What is the number of integer solutions of: $$\frac{1}{x} + \frac{1}{y} = \frac{1}{1000}$$ How to solve these type of problems if am comfortable of solving $x+y=z$. But how to do if multiplicative inverses are involved?

  • $\begingroup$ I guess you mean solutions for integer $x$ and $y$? $\endgroup$ – Simon Feb 25 '12 at 10:10
  • $\begingroup$ yes..obviously. $\endgroup$ – Amol Sharma Feb 25 '12 at 10:39
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    $\begingroup$ Not obviously! I had to make that assumption. $\endgroup$ – fretty Feb 25 '12 at 10:44
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    $\begingroup$ @Amol: Your question was not tagged diophantine-equations and your English is not perfect (not that I hold that against you), so it was not completely obvious. Maybe you meant Gaussian integers... Anyway, click on the question mark in my first comment. $\endgroup$ – Simon Feb 25 '12 at 10:57
  • $\begingroup$ @amol Sharma: The answer depends on whether you mean integer solutions or positive integer solutions. If it is integers, there are $98$, if positive integers it is $49$. $\endgroup$ – André Nicolas Feb 25 '12 at 16:52

Assuming you mean integer solutions, you will be able to rewrite your equation as:

$1000(x+y) = xy$

Then rearranging you will be able to write as:

$(x - 1000)(y - 1000) = 1000^2$

So that your solutions for $x-1000$ and $y-1000$ correspond to divisors of $1000^2$.

  • $\begingroup$ ... and there are $49$ divisors of $10^6$ $\endgroup$ – Henry Feb 25 '12 at 13:48
  • $\begingroup$ thnx...got it...49 solutions is the correct answer for positive values of x and y. $\endgroup$ – Amol Sharma Feb 25 '12 at 16:57

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