MHB How Many Pairs of (x,y) Satisfy the Given Equation?

[Albert1]

$x,y\in N$
$\dfrac {1}{x}+\dfrac {1}{y}=\dfrac {1}{2010}---(1)$

How many pairs of $(x,y)$ we may get to satisfy (1)

 

[lfdahl]

My attempt:

Spoiler

Given the relation:

\[\frac{1}{x}+\frac{1}{y} = \frac{1}{2010}, \: \: \: x,y \in \mathbb{N}.\: \: \: \: \: \: \: \: (1)\]Both $x$ and $y$ must be greater than $2010$.Let $x = 2010 + k$, for some $k \in \mathbb{N}$.Then $y$ can be expressed as: \[y = \frac{2010(2010+k)}{k} = 2010 + \frac{2010^2}{k}\]The question is, how many different natural numbers, $k$, divide the square of $2010$ (including the trivial case $k = 1$)?The prime factorization of the square of $2010$ is: $2010^2 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 67^2$. Thus, the number of divisors, i.e. the number of $(x,y)$-pairs is: $3^4 = 81$.

The answer implies, that a specific pair, e.g. $(2011,2010\cdot 2011)$ and its permutation $(2010\cdot 2011, 2011)$ both count. Otherwise, the answer would be $41$ pairs.

 

[kaliprasad]

Spoiler

We have $2010 y + 2010 x = xy$
or $xy - 2010x - 2010y = 0$
or $(x-2010)(y-2010) = 2010^2= 2^2 * 3^2 * 5^2 * 67^2$
the above has $(2+1)(2+1)(2+1)(2+1) = 81$ factors in natural numbers
number if pairs = $81$
because (x,y) and (y,x) are different