If $a, b \mid c \text { and } \gcd(a, b) = d, \text { then } ab \mid cd $

Question

$a \mid c \implies c = ak \text { and } b \mid c \implies c = bj.$

$ak + bj = 2c = d \implies c \mid d.$

$d \mid a \implies a = dj.$

$c = ak = d(jk) \implies d \mid c.$

So, $c = d.$

$a \mid c \text { and } b \mid c \implies ab \mid cc \implies ab \mid cd.$

Does this make sense to you?

Apply this very standard fact to $a \mid c$ and $(b/d) \mid c$. — Najib Idrissi, Aug 31 '15 at 15:27

score 0 · Answer 1 · answered Aug 31 '15 at 14:20

0

There is a mistake:

$$ak+bj = 2c = d$$

I agree, since $ak=bj=c$, that $ak+bj=2c$, but why would $2c$ be equal to $\gcd(a,b)?$

answered Aug 31 '15 at 14:20

5xum

I thought $ak + bj = d$ by Bezout. Do we need to choose different letters for that like $an + bm = d$? – user263390 Aug 31 '15 at 14:22
@user263390 Of course. The $j,k$ given by Bezout are in general not the same $j,k$ as the ones given by the fact that $a|c$. – 5xum Aug 31 '15 at 14:24
@ 5xum, $ax + by = d \to cax + cby = cd$. Since $a \mid c, a \mid cd \text { and } b \mid c \to b \mid cd.$ Then $ab \mid (cd)(cd). $ Am I going in the right direction with this? – user263390 Aug 31 '15 at 14:56

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