If $m | (8n + 7)$, $m | (6n + 5)$,prove that $m = ± 1$
-We have just starting going over the "divides" notation, and I am aware of a few properties and theorems from my notes. I am; although, a bit lost as to how to properly set up this proof. Any help is appreciated.
So far I have:
$$3(8n + 7) - 4(6n + 5) = (24n + 21) - (24n + 20)$$
This equals 21 - 20 = 1, although I don't know how to prove for ± 1 (is it because the problem doesn't define $m \ge 1$?)
If $m\mid 8n+7$ and $m\mid 6n+5$, then $$m\mid 3(8n+7)-4(6n+5)=1,$$
so $1=mk$ for some $k\in\mathbb Z$, so $m=\pm 1$.