Originally Posted by

**minivan15** yes, I actually was working on it more and reached that conclusion myself

where I went from there was to do my best to narrow down the choices, I said:

let gcd(m + n, m + 7n) = d

as you said, d|6n

gcd(d,6) = 1 or gcd (d,6) does not equal 1

If gcd (d,6) = 1 then by Euclid's Lemma d|n

and if d|n then d|(m + n) iff d|m, but if d|m then d would divide both m and n, contradicting the assumption that gcd(m,n) = 1 for d > 1

so either d = 1 or gcd(d,6) > 1

what I was working on next was to look at the options when gcd (d,6) > 1. I started attempting to show that if d > 6 and gcd(d,6) > 1 then d must be a multiple of b, which seemed simple at first although the more I thought about it the less sure I became...essentially now I am stuck at this part, where should I go from here?