Let $\gcd(b,6) = 1$. Prove that $b^2 \equiv 1 $ (mod 24).
Now I have that as $\gcd(b,6) = 1$, we know that $3\nmid b $ and $2\nmid b$ (else the GCD would be 3 or 2 resp.)
So as $2\nmid b$, $b$ must be odd. Hence $b^2$ is also odd.
Then I'm not sure where to take it from here? Maybe I've gone down the wrong path to start with, and should be looking more along the lines of the Chinese Remainder Theorem? I'm not sure.