Show that for positive integers if $d_1e_1=d_2e_2\ and \gcd(e_1,e_2)=1\Rightarrow\ lcm(d_1,d_2)=d_1e_1=d_2e_2$
We know that: $$lcm(d_1,d_2)gcd(d_1,d_2)=d_1d_2$$ but this doesn't help much!
What's the trick?
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Let $k=\gcd(d_1,d_2)$. Then $d_1=kd_1'$ and $d_2=kd_2'$ where $d_1'$ and $d_2'$ are relatively prime. Thus from the relationship you stated we have $$\text{lcm}(d_1,d_2)k=k^2d_1'd_2',$$ and therefore $$\text{lcm}(d_1,d_2)=kd_1'd_2'.$$ It remains to show that $kd_1'd_2'=d_1e_1=kd_1'e_1$, or equivalently $d_2'=e_1$.
We were told that $d_1e_1=d_2e_2$, or equivalently that $d_1'e_1=d_2'e_2$. Since $e_1$ and $e_2$ are relatively prime, we have $e_1\mid d_2'$. And because $d_1'$ and $d_2'$ are relatively prime, we have $d_1'\mid e_2$.
But because $d_1'e_1=d_2'e_2$, it follows that $e_1=d_2'$ and $d_1'=e_2$.
answered Apr 8 '16 at 5:14
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$\begingroup$ Can you please explain more about the proof of: $e_1=d_2'$ and $d_1'=e_2$? I was a little confused!! $\endgroup$ – Hamid Reza Ebrahimi Apr 8 '16 at 18:03
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1$\begingroup$ In greater detail, instead of the last paragraph, since $e_1\mid d_2'$ we have $d_2'=se_1$ for some $s$. Similarly, $e_2=t d_1'$. So $d_1'e_1=se_1td_1'$, and therefore $st=1$, so $s=t=1$, and we are finished. $\endgroup$ – André Nicolas Apr 8 '16 at 18:10
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1$\begingroup$ You are welcome. There are notationally slicker proofs, I was trying for a proof that would be as "unclever" as possible. No tricks, just chasing things down. $\endgroup$ – André Nicolas Apr 8 '16 at 18:24
