I've started a little reading on quadratic reciprocity, and a reason for this has eluded me. Here's a little of what I came up with so far. I decided I want to show that for all primes $p$, if $p|x^2-2$, then $p$ does not divide $2y^2+3$. Then, by way of contradiction, if $(x^2-2)/(2y^2+3)$ is an integer, then any $p$ such that $p|2y^2+3$ would have to divide $x^2-2$, a contradiction. I see this is true for $p=2$. I want to find all $p$ such that $x^2\equiv 2\pmod{p}$, and since for any odd $p$,

$$\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/8}$$

I see $(2|p)=1$ iff $p\equiv 1,7\pmod{8}$. So only primes of the form $8k+1$ or $8k+7$ divide $x^2-2$. However, I don't see a way to show that primes of the form $8k+1$ or $8k+7$ do not divide $2y^2+3$, so maybe I'm completely off the mark. Does anyone know how to resolve this, or have a better idea of what to do? Thanks!