How to prove that $a+b$ is a multiple of $24$? Let $x$ be an integer one less than a multiple of $24$. Prove that if $a$ and $b$ are positive integers such that $ab=x$, then $a+b$ is a multiple of $24$.

So $ab \equiv -1 \pmod {24} \implies \gcd(a,24) = \gcd(b,24) = 1$. Then let $ab = 24k-1$ so $b = \frac{24k-1}a$ and $a+b = \frac{24k-1+a^2}a$. Since $\gcd(a,24) = 1$, it follows that $a^2 \equiv 1 \pmod {24}$ and we are done. 
