Someone's Revered Grouping Approach

Relevant wiki: Quadratic Diophantine Equations - Solve by Simon's Favorite Factoring Trick

Rearranging the equation gives

$xy = 156x+156y$

This reminds us of SFFT:

$xy-156x-156y+156^{2} = (x-156)(y-156) = 156^{2}$

Each solution of $x-156$ is equivalent to a solution for $x$ and $y$ . Since $156^{2} = 2^{4} \times 3^{2} \times 13^{2}$ has $(4+1)(2+1)(2+1) = 45$ factors, there are a total of $45$ solutions for $x-156$ and thus $\boxed{45}$ pairs $(x,y)$ .

A straightforward approach: Solving for $y$ gives $y=\frac{156x}{x-156}$ $=156+\frac{156^2}{x-156}$ by (not very) long division. For $y$ to be a positive integer, $x-156$ must be positive and divide $156^2=2^4\times3^2\times13^2$ . There are $5\times3\times3=45$ solutions.

The number of ordered pairs $(x,y)$ of the equation $\dfrac {1}{x} + \dfrac {1}{y} = \dfrac {1}{n}$ is given by $\tau(n^2)$ , where $\tau(x)$ denotes the number of positive divisors of $x$ . Hence, the number of ordered pairs $= \tau(156^2) = 45$ . Q.E.D.

Find the solutions of $\frac{1}{x} + \frac{1}{y} = \frac{1}{156}$ .

Suppose $x=\frac{156a}{b}$ ,

Then $\frac{1}{x}=\frac{b}{156a}$ , so that $\frac{1}{y}=\frac{1-\frac{b}{a}}{156}$ and $y=\frac{156a}{a-b}$ .

We are looking for coprime integers $a$ and $b$ such that both $b$ and $a-b$ divide $156a$ .

Because $x$ is a whole number and $a$ and $b$ are coprime, $b$ must divide $156$ .

Because $y$ is a whole number, $a-b$ must divide $156a$ Now the fact that $a$ and $b$ are coprime is equivalent to $a$ and $a-b$ being coprime, so $a-b$ must divide $156$ as well.

Now we will look for coprime pairs of factors of $156=2^2×3×13$ .

• 1 is coprime with 1,2,3,4,6,12,13,26,39,52,78,156 (12 pairs)
• 2 and 4 are coprime with 1,3,13, 39 (2×4=8 pairs)
• 3 is coprime with 1,2,4,13,26,52 (6 pairs)
• 13 is coprime with 1,2,4,3,6,24 (6 pairs)
• 6 and 12 are coprime with 1,13 (2×2=4 pairs)
• 26 and 52 are coprime with 1,3 (2×2=4 pairs)
• 39 is coprime with 1,2,4 (3 pairs)
• 78 and 156 are coprime with 1 (2 pairs)

So there are $12+8+6+6+4+4+3+2=45$ ordered pairs $(b,a-b)$ of coprime factors of $156$ . For each of them we can calculate $x=\frac{156a}{b}$ and $y=\frac{156a}{a-b}$ .

Nice title!