$\ ac≡bc\pmod{\! m}\!\iff\! a≡b\pmod {\!m/d},\ d = \gcd(c,m)\ $ [Congruence Cancellation & Division Rule]

Question

How would you show that if $ac≡bc$ $\mod m$ and $\gcd(c,m)=d$, then $a≡b$ $\mod \frac{m}{d}$?

Any help would be much appreciated!

Answer 1

You have marked this under proof writing. So I will attempt a model proof.

So $ac \equiv bc \pmod{m}$, which means that $m \mid ac-bc$, and hence $m \mid c(a-b)$. Hence, $c(a-b)=km$ for some integer $k$. Now, Given that $\gcd(c,m) =d $, it follows that $c=xd$ and $m=yd$ for some co-prime integers $x$ and $y$. Hence, $$ c(a-b) = km \implies xd(a-b) = kyd \implies x(a-b) = ky $$ Note that $x \mid ky$ from the above, and recall that $x$ is co-prime to $y$. We claim that $x$ divides $k$. (Thanks to @curioushuman for asking for the details).

Indeed, since $x$ divides $ky$, suppose that $xz = ky$ for some integer $z$. By Bezout's theorem, as $x$ is coprime to $y$ there exist $a,b$ integers such that $ax+by=1$. Multiply by $k$ to get $akx+bky=k$, then note that $ky=xz$, so $akx+bzx = k$, the same as $(ak+bz)x = k$. Thus, $x$ divides $k$ since $ak+bz$ is an integer.

Thus, $\frac{k}{x}$ is an integer. Now, $$ (a-b) = \frac{k}{x}y $$ and hence $a-b$ is a multiple of $y$. Hence $a \equiv b \pmod{y}$. But $y = \frac{m}{d}$, hence $a \equiv b \pmod{\dfrac{m}{d}}$.

Answer 2

Assuming integers for all variables, and $c' = c/d$, $m' = m/d$ :

$$\begin{align} ac \equiv bc \pmod m &\iff \exists k ~~ ac - bc = km \\ % &\iff \exists k ~~ ac'd - bc'd \equiv km \\ % &\iff \exists k ~~ a - b \equiv (k/c')(m/d) \end{align}$$

So we have to establish that $k / c'$ is an integer. From $\gcd(c, m) = d$ we can infer $\gcd(c', m') = 1$, so

$$ac - bc = km$$ $$ac'd - bc'd \equiv k(m'd)$$ $$c'(a - b) = km'$$

So $c'$ divides $k$, so $k/c'$ is an integer.

$$\exists j ~ a - b = j(m/d)$$ $$a \equiv b \pmod {m/d}$$

Answer 3

By definition $\ ac\equiv bc\pmod{\! m}\!\iff\! m\mid \color{#0a0}c\:\!(\overbrace{a\!-\!b}^{\textstyle \color{#0a0}n})\!\iff m/(m,c)\mid a\!-\!b,\,$ by

Theorem $ $ (general Euclid's Lemma) $\ \ \ m\mid \color{#0a0}{cn}\iff m/(m,c)\mid n \ \ $

Remark $ $ As is explained carefully here, in fraction language this can be written as

$$ ac\equiv bc\!\!\!\!\pmod{\!m}\iff a\equiv \dfrac{bc}c\!\!\!\!\pmod{\!m}\equiv\!\!\!\!\!\!\!\!\underbrace{\dfrac{bc/\color{#c00}d}{c/\color{#c00}d}\equiv b\!\!\!\pmod{\!m/\color{#c00}d}}_{\textstyle {\rm cancel}\ \color{#c00}d\!=\!(c,m)\,\ \color{#c00}{\rm everywhere}}\qquad\qquad$$

The common divisor $\,\color{#c00}d = (c,m)\,$ must be cancelled $\:\!\rm\color{#c00}{everywhere}\:\!$ (top, bottom and modulus), leaving the new denominator $\,c/d\,$ coprime to the new modulus $\,m/d,\,$ so invertible, so cancellable. Generally, the modular fraction $\,e/c\bmod m\,$ exists $\!\iff (c,m)\mid e,\,$ true here by: $\,d\!=\!(c,m)\mid c\mid bc.\,$ However, when $d>1$ the modular fraction is multi-valued $\!\bmod m,\,$ but single-valued $\!\bmod m/d,\,$ see the prior linked post for details.

