Necessary and sufficient conditions for the sum of two numbers to divide their product How to find necessary and sufficient conditions for the sum of two numbers to divide their product.
Thanks in advance.
 A: There's a solution of this problem in this post on codereview.stackexchange.com. Let me review it, hopefully correctly.
Assume $a, b$ not both zero. Let $d = \gcd(a,b) \ne 0$. Then $a = a' d, b = b' d$, and we have that $a + b = (a' + b') d$ divides $ab = a'b'd^2$, so $a' + b'$ divides $a'b'd$. 
Since $a'$ and $b'$ are coprime, if $p$ is a prime divisor of $a'+ b'$, this must divide $d$. This is because $p$ divides $a'b'd$, so it must divide one of the factors. If $p$ divides $a'$, say, then since $p$ divides $a' + b'$, and then $b'$, against the assumption that $a', b'$ are coprime. Thus $a'+b'$ divides $d$.
So the recipe appears to be the following. Choose any coprime pair $a', b'$, and construct $$a = c (a'+b') a', \qquad b = c (a'+b') b',$$
for arbitrary $c$.
A: If, $\dfrac{ab}{a+b}=n$
Then,
$\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{n}$
I don't know if this is to be considered an answer or not.
A: Dividing $\rm\:(A\!+\!B)n = AB\:$ by $\rm\:d=(A,B)\:$ yields $\rm\: (\color{#C00}{a\!+\!b})n = d\color{#C00}{ab},\:$ for $\rm\:[a,b] = [A,B]/d$
$\rm(a\!+\!b,b)\!=\!(a,b)\!=\!1\!=\!(a\!+\!b,a)\:$ so by Euclid $\rm\:(\color{#C00}{a\!+\!b,ab})\!=\!1\:$ so $\rm\:a\!+\!b\mid d,\,$ so $\rm\,(a\!+\!b)c = d.$
Thus $\rm\ [A,B] = d[a,b] = (a\!+\!b)c[a,b].\:$ Indeed $\rm\:A\!+\!B = (a\!+\!b)^2c\mid (a\!+\!b)^2 c^2 ab = AB.$
A: A simple condition can be found that does not require using the GCD and yet allows you to determine all the solutions. But let me rephrase the problem:
Given any integer $x$, find all integers $y$ such that $x+y \mid xy$.
We know that $x+y \mid x(x+y)$ and so
$x+y \mid x(x+y) - xy = x^2$. The RHS does not depend on $y$ anymore.
Therefore, $y=d-x$ for all $d \mid x^2$
