Prove that $a \mid bc$ implies $a \mid \gcd(a,b)\times \gcd(a,c)$

Question

I want to show that $a \mid bc$ implies $a \mid \gcd(a,b)\times \gcd(a,c)$.

My answer: since $$ a\mid ac \quad \text{ and } \quad a\mid bc,$$ we get that $a \mid \gcd(ac,bc)$ which implies that $$ a \mid |c|\gcd(a,b).$$

However, I don't know the final step.

Do you mean "gcd" which is "greatest common divisor" instead of "gdc"? Or am I just ignorant? — Accelerator, Nov 08 '22 at 20:21
Equivalently (at least, in $\Bbb Z\setminus{0}$), is it true that $x\le y+z$ implies $x\le \min(x,y)+\min(x,z)$ ? The answer is yes. — Sassatelli Giulio, Nov 08 '22 at 20:23
Does this answer your question? Prove if $n\mid ab$, then $n\mid [\gcd(a,n) \times \gcd(b,n)]$ - found in the RHS "Related" section list. — John Omielan, Nov 08 '22 at 20:31

score 1 · Accepted Answer · edited Nov 08 '22 at 20:53

1

Recall that the $gcd$ is a linear combination. i.e. $(a,b)(a,c)$ can be written as

$$(a,b)(a,c)=(am+bn)(ak+cl)=a^2mk+amcl+bnak+bncl$$

$$=a(amk+mcl+bnk) + bc(nl)$$

If $a\mid bc$ then $a$ divides the right hand side.

edited Nov 08 '22 at 20:53

Thomas Andrews

answered Nov 08 '22 at 20:29

David P

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Nov 09 '22 at 10:38
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@BillDubuque It was original to me. I did not search, but I will strive to do so. – David P Nov 09 '22 at 15:06

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