I tried to start by showing that $ \frac{a}{\gcd(a,b)} $ is always an integer, let's call it $d$, because $a$ is always a multiple of $(a,b)$ based on the definition of a g.c.d. I then tried to show that if $a | b$ then $a|bc$ from the theorem that states that $a|b$ implies that $a|bc$ for any integer $c$, but I cant seem to prove that $a | b$ or that $a|bc$ only if $d \mid c$ (where $d = \frac{a}{\gcd(a,b)}$ ). Any help would be much appreciated.


Note that $x|y$ if and only if $(x,y) = x$; then use the properties of the gcd.

\begin{align*} a|bc &\Longleftrightarrow (a,bc) = a\\\ &\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{bc}{(a,b)}\right) = \frac{a}{(a,b)}\\\ &\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{b}{(a,b)}c\right) = \frac{a}{(a,b)}\\\ &\Longleftrightarrow \left(\frac{a}{(a,b)}, c\right) = \frac{a}{(a,b)}&&\mbox{(by Euclid's Lemma)}\\\ &\Longleftrightarrow \frac{a}{(a,b)} \Bigm| c, \end{align*} using Euclid's Lemma ("If $(x,y)=1$, then $x|yz\Leftrightarrow x|z$") and $$1 = \frac{1}{(a,b)}(a,b) = \left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right).$$

The manipulations are consequences of $x(y,z) = (xy,xz)$; see Bill Dubuque's proof here.

  • $\begingroup$ How did you get rid of $\frac{b}{(a,b)}$ in between the third and fourth line? $\endgroup$ – Katie Barosie Feb 2 '11 at 19:24
  • $\begingroup$ @Katie: $(\frac{a}{(a,b)},\frac{b}{(a,b)}) = 1$; so the down-implication follows from Euclid's Lemma mentioned later. $\endgroup$ – Arturo Magidin Feb 2 '11 at 19:31

It is easy: $\rm\ \ a\: |\: bc\: \iff\: a\: |\: ac,bc\: \iff\: a\: |\: (ac,bc) = (a,b)\:c\: \iff\: a/(a,b)\: |\: c\quad$ QED

Or, dually: $\rm\ a\: |\: bc\: \iff\: a,b\: |\: bc\: \iff\: [a,b]\: |\: bc\: \iff\: [a.b]/b\ |\ c\:,\ $ $\rm\ [m,n]\: :=\ lcm(m,n)$

Combining the two yields the GCD $\cdot$ LCM law: $\rm\ [a,b]/b\: =\: a/(a,b)\:,\ $ i.e. $\rm\ (a,b)\ [a,b] =\: ab\:.$

Alternatively, cancelling $\rm\ (a,b)\ $ reduces it to Euclid's Lemma $\rm\ (a,b)= 1,\ a\:|\:bc\ \Rightarrow\ a|c\:.$

Above we used the $\: $ GCD $\: $ distributive law $\rm\: \ (ac,bc)\: =\: (a,b)\,c,\,$ proved here.

More generally the result is the special case $\rm\ a\:|\:bc\ $ of $\rm\ (a,bc) = (a,(a,b)\:c)\:,\:$ which holds true for both GCDs and ideals. As you can see these basic properties are all intimately connected, so it's useful to master these properties to become proficient with GCD arithmetic. For another example of applying these properties to obtain simple proofs see this proof of the Freshman's Dream $\rm\ (A,B)^n = (A^n,B^n)\ $ for GCDs and invertible ideals.


$$a\mid bc \Longleftrightarrow \frac{a}{(a,b)}\mid c$$



$a\mid bc\Rightarrow\exists k\in\mathbb{N}$ such that $bc=ak$$$bc=ak\Rightarrow\frac{bc}{(a,b)}=\frac{ak}{(a,b)}\Rightarrow\frac{b}{(a,b)}c=\frac{a}{(a,b)}k\Longrightarrow$$$$\frac{a}{(a,b)}\mid\frac{b}{(a,b)}c$$

Two theorems that we will use $$\text{If}\;\;x\mid yz\;\;\text{and}\;\;(x,y)=1\Longrightarrow x\mid z\;\;\;\;(\text{Theorem}\;*)\\\text{and}\\(\frac{x}{(x,y)},\frac{y}{(x,y)})=1\;\;\;\;(\text{Theorem}\;**)$$

Returning to the question, we have $$\frac{a}{(a,b)}\mid\frac{b}{(a,b)}c$$ by theorems $*$ and $**$ we have $$\frac{a}{(a,b)}\mid c\;\;\;\;\;\Box$$


$$\frac{a}{(a,b)}\mid c\Longrightarrow a\mid bc$$

$$\frac{a}{(a,b)}\mid c\Longrightarrow\exists k'\in\mathbb{N}\;\;\text{such that}\;\; c=\frac{a}{(a,b)}\cdot k'\\bc=\frac{ab}{(a,b)}\cdot k'\Longrightarrow bc=a(\frac{b}{(a,b)}\cdot k')\Longrightarrow a\mid bc\;\;\;\;\;\Box$$


Here is a modification and simplification to @marcelolpjunior answer with divisibility and gcd rules.

To prove such statement, we need these three propositions:

  • $$a|bc \Longrightarrow \frac{a}{(a,b)}|\frac{bc}{(a,b)} \hspace{5cm} (1)$$
  • $$ (ca,cb) = c(a,b) \hspace{6cm} (2)$$
  • $$a|b, a|c \Longrightarrow a|(b,c) \hspace{5.5cm} (3)$$
  • $$(\frac{a}{(a,b)}, \frac{b}{(a,b)}) = 1 \hspace{5.4cm} (4)$$

since by $(3),(4):$ $$(\frac{a}{(a,b)},\frac{b}{(a,b)}) = 1 \Longrightarrow (\frac{ac}{(a,b)},\frac{bc}{(a,b)}) = c$$ Clearly, $\frac{a}{(a,b)}|\frac{ac}{(a,b)}$ and , from the hypothesis and $(1),\frac{a}{(a,b)}|\frac{bc}{(a,b)}$. Thereby, by $(3),\frac{a}{(b,c)}|c$ as required.

For the converse:

since $\frac{a}{(a,b)}|c \Longrightarrow a|(a,b)c$. But $(a,b)|b$ Thus, $a|bc$ as required.


This might not qualify as a formal proof, but might have some pedagogical value.

We have the prime factorizations

$b = b_1^{n_1} b_2^{n_2} b_3^{n_3} \dots b_j^{n_j}$

$c = c_1^{m_1} c_2^{m_2} c_3^{m_3} \dots c_k^{m_k}$

It is useful to put $b \times c$ in the numerator and $a$ in the denominator of the integer,

$$\frac{b_1^{n_1} b_2^{n_2} b_3^{n_3} \dots b_j^{n_j} \times c_1^{m_1} c_2^{m_2} c_3^{m_3} \dots c_k^{m_k}}{a}$$

If we 'peel off' any prime factor of $a$ all we know for sure is it can be 'put under' some $b_h$ or $c_i$, and perhaps both when the prime number goes into $a$, $b$ and $c$.

We have an algorithm that takes one prime factor at time 'off of' $a$ and, if the prime factor is part of $b$ and if there is still 'room left', puts it in the denominator slot under the matching $b_j^{h_j}$ while increasing the exponent in that slot. If that doesn't work, the prime factor has to fit under $c$, the right side.

When the algorithm completes, the denominator looks like $(a,b) \times \frac{a}{(a,b)}$.


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