Diophantine equation

Number Theory Level 5

Given that $\frac { 1 }{ x } +\frac { 1 }{ y } =\frac { 1 }{ 2015 }$ , $x$ and $y$ are both integers, and $x>y$ , find the number of ordered pairs of $(x,y)$ .

The answer is 26.

1 solution

Alex Zhong
Apr 3, 2015

$\dfrac{x+y}{xy}=\dfrac{1}{2015} \implies xy -2015x-2015y + 2015^2 = 2015^2.$

Use SFFT, we have $(x-2015)(y-2015) = 5^2\times 13^2 \times 31^2.$

The number of ordered pairs: $3\times 3\times 3 -1 = \boxed{26}.$

Moderator note:

The final calculation is somewhat slipshod. It doesn't follow immediately. Let me spell it out:

The number of pairs of solutions arise from factors of $2015 ^ 2$ . There are $3 \times 3 \times 3 = 27$ unordered pairs of positive factors. We also have to allow for negative factors, so there are a total of 54 unordered pairs.
The pair $(-2015, -2015)$ leads to $(0,0)$ so we have to reject it.
The pair $(2015, 2015)$ leads to $x = y$ , so we have to reject it.
Otherwise, every pair would occur twice, so we have to divide by 2. This gives us $\frac{52}{2} = 26$ .

Exactly what I'm looking for! +1

Joel Yip - 5 years, 4 months ago

How do we know $x \gt y$ and you have not included negative solutions?

Rishik Jain - 5 years, 3 months ago

I've added an explanation of how to reach this calculation properly.

Calvin Lin Staff - 4 years, 1 month ago

Diophantine equation

The answer is 26.

1 solution

Moderator note:

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