Number Theory problem

Question

Prove that if $n \equiv 23 \pmod{24}$ then $\sigma (n) \equiv0\pmod{24}$.

Answer 1

Note that $n\equiv -1 \pmod 3$. This implies that $n$ can't be a perfect square.

If $n=d_1d_2$ then one of the factors must be $1\pmod 3$ and the other must be $-1\pmod 3$ which implies that $$d_1+d_2\equiv 0 \pmod 3$$

This implies that $$\sigma(n)\equiv 0 \pmod 3$$

Note too that $n\equiv -1 \pmod 8$ We might have $n\equiv 1\times -1\pmod 8$ or $n\equiv 3\times 5 \pmod 8$ but either way we deduce that$$d_1+d_2\equiv 0 \pmod 8$$

This implies that $$\sigma(n)\equiv 0 \pmod 8$$

Inspection shows that we are done.