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Suppose we have $\gcd(x,z) = 1$, prove that $\gcd(xy, z) = \gcd(y,z)$. This appears very intuitive to me however is there a formal way of proving this?

I could write the obvious statement that since $z$ does not have a common divisor as $x$, then the only common divisor comes from $y$. So this means that $\gcd(xy, z) = \gcd(y,z)$ directly. However, this does not seem mathematical enough, is it?

José Carlos Santos
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oldselflearner1959
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  • Show that $xy,z$ and $y,z$ have exactly the same common divisor. – Azlif Jan 08 '18 at 11:58
  • "However, this does not seem mathematical enough, is it?" The idea is mathematical enough. You just need to write it insome way so that both you and anyone else who reads it are entirely convinced that the idea actually works. – Arthur Jan 08 '18 at 12:00
  • I don't see how this is a duplicate. It's not the same question as the proposed dupe. – G Tony Jacobs Jan 08 '18 at 14:26

2 Answers2

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Hint. We have to show that $xy$, $z$ and $y$, $z$ have exactly the same common divisors. It is clear that if $d$ is a common divisor of $y$, $z$ then $d\mid y$ implies $d\mid xy$. It remains to show that if $d\mid xy$ and $d\mid z$ then $d\mid y$. Note that $\gcd(x,z)=1$ implies that there are $a,b\in\mathbb{Z}$ such that $ax+bz=1$, and therefore $axy+bzy=y$.

Robert Z
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  • Thanks for the hint. I am however stuck at the last part as I'm unable to relate how $axy + bzy = y$ can lead to the second part of the implication. I expressed the 2 gcds in the second part as a linear combination and I ended up with many variables to solve. Also, for the part on $d|y$ etc, are we supposed to use truth tables to prove it? It seems that If the first implication till $d|xy$ is true, then I can prove the second implication? However how does $d|z$ come into play? – oldselflearner1959 Jan 08 '18 at 12:30
  • We have to show that $xy$, $z$ and $y$, $z$ have exactly the same common divisors. In the second part we prove that if $d$ is a common divisor of $xy$, $z$ then $d$ is a common divisor of $y$, $z$. Note that $d$ divides $y=axy+bzy$ because $d$ divides $axy$ and $bzy$. – Robert Z Jan 08 '18 at 13:08
  • Thanks for the reply. I'm wondering how I can show the clear fact you mentioned that $d|y$ and $d|z$ means $d|xy$, because $x$ doesn't seem to be involved over here. Is there a proof for this? It seems that this fact will be very important in showing that they have common divisors. Also, using the fact that $d$ divides $axy + bzy = y$, I can show that $d$ divides $y$. But I'm unable to relate to how $gcd(b,c)$ is equals to $d$, since I can only show $gcd(b,c)$ divides b too. – oldselflearner1959 Jan 08 '18 at 13:24
  • If $d\mid y$ then $y=qd$ therefore $xy=x(qd)=(xq)d$ that is $d$ divides $xy$. – Robert Z Jan 08 '18 at 13:30
  • I can get the idea that $d|y \implies d|xy$ but why is there a need to include $d|z$? This information seems to be not related to proving the implication – oldselflearner1959 Jan 08 '18 at 15:48
  • For the implication $d|y \implies d|xy$ we don't need that $d|z$. $d|z$ is necessary for the implication $d|xy \implies d|y$. – Robert Z Jan 08 '18 at 15:54
  • Sorry I'm a little slow here..It seems to me that if $d|z$ can be any fact that we just put into the expression, then we can also do so for any facts that may not be true, which causes $d|y$ and $d|z$ to be false, in which case any implication will be true, because F => T is always true? – oldselflearner1959 Jan 08 '18 at 16:04
  • A superfluous hypothesis can not invalidate an implication. – Robert Z Jan 08 '18 at 16:19
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Let $a,b\in \mathbb{N}_0$, $a=\prod_{i=1}^np_{i}^{t_i}, b = \prod_{i=1}^np_{i}^{s_i}$, where $p_i$ are primes, $t_i,s_i$ are non-negative integers. Then using the following fact $$ \operatorname{GCD}(a,b) = \prod_{i=1}^np_{i}^{\operatorname{min}\{t_i,s_i\}} $$ you can easily show your equality.

Mikhail Goltvanitsa
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