Pyramids of Egypt

Number Theory Level 5

Let $d(k)$ denote the number of positive divisors of an integer $k$ , including 1 and itself, and let $n$ be a positive integer. In terms of $n$ , how many ordered pairs of integers $(x,y)$ are there such that $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{n}$ ?

d(n)^2

d(n^2)

2d(n^2)

\frac{d(n)}{2}

None of these

2d(n^2)-1

d(n)

d(n)-1

1 solution

Alex Li
Jul 21, 2015

Combining the fractions, we get $\frac{x+y}{xy}=\frac{1}{n}$ . Cross-multiplying and rearranging, $xy-nx-ny=0$ . Now, we use Simon's Favorite Factoring Trick to get $(x-n)(y-n)=n^2$ . To find the number of ordered pairs, we must multiply the number of positive divisors of $n^2$ by $2$ to account for negative numbers, giving us $2d(n^2)$ . However, note that we included $(0,0)$ as a solution, which is extraneous, so we must subtract $1$ for our final answer of $\boxed{2d(n^2)-1}$ .

Moderator note:

Standard result in Diophantine Equations. Very useful to know.

Pyramids of Egypt

1 solution

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