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!

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!

No vote yet

3 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewest\( \frac{1}{x} + \frac{1}{y}= \frac{1}{2013} \implies xy= 2013(x+y) \) 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= \( 54-1= 53 \). – Sreejato Bhattacharya · 4 years ago

Log in to reply

– Hero P. · 4 years 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.Log in to reply

– Anoopam Mishra · 4 years ago

Thank You!Log in to reply