1

I'm trying to find all integral solutions to the equation I found in a competition math packet

"Find the number of integer solutions (x, y) to if, for instance, (x, y) = (2, −4) and (x, y) = (−4, 2) are counted as different integer solutions."

$ \frac{1}{x} + \frac{1}{y}= \frac{1}{4} $

Is there an easier method than just brute-forcing it?

I attempted to find solutions by using some algebra to simplify it

$ \frac{1}{\frac{1}{x}+ \frac{1}{y}} = 4$

$\frac{xy}{x+y} = 4 $

$ xy = 4x + 4y $

$xy -4x - 4y + 16 = 16 $

$(y-4)(x-4) =16 $

In this form, I still need to guess

Samb1221
  • 13
  • 3
  • (1) Please provide some context in the post. Where is this problem from? Why is it interesting and important? (2) Please tell in the post what progress on this question did you achieve? Where were you “stuck”? – Aig Mar 07 '24 at 04:09
  • 4
    You've got the problem in a form $ab=16$, so you don't need to guess, you need to factor $16$. – Sil Mar 07 '24 at 04:32

1 Answers1

3

$$\begin{aligned} \frac1y=\frac14-\frac1x=\frac{x-4}{4x}\implies y = \frac{4x}{x-4} \end{aligned}$$ This is an integer if, and only if, $(x-4)|4x$, which is true if, and only if $$(x-4)|4x-4(x-4) = 16,$$ therefore $x-4\in\{\pm1,\pm2,\pm4,\pm8,\pm16\}$, i.e., $x\in\{-12,-4,0,2,3,5,6,8,12,20\}$. Of course we can't have $x=0$, but the other values produce valid solutions. Here they are: $$\begin{alignat}{2} -\dfrac1{12}+\dfrac13&=\dfrac14\qquad\dfrac13-\dfrac1{12}&=\dfrac14\\ -\dfrac14+\dfrac12&=\dfrac14\qquad\dfrac12-\dfrac14&=\dfrac14\\ \dfrac15+\dfrac1{20}&=\dfrac14\qquad\dfrac1{20}+\dfrac15&=\dfrac14\\ \dfrac16+\dfrac1{12}&=\dfrac14\qquad\dfrac1{12}+\dfrac16&=\dfrac14\\ \dfrac18+\dfrac18&=\dfrac14\\ \end{alignat}$$

Alma Arjuna
  • 3,759