Prove that if $\gcd( a, b ) = 1$ then $\gcd( ac, b ) = \gcd( c, b ) $

Setup: Let $d_1 = \gcd(c,b)$ and $d_2 = \gcd(ac,b)$.

So we have $cx_1 + by_1 = d_1$, $acx_2 + by_2 = d_2$, and $ax + by = 1$ by Bezout.

Step 1: Multiply $ax+by = 1$ by $d_1$ (using $d_1 := cx_1 + by_1$) and rearrange to show that $d_2|d_1$.

$$\begin{aligned} d_1 (ax+by &= 1)\\ \implies ax(cx_1 + by_1) + bd_1y &= d_1 \\ \implies a c (x x_1) + b(a x y_1 + d_1 y) &= d_1 \end{aligned}$$

Since we know that $d_2 = \gcd(ac,b)$ divides any integer linear combination of $ac$ and $b$, we have $d_2 | d_1$.

Step 2: By a similar argument, multiply $ax+by = 1$ by $d_2$ (using $d_2 := acx_2 + by_2$) and rearrange to show that $d_1|d_2$.

$$\begin{aligned} d_2 (ax+by &= 1) \\ \implies ax(acx_2 + by_2) + bd_2y &= d_2 \\ \implies c (a^2 x x_2) + b(a x y_2 + d_2 y) &= d_2 \end{aligned}$$

Since we know that $d_1 = \gcd(c,b)$ divides any integer linear combination of $c$ and $b$, we have $d_1 | d_2$.

Conclusion: Finally, because we have $d_1 | d_2$, $d_2 | d_1$, and $d_1$ and $d_2$ are non-negative (since they are the gcd of two integers), we conclude that $d_1 = d_2$. Thus, $gcd(ac,b)=gcd(c,b)$.

This form of Euclid's Lemma follows easily from basic laws of GCD arithmetic. First I will present the proof using the standard notation $\rm\: (a,b)\:$ for $\rm\: gcd(a,b),\: $ immediately followed by a proof employing a more suggestive arithmetical notation, denoting $\rm\:\gcd(a,b)\:$ by $\rm\ a \dot+ b\:.\:$ Because the arithmetic of GCDs shares many of the same basic laws of the arithmetic of integers, the proof becomes more intuitive using a notation that highlights this common arithmetical structure.

The proof below uses only said basic GCD laws: the associative law, $ $ the commutative law, the distributive law $\rm\, (a,b)c = (ac,bc)\, $ and the GCD-specific law $\rm\: \color{#C00}{(c,1) = 1}.\: $

Lemma $\rm \ \ (ac,b) = ((a,b)c,b)\,\ [=\, (c,b)\ \ {\bf if}\ \ (a,b)=1]$

$\begin{align}\rm {\bf Proof}\ \ \ ((a,\,b)c,\ b) &\rm =\, (ac,\,bc,\,b) = (ac,\ b\color{#c00}{(c,\ 1)}) = (ac,b)\\[.3em] \rm (a\dot+b)c\dot+b\ \:\! &\rm =\,\ ac\dot+bc\dot+b\, = \ \, ac\dot+b\color{#c00}{(c\dot+1)}\ =\ ac\dot+b \end{align}$

Notice how the suggestive notation in the second proof invites us to exploit our well-honed arithmetical intuition regarding the associative, commutative and distributive laws of integer arithmetic during analogous GCD arithmetic proofs. $ $ For a less trivial example see a similar proof of the Freshman's Dream $\rm\ (A,B)^n\, =\ (A^n,B^n)\ $ for GCDs and cancellable ideals.

The motivation behind this powerful abstract axiomatic approach becomes much clearer should you go on to study ideal theory and/or divisor theory (and you should, for there lies much beauty).

For further discussion and generalizations see this post and see my menagerie of posts on GCDs.

My attempt at a Bill Dubuquesque argument: \begin{align*} r|ac,\;b &\Longleftrightarrow r|ac,\;bc,\;b\\ &\Longleftrightarrow r|\gcd(ac,bc),\;b\\ &\Longleftrightarrow r|\gcd(a,b)c,\;b &\quad&\mbox{(since $\gcd(ac,bc)=\gcd(a,b)c$)}\\ &\Longleftrightarrow r|c,\;b \end{align*}