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Prove that if $xy=n(x+y)$ with $\gcd(x,y,n)=1$ has a solution in natural numbers $x,y,n\in \mathbb{N}$, then it follows necessarily that:

  1. $n = ab, \gcd(a,b)=1$
  2. $x = n+a^2$
  3. $y = n+b^2$

I have a proof for this, but it would be interesting to see other solutions.

  • Yes, thanks for your comment. I will edit the question. –  Aug 02 '17 at 07:00
  • Just to clarify what you mean by the triple $\gcd$, we do have $\gcd(6, 10, 15) = 1$, right? – Arthur Aug 02 '17 at 07:00
  • yes, $\gcd(x,y,z) = \gcd(x,\gcd(y,z))$ –  Aug 02 '17 at 07:02
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    This problem has been dealt with before. [See this] for example.(https://math.stackexchange.com/questions/1166999/find-all-integers-x-y-and-z-such-that-frac1x-frac1y-frac) – Steven Alexis Gregory Aug 02 '17 at 07:05
  • Thank you for your comment. That seems like the same problem but not with $\gcd(x,y,n)=1$. –  Aug 02 '17 at 07:08

1 Answers1

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We get $\frac{1}{n} = \frac{1}{x} + \frac{1}{y}$ so $n \lt x$ and $n \lt y$

Now rewrite this as

$$(x-n)(y-n) = n^2$$

Now if a prime $p$ divides $x-n$ and $y-n$ and $n$, then the gcd won't be $1$.

Thus both $x-n$ and $y-n$ are both perfect squares, with no common factor.

Aryabhata
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