The number of solutions to the equations $x^2-2y^2=\pm p$ Knowing that the two equations $x^2-2y^2=\pm1$ have infinitely many solutions in $\mathbb{Z}\times \mathbb{Z}$ (i.e., that there are infinitely many elements with norm $\pm 1$ in $\mathbb{Z}[\sqrt{2}]$), how does one prove that the two equations $x^2-2y^2=\pm p$ also have infinitely many solutions in $\mathbb{Z}\times \mathbb{Z}$ for every prime $p$ satisfying $p\equiv_8 \pm 1$? 

Let $ K = \mathbf Q(\sqrt{2}) $. First, observe that if we exhibit only one solution, we automatically have infinitely many solutions. Now, exhibiting a divisor of a prime $ p $ in $ \mathcal O_K $ is the same thing as exhibiting a prime ideal of $ \mathcal O_K $ lying above $ p $, since $ K $ has class number $ 1 $. Hence it suffices to identify the inert primes in $ \mathcal O_K $. Dedekind's factorization criterion tells us that the inert primes are precisely the primes $ p $ such that $ X^2 - 2 $ is irreducible modulo $ p $, in other words the primes $ p $ such that
$$ \left( \frac{2}{p} \right) = -1 $$
Quadratic reciprocity identifies these primes for us: they are precisely the primes that are $ 3 $ and $ 5 $ modulo $ 8 $. Thus, the primes that are $ 1 $ and $ 7 $ modulo $ 8 $ are split in $ \mathcal O_K $.
