Show that $a^2 - 15b^2 =3$ has no integer solutions.

Question

Show that $$a^2 - 15b^2 =3$$ has no integer solutions.

I'm not overly experienced with number theory nor Diophantine equations, but upon looking around a bit I've realised this is a Pell-type equation. I can find a lot of information about if a solution exists, but not much on the criteria for existence.

I also need to show this for $-3$ and $\pm5$ instead of $3$ but I imagine the proof would be similar.

Answer 1

Often, when trying to prove a Diophantine equation has no solution, it helps to try to prove it has no solution modulo some number $n$. It's easier because there are then finitely many cases to check.

Take the equation modulo $4$:

$$a^2+b^2 \equiv 3 \pmod{4}$$

But a square is $0$ or $1$ mod $4$ so the sum of two squares cannot be $3$, modulo $4$.

Answer 2

Simply look at this modulo $5$. You obtain that $3$ is a square modulo $5$, whereas the only squares modulo $5$ are congruent to $0$ or $\pm 1 \pmod{5}$.