Prove that for any integer $n > 3$ If $n$ is a prime number then $n$ $\in$ $[1]_6$ or $n$ $\in$ $[5]_6$

Question

I am not sure how to proceed with the proof. That is, I’m trying to prove the contapostive, but I don’t know how to prove that a composite number greater than 3 will not be in either set. Insight would be appreciated.

Answer 1

" I don’t know how to prove that a composite number greater than 3 will not be in either set. "

That isn't the contrapositive and it isn't true.

The contra positive of: If $p$ is prime and $p > 3$ $\implies$ $p\in[1]_6$ or $p\in [5]_6$. then the contra positive is

$p\not \in [1]_6$ and $p\not \in [5]_6$ $\implies$ $p$ is not prime or $p \le 3$.

So prove that if $p\in [0]_6, [2]_6, [3]_6, [4]_6$ then eithe $p$ is composite of $p \le 3$.

.....

Might be worth noting. If $m \in [k]_6$ then $\gcd(k,6)|m$. Do you see why that would be true?

Answer 2

Well:

$$6 | 6n$$ $$2|(6n+2)$$ $$3|(6n+3)$$ $$2|(6n+4)$$