Assume $\gcd(r,s)=\gcd(r',s')=1$. How do I show that also $\gcd(rs'+r's,ss')=1$?
Show certain property of a greatest common divisor
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elementary-number-theory
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5You don't. Suppose $s=s'=2$, say. With $r,r'$ odd, of course. – 2017-01-03
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0Hmm you are right. Strange. I have to use it for an exercise. – 2017-01-03
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1You can additionally impose $(s,s') = 1$. This suffices. – 2017-01-03
1 Answers
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This isn't true. Let $r=r'=3$ and $s=s'=2$. Then you have $$\gcd(rs'+r's,ss')=\gcd(3\cdot2+3\cdot 2,2\cdot2)=\gcd(12,4)=4$$