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How to find necessary and sufficient conditions for the sum of two numbers to divide their product.

Thanks in advance.

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  • $\begingroup$ By numbers, do you mean natural numbers? $\endgroup$
    – Aang
    Mar 8 '13 at 8:35
  • $\begingroup$ @Avatar .integer number $\Bbb Z$ $\endgroup$
    – elham
    Mar 8 '13 at 8:52
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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 $c$.

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    $\begingroup$ Nice! Yours is much better. $\endgroup$ Mar 8 '13 at 11:58
  • $\begingroup$ why you let $d=a'+b'$? $\endgroup$
    – agustin
    Mar 8 '13 at 14:14
  • $\begingroup$ @agustin, you're right, bad notation, just fixed, thanks. $\endgroup$ Mar 8 '13 at 14:33
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    $\begingroup$ This seems to be arguing that "every prime p dividing a'+b' divides d" implies "a'+b' divides d". That doesn't hold, but I believe it can be fixed simply by strengthening the antecedent to "every prime power p^k dividing a'+b' divides d". So I made a suggested edit to change "if p is a prime divisor of" to "if p^k is a prime power dividing". The edit was rejected (harshly), so I'm framing it as a comment instead. $\endgroup$
    – Don Hatch
    Nov 12 '14 at 0:50
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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.

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  • $\begingroup$ looks necessary and sufficient so looks like an answer to me $\endgroup$
    – oks
    Mar 8 '13 at 9:42
  • $\begingroup$ But it's not particularly elegant, is it? $\endgroup$ Mar 8 '13 at 9:43
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    $\begingroup$ It's not a matter of elegance, it is just that it does not give a constructive recipe to find all such $a, b$. I have given one in my answer, quoting a post on codereview.stackexchange.com $\endgroup$ Mar 8 '13 at 10:56
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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 $\rm\:A\!+\!B = (a\!+\!b)^2c\mid (a\!+\!b)^2 c^2 ab = AB.$

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  • $\begingroup$ Above, vector equations $\rm\ [x,y] = [u,v]\ $ mean $\rm\ x=u,\ y=v,\ $ and $\rm\ c[u,v] = [cu,cv].\ \ $ $\endgroup$
    – Math Gems
    Mar 8 '13 at 16:20
  • $\begingroup$ You had me fooled that you were Bill Dubuque with this formatting. $\endgroup$ Jun 15 '16 at 20:31
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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 $d \mid x^2$

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