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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?

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2 Answers

up vote 2 down vote accepted

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)

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

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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
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