Showing there are Infinitely Many Pairs that Satisfy Certain Criteria...

Question

I don't quite understand how to complete the following proof. Something tells me the solution is actually quite obvious, but for whatever reason I just can't crack it. Any help would be appreciated!

• In this problem, we discuss pairs of consecutive whole numbers satisfying the property that one of the numbers is a perfect square, and the other is a double of a perfect square.
• The task: Show that there are infinitely many pairs of the form $$(2a^2,b^2)$$ where the smaller number is the double of a perfect square satisfying the given property.

Thanks in advance!

Answer 1

The question is equivalently phrased as follows:

Prove that there are infinitely many solutions to the equation $$b^2 - 2a^2 = 1,$$ where $a$ and $b$ are positive integers.

Then, consider that if you can find one such solution, how can you generate an additional solution from it? Can you devise a formula that says if $a = a_0$ and $b = b_0$, then you can find another solution $a = a_1$, $b = b_1$ from $a_0$ and $b_0$ such that $a_1$ and $b_1$ are also positive integers?

Answer 2

If a pair $(a,b)$ satisfies the equation $2a^2+1 = b^2$, then the pair $(2ab,4a^2+1)$ does as well (check it!). And you already know of 1 such pair: (2,3). So, there are infinitely many.