Let p be a prime number, show that if p ≡ 1 mod 3, then p ≡ 1 mod 6

Question

I have the following assignment: Let $p$ be a prime number, show that if $p\equiv_{3}1$, then $p\equiv_{6}1$

I am having trouble making an approach to the solution.I appreciate any help. I can´t make the modulo to change to show the equivalence. I only proved that $p^2=3L+1$, equivalent to $p^2\equiv_{3}1$.

Thank you in advance.

Answer 1

Let $p$ be a prime with $p\equiv 1\pmod 3$. Suppose toward a contradiction that $p\not\equiv 1\pmod 6$. Since $p\equiv 1\pmod 3$ we must then have that $p\equiv 4\pmod 6$. But then $p$ is even and since $p$ is prime $p = 2$. But $2\not\equiv 1\pmod 3$, a contradiction.

Answer 2

Hint: $ $ it's easy as $\,a,m,n = 1,2,3\,$ below, for $\,p\,$ any integer.

CCRT $ $ For coprime $\,m,n\!:\,\ p\equiv a\pmod{\!mn}\iff \begin{align}&p\equiv a\!\!\pmod{\!m}\\ &p\equiv a\!\!\pmod{\!n}\end{align}$

since $\, mn\mid p\!-\!a\iff m,n\mid p\!-\!a\,$ by $\,m,n\,$ coprime (so lcm = product, or by Euclid's Lemma)

Remark $\ $ This is CCRT = Constant case of CRT = Chinese Remainder Theorem - a ubiquitous result well worth commiting to memory. You can find a handful of complete proofs in the linked answer if the above hint doesn't suffice.