Divisibility proof problem

Question

I need assistance with the following proof.

Let a,b,c,m be integers, with m $\geq$ 1. Let d = (a,m). Prove that m divides ab-ac if and only if $\frac md $ divides b-c.

Alright, I know that since d = (a,m) there exists an r and t such that $ar + mt = d$

I figure since we're trying to prove m divides ab-ac iff $\frac md$ divides $b-c$, we look to $\frac md$ dividing $b-c$

Only I get stuck when trying to work with $\frac md$ dividing $b-c$ algebraically.

I've tried relating d = (a,m) to $\frac md$ but I'm still stuck.

Answer 1

Let $k=b-c$. We want to show that $m$ divides $ak$ if and only if $\frac{m}{d}$ divides $k$.

Let $a=a_1d$ and let $m=m_1 d$. Note that $\gcd(a_1,m_1)=1$.

One direction: We show that if $\frac{m}{d}$ divides $k$, then $m$ divides $ak$. By assumption we have $k=\frac{m}{d}q=m_1q$ for some $q$. Thus $ak=(a_1 d)(m_1q)=(a_1q)m$, so $m$ divides $ak$.

The other direction: We show that if $m$ divides $ak$, the $\frac{m}{d}$ divides $k$.

We have $m_1d$ divides $(a_1d)k$, so $m_1$ divides $a_1k$. But since $d=\gcd(a,m)$, we have $\gcd(a_1,m_1)=1$. Then from $m_1$ divides $a_1k$ we can conclude that $m_1$ divides $k$.

Remark: We have used without proof the fact that if $u$ divides $vw$ and $u$ and $v$ are relatively prime, then $u$ divides $w$. That's because it has probably been proved in your course. For a proof, we can use the Bezout argument you began. There are integers $x$ and $y$ such that $ux+vy=1$. Multiply through by $w$. We get $u(xw)+(vw)y=w$. By assumption $u$ divides $vw$, so $u$ divides $u(xw)+(vw)y$, that is, $u$ divides $w$.

Answer 2

Hint: This comes down to the following: We have $m|xy$, if and only if $$\left.\frac{m}{\gcd(m,x)}\right|y$$ In words: every factor of $m$ that is not in $x$ must be in $y$ for $xy$ to be divisible by $m$.

Answer 3

$ m\mid a(b\!-\!c)\overset{{\rm cancel}\ d\,}\iff\! \frac{m}d\mid \frac{a}d(b\!-\!c)\overset{\rm Euclid}\iff \frac{m}d\mid b\!-\!c,\ $ since $\,\left(\frac{m}d,\frac{a}d\right) = 1,\ $ by $\ (m,a)=d.$