Here is similar solution to @neyft's, but more precise. If $p=2$, then take $b$ even and we are done. From now assume that $p$ is odd prime. Assume that $gcd(a,p)=gcd(b,p)=1$. We will use the following $2$ facts related to quadratic residues. Fact 1: $(\frac{a}{p})(\frac{b}{p})=(\frac{ab}{p})$ Fact 2: If $a \equiv b \pmod p$, then $(\frac{a}{p})=(\frac{b}{p})$ Since $m^2 -2 $ is divisible by $p$, then $m^2 \equiv 2 \pmod p$, which implies that $(\frac{2}{p})=1$. Similarly we get that $(\frac{2-m}{p})=1$. Using fact $1$ we get that: $$(\frac{2-m}{p})(\frac{2+m}{p}) = (\frac{4 - m^2}{p}) =(\frac{2+m}{p})$$Note that $m^2 \equiv 2 \pmod p$, which implies that $4-m^2 \equiv 2 \pmod p$. Using fact $2$ we get that: $$ (\frac{2+m}{p})=(\frac{4 - m^2}{p}) =(\frac{2}{p})=1$$Since we obtained that $(\frac{2+m}{p}) = 1$, we conclude that there exists positive integer $b$ such that $b^2 - m -2$ is divisible by $p$