I have read through a lot of similar posts so I am not trying to re ask a question but just seeking some clarity.

I am looking at the Pell and Pell like equations:

$x^2-2y^2=1$ and $x^2-2y^2=k$ where $k=6n+1$ for $n \in \mathbb N$. I am just curious to know if all solutions have either $x$ or $y \equiv 0 \mod 3$ $\quad$ or if I am thinking about this completely wrong.

So for a problem that I am looking at the smallest solution would be $(3,2)$ because $(1,0)$ would not make sense. So when I used the seed it gave solutions with either $x,y$ divisible by 3 but I am sure my thinking is flawed.


Taken mod $3$, any equation of the form $x^2-2y^2=6n+1$ becomes $x^2+y^2\equiv1$ mod $3$. If $xy\not\equiv0$ mod $3$, then $x^2\equiv y^2\equiv1$ mod $3$, in which case $x^2+y^2\equiv2\not\equiv1$ mod $3$. So we must have $xy\equiv0$ mod $3$, which is to say $3\mid x$ or $3\mid y$ (since $3\mid xy$ and $3$ is prime).

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