Answer 4

Alternatively, in two steps: $ $ first cancel $\,d = (c,m)\,$ $\rm\color{#c00}{everywhere}$ to reduce to the case of $\rm\color{#0a0}{coprime}$ $\,\bar c,\bar m = c/d, m/d,\,$ then cancel $\,\bar c\,$ by scaling by $\,\bar c^{-1}$ (exists $\!\bmod \bar m$ by $\,\bar c\,$ $\rm\color{#0a0}{coprime}$ to $\,\bar m)$

$\qquad\quad\ \ \begin{align} ac &\equiv bc\!\!\pmod{m}\\[.2em] \smash[t]{\overset{\bf\color{#0af}{L1}\!}\iff}\ \ \ a\bar c&\equiv b\bar c\!\!\!\pmod{\bar m}\ \ \ {\rm via\ cancel\ } d = (c,m)\ \,\&\,\ \rm {\bf\color{#0af}{L1}}\ below\\[.2em] \iff\ \ \ \ a &\equiv b\pmod{\bar m}\ {\rm\ \ by\ scaling\ by\ }\, \bar c^{-1} \end{align}$

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$\bf\color{#0af}{L1}$ $ $ If $\, d\mid c,m\!:\ ac\equiv bc\pmod{\!m}\!\!\iff\! {ac/\color{#c00}d \equiv bc/\color{#c00}d\pmod{\!m/\color{#c00}d}}\ $ [cancel $\,d\ \rm\color{#c00}{everywhere}$]

$\!\begin{align}{\bf Proof}\quad \exists\:\! k\!:\quad\ ac &= bc + km\\[.2em] \iff\ \exists\:\! k\!:\ ac/d &= bc/d \,+\, k\:m/d,\ \ \text{via cancel $\,d$}\\[.2em] \iff\ \qquad ac/d &\equiv bc/d\!\!\!\pmod{\!m/d} \\[.2em] {\rm i.e.} \ \ \ \ \ a\bar c\, &\equiv\, b\bar c\,\pmod{\!\bar m}\end{align}$

Remark $\ $ The (two-step) cancellation has a natural view as modular fraction reduction as explained in the remark in my other answer here. The first step reducing to coprime $\bar c,\bar m\,$ is a special case of homogeneous reduction as explained here.

Answer 5

By definition, $ac\equiv bc \pmod{m}$ means that there exists a $k\in\mathbb{Z}$ such that $ac=bc+km$. If $d=\gcd(c,m)$, let $c=sd$ and $m=td$ (note now that $\gcd(s,t)=1$). This means that $$ \begin{aligned} ac &\equiv bc \pmod{m} \Longleftrightarrow \\ ac &= bc+km \Longleftrightarrow \\ asd &= bsd+ktd \Longleftrightarrow \\ as &= bs+kt, \end{aligned} $$ so $as\equiv bs \pmod{t}$, or in other words $as\equiv bs \pmod{m/d}$. But since $\gcd(s,t)=1$ we can conclude that $a\equiv b \pmod{m/d}$, and we are done.

Answer 6

Theorem.

$$\forall a, b, c \in \mathbf{Z}, n \in \mathbf{Z}^*, \quad ac \equiv bc \pmod n \iff a \equiv b \pmod{\tfrac{n}{\gcd(c, n)}}.$$

Proof. Let $a, b, c \in \mathbf{Z}, n \in \mathbf{Z}^*$.

$$ \begin{align} ac \equiv bc \pmod n & \iff \exists k \in \mathbf{Z}, ac - bc = kn \\ & \iff \exists k \in \mathbf{Z}, (a - b)c = kn \\ & \iff \exists k \in \mathbf{Z}, (a - b) \frac{c}{\gcd(c, n)} = k\frac{n}{\gcd(c, n)}. \end{align} $$

Therefore, $\frac{c}{\gcd(c, n)}$ divides $k\frac{n}{\gcd(c, n)}$, and since $\frac{c}{\gcd(c, n)}$ is coprime to $\frac{n}{\gcd(c, n)}$, it follows from Euclid’s lemma that $\frac{c}{\gcd(c, n)}$ divides $k$, hence

$$ \begin{align} ac \equiv bc \pmod n & \iff \exists k \in \mathbf{Z}, a - b = \frac{k}{\frac{c}{\gcd(c, n)}} \frac{n}{\gcd(c, n)} \\ & \iff \exists l \in \mathbf{Z}, a - b = l \frac{n}{\gcd(c, n)} \\ & \iff a \equiv b \pmod{\tfrac{n}{\gcd(c, n)}}. & \blacksquare \end{align} $$