Necessary and sufficient conditions for the sum of two numbers to divide their product

Question

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

Thanks in advance.

By numbers, do you mean natural numbers? – Aang Mar 08 '13 at 08:35 — Aang, Mar 08 '13 at 08:35
@Avatar .integer number $\Bbb Z$ – elham Mar 08 '13 at 08:52 — elham, Mar 08 '13 at 08:52

score 6 · Accepted Answer · edited Apr 13 '17 at 12:41

6

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 $c$.

edited Apr 13 '17 at 12:41

Community

1

answered Mar 08 '13 at 10:48

Andreas Caranti

68,873

1

Nice! Yours is much better. – Ishan Banerjee Mar 08 '13 at 11:58
why you let $d=a'+b'$? – agustin Mar 08 '13 at 14:14
@agustin, you're right, bad notation, just fixed, thanks. – Andreas Caranti Mar 08 '13 at 14:33
1

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. – Don Hatch Nov 12 '14 at 00:50

score 3 · Answer 2 · answered Mar 08 '13 at 08:32

3

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.

answered Mar 08 '13 at 08:32

Ishan Banerjee

6,459

looks necessary and sufficient so looks like an answer to me – oks Mar 08 '13 at 09:42
But it's not particularly elegant, is it? – Ishan Banerjee Mar 08 '13 at 09:43
1

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 http://codereview.stackexchange.com/ – Andreas Caranti Mar 08 '13 at 10:56

score 2 · Answer 3 · edited Jul 20 '23 at 22:21

2

Dividing $\rm\:(A\!+\!B)n = AB\:$ by $\rm\:d=(A,B)\:$ yields $\rm\: (\color{#C00}{a\!+\!b})n = d\color{#C00}{ab},\:$ for $\rm\,a,b = A/d,B/d$.

$\rm(a\!+\!b,b)\!=\!(a,b)\!=\!1\!=\!(a\!+\!b,a)\:$ so by Euclid $\rm\:(\color{#C00}{a\!+\!b,ab})\!=\!1\:$ so $\rm\:a\!+\!b\mid d,\,$ so $\rm\,(a\!+\!b)c = d.$

So $\rm\, A,B = da,db = (a\!+\!b)ca,(a\!+\!b)cb,\,$ and $\rm\:A\!+\!B = (a\!+\!b)^2c\mid (a\!+\!b)^2 c^2 ab = AB.$

edited Jul 20 '23 at 22:21

Bill Dubuque

272,048

answered Mar 08 '13 at 16:16

Math Gems

19,574

Above, vector equations $\rm\ [x,y] = [u,v]\ $ mean $\rm\ x=u,\ y=v,\ $ and $\rm\ c[u,v] = [cu,cv].\ \ $ – Math Gems Mar 08 '13 at 16:20
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You had me fooled that you were Bill Dubuque with this formatting. – Mario Carneiro Jun 15 '16 at 20:31

score 0 · Answer 4 · answered Mar 27 '21 at 03:54

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 $d \mid x^2$

Necessary and sufficient conditions for the sum of two numbers to divide their product

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