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We also know that $x$ and $y$ are positive integers. I understand that if $x+y$ is prime, we know that it must divide either $x$ or $y$. But when it isn't, I can't find any other ways to find properties of these numbers.

If the given is true, then $N \in \mathbb{Z}, N(x+y) = xy$ should also be true. As far as I can tell, either $x$ or $y$ must be a multiple of $N$ and $x+y$ can't be coprime to one of $x$ or $y$, and maybe both. The problem is that I'm not sure if this is true in all cases and can't think of ways to prove it.

Thanks for your help!

Kip M.
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    Given a prime factor $p$ of $x+y$ we see that also $p\mid xy$. So $p$ is a factor of either $x$ or $y$. Consequently it is a factor of both of them. Then what can we say about prime power factors of $x+y$? Hmm... – Jyrki Lahtonen Dec 30 '19 at 05:54
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    Note that your question is equivalent to characterising the integral solutions to the equation $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$. – Clement Yung Dec 30 '19 at 06:07
  • Following up on @ClementYung's observation I would advice you to take a look at this thread. – Jyrki Lahtonen Dec 30 '19 at 06:09

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Using @BarryCipra's solution we proceed as follows.

Note first that if $a,b,d$ are any positive integers, then $$x=da(a+b),y=db(a+b)\quad (*)$$ gives a solution, because $x+y=d(a+b)^2$ divides $d^2ab(a+b)^2=xy$

Now suppose $x.y$ are any positive integers satisfying the condition that $x+y|xy$. Let $x=ga,y=gb$ where $a,b$ are coprime. Then $xy=g^2ab,x+y=g(a+b)$, so $x+y|xy$ implies that $a+b|gab$. But $a+b,ab$ must be relatively prime (since $a,b$ are), so we have $g=d(a+b)$ for some $d$. Hence $$x=da(a+b),y=da(a+b)$$ So (*) gives us all solutions.

almagest
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  • In the link, it says that all this means $N = abd$. For some $N$ is there a way to generate all the $a$, $b$ and $d$ so we can get all the $x$ and $y$ solutions? I'm not really sure how to get two coprime factors and another factor easily. – Kip M. Dec 30 '19 at 17:20
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    Pick any $a$ dividing $N$. Then pick any $b$ dividing $N/a$ which is not divisible by any prime dividing $a$. Finally take $d=N/(ab)$. For example, suppose $N=60$. You could take $a=10$, then $b=3$, then $d=2$. Or you could take $a=10$, then $b=1$, then $d=6$. – almagest Dec 30 '19 at 17:51