Three distinct positive integers $x,y,z$ are pairwise relatively prime, and the sum of any two is a multiple of the third one. Find $xyz$.

I don't know how to start.

  • $\begingroup$ Hint: It's easy to Check $x=1,y=2,z=3$ and all permutation of these satisfies ... Then Prove that these are only possible solution :) $\endgroup$ – Rezwan Arefin Feb 12 '17 at 17:33
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    $\begingroup$ These downvotes are so annoying. Just because you have 5k rep doesn't mean you can't ask elementary number theory questions $\endgroup$ – suomynonA Feb 12 '17 at 17:35
  • $\begingroup$ @suomynonA, keep in mind that downvotes on questions are free, so you should not allow an isolated downvote without any comments to affect your morale. $\endgroup$ – Andreas Caranti Feb 12 '17 at 17:39
  • $\begingroup$ I know...thanks for the answer and for the support. $\endgroup$ – suomynonA Feb 12 '17 at 18:00
  • $\begingroup$ There's a slight generalization of this problem. You just ask for a triple $(x, y, z)$ of positive integers, normalized to $x \le y \le z$, say, such that each of them divides the sum of the other two. Then I think you can show that they are of the form $(a, a, a)$, $(a, a, 2a)$ or $(a, 2 a, 3 a )$. $\endgroup$ – Andreas Caranti Feb 13 '17 at 8:12

Suppose WLOG $x < y < z$. Then $z \mid x + y < 2 z$, and there are no other multiples of $z$ between $z$ and $2 z$. It follows that $x + y = z$.

Now $x \mid y + z = x + 2 y$, so $x \mid 2 y$, and since $\gcd(x, y) = 1$, we have $x \mid 2$, and thus $x = 1$ or $x = 2$.

Similarly $y \mid x + z = 2 x + y$, so as above $y = 1$ or $y = 2$. Since $x < y$, we must have $x = 1$, $y = 2$, and thus $z = x + y = 3$.


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