I'm having a hard time proving the following claim:

if $ab\mid cd$ and $a\mid c$ and $ab\nmid c$ then $b\mid d$

Any help would be appreciated


closed as off-topic by Eevee Trainer, Clayton, Vinyl_cape_jawa, Shaun, RRL Mar 12 at 22:52

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  • $\begingroup$ What have you tried so far? $\endgroup$ – Vinyl_cape_jawa Mar 12 at 20:26
  • $\begingroup$ $ab\nmid c\,$ doesn't suffice, we need $(b,c/a)=1$ or equivalently $(ab,c)=a\ $ $\endgroup$ – Bill Dubuque Mar 12 at 21:07

Counterexample: $a=1,\,b=4,\,c=d=2$.


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