Is there something faulty about this statement? Show any prime of the form $3k+1$ is of the form $6k+1$. 
I came up with my own solution that made perfect sense to me, but when I read the text's solution, it argued that for the primes that are of the particular form are $6k+1 = 3(2k)+1$. But doesn't that really say the primes in the form of $3k+1$ are in the form of $6m+1$? It seems to me as though there's some misuse of notation here -- allowing $k = 2k$. So should the exercise be phrased as $6m+1$ instead? 
 A: I think it would have made more sense to pose the problem as:

"Show that any prime of the form $3n+1$ is of the form $6k + 1$" 

to distinguish the integers $n, k$, and allow for subsequently proving this is the case for $n = 2k$: primes of the form $3n + 1 = 3(2k) + 1$ are thus of the form $6k + 1$.
But just like indexing variables, I suspect that "$k$" as used in the actual problem statement was intended to be a "dummy" variable standing in for "some integer", much like $x$ in the expressions "$\forall x P(x)$, and $\forall x Q(x)$" each use $x$ independently of its use in the corresponding assertion. But this is not standard.
A: Another way to phrase it is
"If k is a positive integer such that 3k+1 is prime
then k is even".
The proof, of course, is easy:
If k is odd, then k=2h+1 for some integer h.
But 3k+1 = 3(2h+1)+1 = 6h+4 is even
and therefore not prime.
A: The only $k$ for which $3k+1=6k+1$ is $k=0$, which doesn't even give a prime. The statement means "if a prime is a multiple of three plus one, then it is a multiple of six plus one". The $k$ is a dummy variable; although it is not proper to us $k$ twice, the statement is not much clear if you use two different variables. 
A: $3n+1=6k+1$ iff $n=2k.$ If $n$ is even, this will always work, but if $n$ is odd, then $3n+1$ is even and thus not prime and there are no solutions and so the statement is true.
You can of course extend this to the rational numbers, and then it gives every prime, but that's not very interesting.
