Let p be a prime number greater than or equal to 3. Show that if p ≡ 1 (mod 3), then p ≡ 1 (mod 6).

Question

I'm not sure how p being a prime number and greater than or equal to three plays into the proof.

Answer 1

If is a prime then it can’t be multiple of 2. Or even. Given that remainder is 1 when dividing by 3, then quotient is even. Hope this helps.

Answer 2

If $p\equiv1\pmod3$, then $p\equiv1\pmod6$ or $p\equiv4\pmod6$.

But $p\equiv4\pmod6$ won't work, because that would mean $2\mid p$,

($6\mid p-4\implies 2\mid p-4\implies 2\mid p$),

so we're done.