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Suppose that $x, y, z$ are positive integers.

Let $\sigma(x)$ be the sum of the divisors of $x$, and let $\gcd(y, z)$ be the greatest common divisor of $y$ and $z$.

Here is my question:

If the following divisibility constraints hold: $$\gcd(X, Y) = 1$$ $$\gcd(X, \sigma(X)) = 1$$ $$\gcd(Y, \sigma(Y)) = 1$$ $$\gcd(XY, \sigma(XY)) = 1$$ does it follow that either $$\gcd(X, \sigma(Y)) = 1$$ or $$\gcd(Y, \sigma(X)) = 1?$$

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  • $\begingroup$ one or the other or is this two question? X = 7 y = 13 has gcd(X,sumY) = 7 but it has gcd(y, sum X) = 1. $\endgroup$ – fleablood Jun 1 '16 at 19:55
  • $\begingroup$ @fleablood, thank you for your comment. For $X = 7$ and $Y = 13$, the fourth condition in the hypothesis: $$\gcd(XY, \sigma(XY)) = 1$$ is not satisfied, as $$\gcd(7\cdot{13}, \sigma(7\cdot{13})) = \gcd(7\cdot{13}, 8\cdot{14}) = 7 \neq 1.$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 1 '16 at 20:00
  • $\begingroup$ Any way, to answer your question, I am looking for a proof or a disproof of the disjunction $$\{\gcd(X,\sigma(Y))=1\} \lor \{\gcd(Y,\sigma(X)) = 1\}$$ given that the four conditions in the hypothesis are satisfied. $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 1 '16 at 20:02
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    $\begingroup$ You have given the appearance that you did not check this for small $x,y$ with a computer program. You should learn to do that. I just wrote a program for $1 \leq x < y \leq 2500$ and both gcd's are always 1. This is such a strong outcome that a proof may be possible. Next is to see if there are any genuine restrictions on the factorization of, say $y.$ $\endgroup$ – Will Jagy Jun 1 '16 at 20:15
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    $\begingroup$ Okay, agreed @WillJagy. $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 1 '16 at 20:19
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From the Wikipedia page on the greatest common divisor, we use the following property:

Property G $$\gcd(a, b \cdot c) = 1 \iff \{\gcd(a, b) = 1 \land \gcd(a, c) = 1\}.$$

In particular, we obtain $$\gcd(XY, \sigma(XY)) = 1 \iff \gcd(XY, \sigma(X)\sigma(Y)) = 1$$ (since $\gcd(X,Y) = 1$ and $\sigma$ is weakly multiplicative) $$\iff \{\gcd(XY, \sigma(X)) = 1 \land \gcd(XY, \sigma(Y)) = 1\}$$ (using Property G) $$\iff \{\gcd(X, \sigma(X)) = 1\} \land \{\gcd(Y, \sigma(X)) = 1\} \land \{\gcd(X, \sigma(Y)) = 1\} \land \{\gcd(Y, \sigma(Y)\} = 1\}$$ (using Property G) $$\iff \{\gcd(Y, \sigma(X)) = 1\} \land \{\gcd(X, \sigma(Y)) = 1\}$$ (since $\gcd(X,\sigma(X)) = 1$ and $\gcd(Y,\sigma(Y)) = 1$).

QED

I hope that everything that I have written out is correct! =)

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    $\begingroup$ yes, that works. Before I went to the post office, i deliberately ran the program without the final condition (out of four) and it quickly gave examples, such as $x=2, y=3,$ where things went wrong. $\endgroup$ – Will Jagy Jun 1 '16 at 21:10
  • $\begingroup$ Thank you very much for the confirmation, @WillJagy! =) $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 1 '16 at 21:16
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    $\begingroup$ Notice that you do not need the second and third of your four hypotheses. $\endgroup$ – Will Jagy Jun 1 '16 at 21:19

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