<s>[INST] Using quadratic residue show that $2b^2+ 3$ cannot be a divisor of $a^2-2$
I have tried working with residue of $2$ modulo $p$ where $p$ is a divisor of $2b^2+3$ but couldn't stablish the result.
 [/INST] You have the right idea. Consider there are integers $a$ and $b$ with $2b^2 + 3 \mid a^2 - 2$. With $2b^2 + 3$, it's always odd and $\gt 1$, so it has at least one prime factor and all of these prime factors are odd. Let $p$ be any of these prime factors. Since $p \mid 2b^2 + 3$, then $p \mid a^2 - 2$. Thus, as you stated, $2$ must be a quadratic residue modulo $p$. The table in the Law of quadratic reciprocity section shows $2$ is a quadratic residue mod $p$ if and only if
$$p \equiv 1, 7 \pmod{8} \tag{1}\label{eq1A}$$
Note since $7 \equiv -1 \pmod{8}$, the product of any number of these factors will only be congruent to either $1$ or $7$ mod $8$.
However, if $b$ is even, then $2b^2$ has a factor of $2 \times 2^2 = 8$, so $2b^2 + 3 \equiv 3 \pmod{8}$. Alternatively, if $b$ is odd, then $b^2 \equiv 1 \pmod{8}$, so $2b^2 + 3 \equiv 2 + 3 \equiv 5 \pmod{8}$. In both cases, it's not congruent to $1$ or $7$ mod $8$, meaning $2b^2 + 3$ must have at least one odd prime factor for which $2$ is not a quadratic residue. Thus, this means $p \not\mid a^2 - 2$, so
$$2b^2 + 3 \not\mid a^2 - 2 \tag{2}\label{eq2A}$$
</s>