# Finding/setting numbers that satisfy $ab=xy$ and $a+b=cd$

All numbers are assumed nonzero natural numbers. All numbers are assumed unequal.

Suppose we have $ab = xy$, and $a+b = cd$.

$x$ is set to be fixed from beginning, and it is assumed that one knows the prime factorization of it. $d$ is not fixed, and $d$ is a factor of $x$.

What I want to know is how one finds/sets the values of $a,b,c,d,y$ that satisfies the constraint with the given $x$.

Edit: I edited the question so that $d$ is now a factor of $x$.

Without brute-forcing, what would be an easy way to find a possible way of setting values?

-

The complete solution space consists of all possible results that can be obtained by the following procedure, then rejecting any solutions that have a repeated value:

• Choose any value for $y$
• Obtain the prime factorization of $xy$
• Choose $a$ and $b$ by distributing the primes in $xy$ between them
• Obtain the prime factorization of $a+b$
• Choose $c$ and $d$ by distributing the primes in $a+b$ between them

For example, if $x$ is $6$, then one solution obtained by this procedure is

• Let $y = 10$
• $xy = 2 \cdot 2 \cdot 3 \cdot 5$
• Let $a = 2 \cdot 2 \cdot 5$ and $b = 3$
• $a+b = 23$
• Let $c = 1$ and $d = 23$

For the revised problem, the following procedure works for all solutions.

• Choose any value for $d$ dividing $x$
• Choose any factorization $x = uv$
• Obtain the complete solution space $(s,t)$ for the equation $$us + vt \equiv 0 \pmod d$$
• Choose any positive values for $s,t$
• Let $a = us$
• Let $b = vt$
• Let $c = (a+b) / d$
• Let $y = st$

(Note this procedure is redundant: I believe the same solution can be obtained in multiple ways)

-
I apologize.. I just edited the question. Would you like to have a look at it? Thanks. – La Ventana Nov 3 '12 at 10:02
Thank you very much! – La Ventana Nov 3 '12 at 11:32

You ask for one possible way of setting values. Here's one: $$a=b=y=d=x,\quad c=2$$

-
So the easy way of finding the possible way of setting values is reading your answer? ;) – tomasz Nov 3 '12 at 12:03
@tomasz, the wording of the problem was altered after I posted my answer. – Gerry Myerson Nov 3 '12 at 22:56