How to use Legendre symbol to find a prime which divides $ax^2+b$? I'm trying to prove that $\dfrac{x^2-2}{2y^2+3}$is never an integer if $x,y\in\mathbb{Z}$. 
It can be proven if $\forall p\in\mathbb{P}\:$doesn't suffice both of the following congruences: 
$$\: x^2\equiv 2 \pmod{p}$$
$$2y^2\equiv -3\pmod{p}$$
If I could know that which prime $p$ suffices the congruence $ax^2+b\equiv 0\pmod{p}$ where $a\not\equiv 0\pmod{p}$ (in this case, $2y^2\equiv -3\pmod{p}$), I can prove the claim. 
Could you give me any idea? 

There are odd primes $p$ such that the congruences $x^2\equiv 2\pmod{p}$ and $2y^2\equiv -3\pmod{p}$ have solutions. For example, $p=7$ works ($x=y=3$).
OP has asked for which (odd) primes $p$ does the congruence $2y^2\equiv -3\pmod{p}$ have a solution.  It is not difficult to show that this  congruence has a solution if and only if the congruence $z^2\equiv -6\pmod{p}$ has a solution.
Let $p\gt 3$. We want to find the primes such that the Legendre symbol $(-6/p)$ is equal to $1$.  Since $(-6/p)=(2/p)(-3/p)$, there are two cases to consider, $(2/p)=1$ and $(2/p)=-1$.
In the first case, we want $(-3/p)=1$, and in the second case we want $(-3/p)=-1$.
We deal with the first case only. The analysis of the second case is very similar.
Because $(2/p)=1$, we want $p\equiv \pm 1\pmod{8}$.  We have $(-3/p)=1$ if and only if $p\equiv 1\pmod{6}$, that is, $p\equiv 1$ or $7$ or $13$ or $19$ modulo $24$. The condition $p\equiv \pm 1\pmod{8}$ rule out the last two. So the first case holds if $p\equiv 1\pmod{24}$ or $p\equiv 7\pmod{24}$.
Remark: Finding for which $p$ we have $(-3/p)=1$ is straightforward. We divide into $p\equiv 1\pmod{4}$, which makes $(-1/p)=1$, and $p\equiv 3\pmod{4}$, which makes $(-1/p)=-1$.
For the case $p\equiv 1\pmod{4}$, we get $(3/p)=(p/3)$, so we want $p\equiv 1\pmod{3}$. In the case $p\equiv 3\pmod{4}$, we have $(3/p)=-(p/3)$, so again we end up wanting $p\equiv 1\pmod{3}$.
