > 10^{12}$ seems like an artificial restriction. It is also unnecessary to say in advance that you want $a$ to be odd, since when $x^2 - 2a^2 = -1$, we can prove $a$ must be odd if it exists at all: if $a$ is even and $x^2 - 2a^2 = -1$ then $x^2 \equiv -1 \equiv 7 \bmod 8$ and that has no solution. — KCd, Dec 27 '23 at 16:10
@CalvinLin Pell's equation was the best solution. Thank you! — Mykola Niemtsov, Dec 29 '23 at 04:34

Dominique · Answer 1 · 2023-12-27T16:37:43.850

1

First thing which comes in my mind, is this:

$$x^2 = 2a^2-1 \iff x^2-a^2=a^2-1 \iff (x-a) \cdot (x+a) = (a+1) \cdot (a-1)$$

Now it's a game of finding numbers, who can be written as $pq=rs$, who obey the mentioned equation.

edited Dec 27 '23 at 16:37

answered Dec 27 '23 at 16:27

Dominique

3,267

$7^2=2\cdot5^2-1$ is a solution besides $x=\pm1$ – J. W. Tanner Dec 27 '23 at 16:35
@J.W.Tanner: Nice one: $(7-5) \cdot (7+5) = (5-1) \cdot (5+1)$. (I'll update my answer) – Dominique Dec 27 '23 at 16:37

Find big perfect squares in the form 2x^2 - 1

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