I recently learned a simple theorem about GCDs. It's intuitive and quick to prove. My question is whether I can generalize it in a way I will explain below. I did look on the search engines, but I couldn't find the right way to put the question.
Here's the theorem: If \(\displaystyle a|bc\) and \(\displaystyle (a,b) = 1\), then \(\displaystyle a|c\). Intuitively, if a and b have no factors in common, but a and bc do, then the common factor must be between a and c. That's not hard. The mathy proof uses Bezout's identity, which says that for some integers u and v, \(\displaystyle \left( {a,b} \right) = 1 \to au + bv = 1\). We multiply this by c to get \(\displaystyle acu + bcv = c\). But \(\displaystyle a|bc\), so that for some r we have \(\displaystyle bc = ar\). Therefore, \(\displaystyle c = acu + bcv + \left( {ar} \right)v = a\left( {cu + rv} \right) \to a|c\).
That's just a little review.
Here's my question. Still assuming that \(\displaystyle (a,b) = 1\), and given that \(\displaystyle am = bc\) where m is some integer, is it still true that \(\displaystyle a|c\)? Does that extra factor of m prevent a from dividing bc?
I cannot prove this, but I can't find a counterexample either. Suppose a = 3 and b = 7. Then I must find m and c such that \(\displaystyle 3m = 7c\). I can find many such pairs, but none where 3 does not divide c.
No, this is not for HW or any exam. I'm not taking a math class right now. But I'd really appreciate any help.
Here's the theorem: If \(\displaystyle a|bc\) and \(\displaystyle (a,b) = 1\), then \(\displaystyle a|c\). Intuitively, if a and b have no factors in common, but a and bc do, then the common factor must be between a and c. That's not hard. The mathy proof uses Bezout's identity, which says that for some integers u and v, \(\displaystyle \left( {a,b} \right) = 1 \to au + bv = 1\). We multiply this by c to get \(\displaystyle acu + bcv = c\). But \(\displaystyle a|bc\), so that for some r we have \(\displaystyle bc = ar\). Therefore, \(\displaystyle c = acu + bcv + \left( {ar} \right)v = a\left( {cu + rv} \right) \to a|c\).
That's just a little review.
Here's my question. Still assuming that \(\displaystyle (a,b) = 1\), and given that \(\displaystyle am = bc\) where m is some integer, is it still true that \(\displaystyle a|c\)? Does that extra factor of m prevent a from dividing bc?
I cannot prove this, but I can't find a counterexample either. Suppose a = 3 and b = 7. Then I must find m and c such that \(\displaystyle 3m = 7c\). I can find many such pairs, but none where 3 does not divide c.
No, this is not for HW or any exam. I'm not taking a math class right now. But I'd really appreciate any help.