# Find all $(x,y)$ such that $1/x+1/y=1/7$.

While working in a workbook, I got stuck on the following problem: Find all the ordered pairs $(x,y)$ such that they satisfy the equation $1/x + 1/y = 1/7$. I made a bit of progress, but got stuck. So far, I have multiplied the entire equation by $7xy$ to get $7y+7x=xy$, then subtracted $7y$ and $7x$ from both sides to get $xy-7x-7y=0$. From here, I'm not sure how to progress, maybe with SFFT? Could someone guide me in what to do next? Thanks!

• If x and y are supposed to be real numbers, then there are certainly an infinite number of pairs. If not you should specify what they are. Commented May 5, 2016 at 1:58
• @vik78 It is likely that the question asks for integral solutions. Commented May 5, 2016 at 1:59
• Add $49$ to both sides of your result and factor... $(x-7)(y-7)=49$... Commented May 5, 2016 at 1:59
• Can you solve what you have for $y$? Commented May 5, 2016 at 2:00

Assuming that $x$ and $y$ are supposed to be integers... Abiessu's solution in the comments is very good and a great approach. This answer provides a more elementary solution.

There are two cases to consider, either $x$ and $y$ are both positive, or one is positive and the other is negative. Let's deal with both cases separately.

• If $x$ and $y$ are both positive, then let's assume that $x\leq y$ (the other case is symmetric). Therefore, $\frac{1}{x}\geq\frac{1}{y}$. Since $\frac{1}{x}+\frac{1}{y}=\frac{1}{7}$, it must be that $\frac{1}{7}>\frac{1}{x}\geq\frac{1}{14}$. If $\frac{1}{x}$ were larger than $\frac{1}{7}$, then $\frac{1}{y}$ would have to be negative. If $\frac{1}{x}$ were smaller than $\frac{1}{14}$, then $\frac{1}{y}$ would be larger. This gives $7$ possible values for $x$, $x=8,\cdots,14$. You can check each one for a possible $y$-value.

• If one of $x$ and $y$ is negative, assume that $y$ is negative and $x$ is positive (the other case is symmetric). Then, $1\geq\frac{1}{x}>\frac{1}{7}$. The LHS inequality occurs because $x$ is positive, while the RHS inequality occurs because $\frac{1}{x}$ must be larger than $\frac{1}{7}$ so that we can subtract $\frac{1}{|y|}$ from it. This leads to $6$ cases for $x$, $x=1,\cdots,6$. You can check each one for a possible $y$-value.

A simple way is to take $7y+7x=xy$, subtract $7y$ from both sides, and then factor out the $y$ on the right, then divide both sides by $x-7$ to obtain the equation $y=7x/(x-7)$. $x=7$ isn't a solution to the original equation so you don't have to worry about the domain restriction.

If you are indeed looking for the integer values, (-42,6) is the lowest value of x that can give an integer value for y and (56,8) is the highest. Checking the others finds (0,0), (6,-42), (8,56), and (14,14), but checking in the original equation shows that (0,0) is not a solution. This happened because multiplying by 7xy is zero when x or y are zero.

• Yes sorry edited to reflect that then saw your comment. Commented May 5, 2016 at 2:45