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# Algebra Question

Find all the solutions of 1/x + 1/y = 1/2013.

I have found 8 solutions but one of my friend claims that it has 14 solutions. He told me that there is a trick for this.

Do any of you know the trick or help me find 14 solutions?

Thanks everyone!

Note by Anoopam Mishra
4 years, 2 months ago

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$$\frac{1}{x} + \frac{1}{y}= \frac{1}{2013} \implies xy= 2013(x+y)$$ This implies $$(x-2013)(y-2013) = xy - 2013(x+y) + 2013^2 = 2013^2$$.
Now note that $$2013^2= 3^2*11^2*61^2$$. So number of divisors (positive and negative both included) of $$2013^2$$ is $$2(2+1)(2+1)(2+1) = 54$$. For each divisor of $$2013^2$$ we get 1 possible value of $$((x-2013), (y-2013))$$. However note that $$(x, y)= (0, 0)$$ satisfies $$xy= 2013(x+y)$$, but not the original equation. So this case also has to be excluded. Thus from the total number of solutions we should subtract $$1$$. That would give the total number of solutions= $$54-1= 53$$. · 4 years, 2 months ago

There are actually 53 distinct integer ordered pairs $$(x,y)$$ that satisfy the equation, not 52. You might want to double check your reasoning. · 4 years, 2 months ago