Suppose p is a prime in the progression 3n+1, n=1, 2, 3... Then there

exists a positive integer k such that:

p=3k+1.

Now if p>=2, then p is odd as the only even prime is 2. Therefore p-1 is

even, so 3k is even, so k is even, so there exists an integer m>0 such that:

p-1=3.2.m=6m,

so for some m>0, p=6m+1, therefore p is in the progression 6n+1, n=1, 2, ..

RonL