Prove and extend of disprove and salvage this statement.
I posted a similar post a few minutes ago. I've been stuck on these two problems for a while and I have no idea what to do. Any help would be greatly appreciated!
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Write 1 as a linear combination of $a$ and $b$ with integer coefficients. That's usually a good first step. – Vik78 Sep 16 '16 at 13:00
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I hope you have deleted the similar post. You should start by working out some examples, to decide whether to prove or disprove the statement. Please show some of your work and we may be able to help. – Ethan Bolker Sep 16 '16 at 13:00
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Other duplicates: If $\gcd(a,b)=1$ and $a$ and $b$ divide $c$, then so does $ab$, and If $a \mid c, b \mid c, \gcd (a,b)=1$ then $ab \mid c.$, or Showing that if $(a,b)=1$ and if $a\mid c$ and $b\mid c$ then $ab \mid c$, in GCD domains, and many others – Dietrich Burde Sep 16 '16 at 13:03
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Use bézout identity (as you your previous post): https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity – Surb Sep 16 '16 at 13:03
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It's not that hard. If $(a,b)=1$ then $\exists r,s\in Z$ st $ar+bs=1$, so $m=mar+mbs$. Because $a|m$ and $b|m$ then the right side of the equation is divisible by $ab$. Done
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