Show that if p is a prime in the arithmetic progression 3n + 1, n=1, 2, 3,..., then it is also in the arithmetic progression 6n + 1, n=1, 2, 3,......
Thanks very much for your help!!!
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