Let a and x be positive integers. Prove that if $x^2 \mid 2016$ and $2016 \mid a$, then gcd(a,x) = |x|

Asked Jul 13 '22 at 18:05

Active Jul 13 '22 at 18:15

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Basically what I know so far is that $$x^2 \mid 2016$$ means there exists an integer q where $$2016 = qx^2$$

As well, $$2016 \mid a$$ means that there exist an integer r where $$a = r2016$$

I can substitute 2016 with $qx^2$ and that can yield $$a = r(qx^2) \implies a = (rq)x^2$$

And since rq is an integer I can conclude from what I've done so far that $$x^2 \mid a$$

Can someone help me on getting to the last part where I can show that gcd(a,x) = |x|?

edited Jul 13 '22 at 18:11

Bill Dubuque

asked Jul 13 '22 at 18:05

Andrew Ranger

You're basically there. If $d$ is the largest number that divides $x$ but $x$ is negative, what value does $d$ have? – abiessu Jul 13 '22 at 18:08
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$x^2,|,a\implies x,|,a$. – lulu Jul 13 '22 at 18:08
$\color{#c00}x\cdot x\mid 2016\mid a\Rightarrow \color{#c00}x\mid a,$ by the linked dupe. – Bill Dubuque Jul 13 '22 at 18:11
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@BillDubuque: I think this question is about the more subtle absolute value determination... – abiessu Jul 13 '22 at 18:11
$gcd(a,x)\leq x$, $x$ positive. – Bob Dobbs Jul 13 '22 at 18:14
@abiessu Please be patient - it takes time to populate the dupe list since SE only allows one dupe on initial closure. It then needs to be edited to add more. – Bill Dubuque Jul 13 '22 at 18:16
@BillDubuque: ah, missed that second one, I wasn't aware more than one would show. Thanks for the heads up! – abiessu Jul 13 '22 at 18:17
Why was this question upvoted? It seems awfully easy and straightforward. -1 – Mike Jul 13 '22 at 18:18

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