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?
Number of integer solutions of $\frac{1}{x} + \frac{1}{y} = \frac{1}{1000}$
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algebra-precalculus
diophantine-equations
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0I guess you mean solutions for integer $x$ and $y$[?](http://www.wolframalpha.com/input/?i=1%2Fx+%2B+1%2Fy+%3D+1%2F1000) – 2012-02-25
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0yes..obviously. – 2012-02-25
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3Not obviously! I had to make that assumption. – 2012-02-25
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2@Amol: Your question was not tagged [tag: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. – 2012-02-25
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0@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 $49$. – 2012-02-25
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0@AndréNicolas yes i agree.....49 for positive integer and 98 else. – 2012-02-25
1 Answers
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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 $1000^2$.
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0... and there are $49$ divisors of $10^6$ – 2012-02-25
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0thnx...got it...49 solutions is the correct answer for positive values of x and y. – 2012-02-25