What must the value of $y$ be so that $a$ and $b$ are integers?

Question

I have made some sort of conjecture that in the equation: $$a^2 = b^2 + 2y^2.$$ $y$ must equal $(2n + 1)\sqrt {8}$ such that $n \in \mathbb{W}$ for $a$ and $b$ to be integers, or at least Pythagorean Triples.

All I believe that is most correct is that for every value of $n$ there are at least four solutions to satisfy the equation. And that for some other integer $x$, if $a^2 = x + y$ and $b^2 = x - y$ then this will work out, but it doesn't seem to be that $x + y$ and $x - y$ can both be squared numbers $\in \mathbb{W}$.

Could somebody please help me prove/disprove this? I am a little bit stuck. Thanks.

Answer 1

$y$ doesn't even need to have a $\sqrt{}$ in it at all. For any $n\in\mathbb N$ there is a solution $y=2n, a=2n^2+1, b=2n^2-1.$ This is because

$$a^2=(2n^2+1)^2=4n^4+4n^2+1$$ $$b^2+2y^2=(2n^2-1)^2+2(2n)^2=4n^4-4n^2+1+8n^2=a^2$$

Answer 2

If $y=\frac{3\sqrt{2}}{2}$, then $2y^2=9$, giving rise to the Pythagorean triple $$ 9+16=25 $$ with $a=5$ and $b=4$ in your notation. It is straightforward to see that $y$ cannot be written of the form $(2n+1)\sqrt{8}$ for $n$ a positive integer, so this does not seem to be true.

Answer 3

$$y\in\left\{\sqrt{\frac{a^2-b^2}{2}}|a^2\geq b^2, \{a,b\}\subset \mathbb Z\right\}$$