I have tried to solve the problem. I have started by assigning m=(ab)/d, where d is the greatest common divisor of a and b. Then I set up c = aq. That is where I am stuck. I am really not sure what to do next.
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1$c$ divides $m$? Don't you mean, prove $m$ divides $c$? – Gerry Myerson Jan 25 '19 at 03:56
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1I am sorry... you are right it is m divides c.!! – Kbiir Jan 25 '19 at 03:57
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1Let $c=mq+r$ with $0\le r<m$ and show that $r$ is also a common multiple. – Gerry Myerson Jan 25 '19 at 04:01
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Hint: $c = rm+d$ where $r$ is some integer and $d < m$. So $d = c-rm$. As $a$ divides $c$ and $m$, and $-r$ is an integer, $a$ divides $d$. Likewise as $b$ divides both $c$ and $m$, it follows that $b$ divides $d$. So both $a$ and $b$ divide $d$. But $m$ is LCM of $a$ and $b$ and $d< m$. What can you say about $d$?
Mike
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