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Find the number of positive integers less than $1000$ of the form

$$\frac{(x+y+z)^2}{xyz}$$

where $x,y,z$ are positive integers.

Usually I have lots of ideas on how to solve a problem, and I include them in the post to show that I have thought a bit about the problem and so that others can build off my ideas if possible. But on this problem I must admit I am stuck. I can think of two possible approaches: (1) treat this as a Diophantine equation to characterize all triples $(x,y,z)$ such that $xyz|(x+y+z)^2$, or (2) to play around with the problem, try to find a class of numbers that are representable. I have made only trivial observations: If $n$ is representable, then scaling $x,y,z$ gives that all factors of $n$ are also representable. Also, if a prime $p|x$ and $p|y$ then $p|z$, so we may assume that $x,y,z$ are pairwise relatively prime. I believe bounding the numerator or denominator may be the way to solve this.

Micah
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math_lover
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  • Do you have anythoughts? – MT_ Mar 10 '15 at 02:22
  • Let me add them to the problem. – math_lover Mar 10 '15 at 02:24
  • This is from a mock aime contest (high school computational math contest). It should involve only elementary methods, although its proving to be quite difficult. – math_lover Mar 10 '15 at 02:41
  • As you've proved that $(x,y,z) = 1$, the problem can come down to finding the number such that $x \mid (y+z)^2$, $y \mid (x+z)^2$ and $z \mid (x+y)^2$, but I haven't much made progress after this. – Stefan4024 Mar 10 '15 at 02:42
  • @Stefan4024: I never said that the gcd of all three numbers is 1. I said all other solutions may be derived from one in which the gcd is 1. – math_lover Mar 10 '15 at 02:50
  • Have you tried Vietta jumping? – Asinomás Mar 10 '15 at 02:52
  • Ahhhhhhh. I totally forgot about that old trick! Yes that is very well suited for these kinds of problems, which are otherwise hard to crack. Let me try that... – math_lover Mar 10 '15 at 02:55
  • Sorry did not realize it was a duplicate. Thankfully I just solved it with vieta jumping before looking at any solutions! – math_lover Mar 10 '15 at 03:16

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