Show that $3$ divides $p-2$ if and only if $3$ divides $p+1$, where $p$ ia a prime different from $3$..

Question

I can observe this but unable to prove this result. If we start by taking the values of $p$ when $3|p+1$ and $p\neq 3$. We get the values of such $p$'s: $5, 11, 17, 23, 41, 47, 53, 59, 71, 83, 89, \cdots $

In each case $3$ divides $p-2$. It also holds in the converse sense.

Please help me to prove this. Thank you.

Answer 1

As $p+1-(p-2)=3\iff p+1=3+p-2$ or $p-2=p+1-3$

So, if $3|(p-2)\iff 3|(p+1)$

Answer 2

The result follows from $$p-2 \equiv 0\bmod 3 \Leftrightarrow p+1\equiv 0\bmod 3 $$