Can anyone please review/verify this proof of assertion?

[Math100]

Homework Statement

Prove the assertion below:
Any prime of the form 3n+1 is also of the form 6m+1.

Relevant Equations

None.

Proof: Suppose that any prime of the form 3n+1
is also of the form 6m+1.
Note that 2 is the only even prime number
and it is not of the form 3n+1.
This means any prime of the form 3n+1 must be odd.
Since 3n+1 is odd, it follows that 3n must be even.
Then we have n=2m for some integer m.
Thus 3n+1=3(2m)+1
=6m+1.
Therefore, any prime of the form 3n+1 is also of the form 6m+1.

Above is my proof for this assertion. Can anyone please review/verify to see if it's correct?

 

[fresh_42]

It is correct. Nothing to complain about.

You can also write it in formulas, after you excluded $p=2$ as you did:
$p=3n-1 \Longrightarrow 3\,|\,(p-1)$ and, as you correctly observed, $p$ is odd, so $p-1$ is even, i.e.
$2\,|\,(p-1).$ Because $2$ and $3$ are coprime (no common divisor), we get $2\cdot 3\,|\,(p-1)$ which is $6m=p-1$ or $p=6m+1$.

 

[Math100]

It is correct. Nothing to complain about.

You can also write it in formulas, after you excluded $p=2$ as you did:
$p=3n-1 \Longrightarrow 3\,|\,(p-1)$ and, as you correctly observed, $p$ is odd, so $p-1$ is even, i.e.
$2\,|\,(p-1).$ Because $2$ and $3$ are coprime (no common divisor), we get $2\cdot 3\,|\,(p-1)$ which is $6m=p-1$ or $p=6m+1$.

Thank you!

 

[fresh_42]

Thank you!

Note that coprime is important here! If two numbers have a common divisor, say $4$ and $6$, then both divide $12$ but $4\cdot 6$ does not!

 

[FactChecker]

Your logic is correct, but I would recommend a little rewording.

Homework Statement:: Prove the assertion below:
Any prime of the form 3n+1 is also of the form 6m+1.
Relevant Equations:: None.

Proof: Suppose that any prime of the form 3n+1
is also of the form 6m+1.

You do not want to "suppose" this. That would be assuming the fact that you want to prove.
You should start with something like:
Proof: Suppose we have a prime, p, of the form p=3n+1.

Note that 2 is the only even prime number
and it is not of the form 3n+1.

So p is not 2.

This means any prime of the form 3n+1 must be odd.
Since 3n+1 is odd, it follows that 3n must be even.
Then we have n=2m for some integer m.
Thus 3n+1=3(2m)+1
=6m+1.

p = 6m+1.
$\blacksquare$

 

[Math100]

Your logic is correct, but I would recommend a little rewording.

You do not want to "suppose" this. That would be assuming the fact that you want to prove.
You should start with something like:
Proof: Suppose we have a prime, p, of the form p=3n+1.

So p is not 2.

p = 6m+1.
$\blacksquare$

Thank you!