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From the assumption I was able to gather the following: $bc= ak_1$. Let $p=gcd(a,b)$ thus $p=dk_2$. Well since $p|a$ and $p|b$ I have the following, $a= pr_1$ and $b=pr_2$. I have been trying to substitute one of the equations into the other and here's what I've done: I write $p$ as $b/r_2$. Well we know $p=dk_2$ so then I substitute for $p$, solve for $b$ and arrive at $b=dk_2r_2$. I then plug this value for $b$ into $bc= ak_1$. I end up with $dcr_2k_2 = ak_1$. Can I relate $r_2k_2$ to $k_1$ or have I gone the wrong way of proving this? Thank you for your time.

Anurag A
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ponderingdev
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    The result you are trying to prove is not true. There has to be another condition that you are missing. For example, take $a=6, b=2$ and $c=3$. Then $a | bc$ and $\gcd(a,b)=\gcd(6,2)=2$. Thus there are two possible values of $d$, namely $1,2$. With $d=1$, the conclusion is not satisfied. – Anurag A Mar 27 '15 at 02:44
  • i apologize i misread the question, it's let d = gcd(a,b) instead of divide gdc(a,b). maybe why i wasn't going anywhere... – ponderingdev Mar 27 '15 at 03:15

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As Anurag explains above, simply picking d=1 turns the statement into "let a,b,c be integers so that a|bc. Prove a|c" which is more clearly not always true.

You may want a stronger condition on d since it's used in the statement to be proved.

Nate 8
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  • i apologize i misread the question, it's let d = gcd(a,b) instead of divide gdc(a,b). maybe why i wasn't going anywhere... – ponderingdev Mar 27 '15 at 03:15
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with the modification $d=\gcd(a,b)$ you can prove it as follows: Since GCD is a linear combination therefore $\exists x,y \in \mathbb{Z}$ such that $$d=ax+by.$$ Now multiply this by $c$ to get $$cd=cax+bcy.$$ Since we are given that $a|bc$ (hence it divides $bcy$ as well) and for a fact $a|acx$, therefore $a | cd$.

Anurag A
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It is an immediate consequence of the gcd Distributive Law $\,\rm\color{#c00}{DL}$

${\bf Lemma}\ \ \ a\mid bc\iff a\mid (a,b)c$

$\smash[t]{{\bf Proof}\quad\ \ a\mid bc\iff a\mid ac,bc\iff a\mid (ac,bc)\overset{\rm\color{#c00}{DL}}=\, \overbrace{(a,b)}^{d}\,c}$

See here for a few proofs of this law (by Bezout, universal property, and prime factorizations).

Community
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Bill Dubuque
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