Question

Prove that any prime of the form 3m+1 for some m in the set of natural numbers is also of the form 6n+1 for some n in the set of natural numbers

Answer 1

if it is prime of the form
\[3m+1\] then it is odd, so it is also of the form
\[2k+1\]

Answer 2

and then 2k+1 = 6m.... ? would that be the next part?

Answer 3

does that help? because it tells you that
\[3m+1=2k+1\] so
\[3m=2k\]

Answer 4

say basically that m is even

Answer 5

if you want to be more formal say that since
\[3m=2k\] you must have
\[2| m\] since clearly 2 does not divide 3. therefore
\[m=2n\] for some n and so
\[3m+1=6n+1\]

Answer 6

think that is as formal as you can be in this problem

Answer 7

Oh ok, I see. Thank you, I always have trouble knowing how to start a proof.