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If we know that $cb'=da'$ and if we know that $\gcd(a',b')=1$ then show that $c=ka'$ and $d=kd'$ for some $k\in\mathbb{N}$.

My attempt : since $\gcd(a'b')=1\implies a'u+b'v=1$ by multiplying by $c$ we get that $a'cu+b'cv=c$. Which indcates that that $a'$ divides $c$. Is this a valid proof for $c=ka'$?

Sumanta
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  • You still need to show $d = kb'$ for the same $k = cu + dv$. But that part is easy. – player3236 Sep 24 '20 at 15:47
  • Does this indicate that my proof is so far valid? I still do not know how $cb'=da'$ would be implemented in the proof. –  Sep 24 '20 at 15:48
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    Yes. To show that $a'$ divides $c$, you used $cb'=da'$. You will use that again in the second part. – player3236 Sep 24 '20 at 15:49
  • If you wish to write your comments as an answer it would be better so that I mark your answer as correct. –  Sep 24 '20 at 15:52

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From $a'cu + b'cv = c$ and $cb'=da'$ we have

$$c = a'cu+da'v = a'(cu+dv)$$

so we can take $k = cu+dv$.

It remains to show that $d=kb'$.

This is trivial since

$$kb'=b'cu+b'dv=da'u+db'v=d(a'u+b'v)=d$$

player3236
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