Does $8a+5$ ever divide $b^2+8$? For natural $a,b$, does $8a+5$ ever divide $b^2+8$ ?
It doesn't for $b$ up to $10^7$.
Couldn't find congruence obstructions for moduli up to $500$.
$b^2+8$ can be even.

Using the fact that $\mathbb{Z}[\sqrt{-2}]$ has unique factorisation, we can write:
$$ 8a + 5 \; \; | \; \; (b + 2 \sqrt{-2})(b - 2 \sqrt{-2}) $$
Now, the $\gcd$ of $b \pm 2 \sqrt{-2}$ divides $4\sqrt{-2}$. On the other hand, $8a + 5$ is not divisible by $\sqrt{-2}$, so any prime divisor of $8a + 5$ cannot divide both $b \pm 2 \sqrt{-2}$.
Consequently, $8a + 5$ has no real prime divisors, so we can write:
$$ 8a + 5 = b^2 + 2c^2 $$
Working modulo $8$, this is an immediate contradiction.
