Solving for positive integral values of an equation

Question

The question I originally came across is this,

Find positive, non-equal ,integrs such that the sum of their reciprocals is $\frac12$

I got an equation,

$$\frac1{x}+\frac1{y}=\frac12$$

Where $x$ and $y$ are the positive integers. Which reduces to, $$x+y=\frac{xy}{2}$$

I have come across many similar equations earlier too, had the term of $xy$ been not there I would have solved it.

But now I don't know how to do it, I seek not only the solution of theis equation but also of equations that have a similar general form.

Answer 1

If $x,y>4$, then $(x,y)$ will not be a solution since $\frac{1}{x}+\frac{1}{y}<\frac{1}{2}$.

Without loss of generality, $x\le y$, so $x=1,2,3$ or $4$. Then find $y$ in each of the four cases and see if it is a positive integer.

It turns out there is only one solution $$x=3,y=6$$

If we look for ordered pairs then there are two solutions.

Answer 2

Solving $$y=\frac{2x}{x-2}$$ try $x=1,3$ or ingeneral we want $x-2\mid 2x$

Answer 3

WLOG $\dfrac1x<\dfrac1y,$

$\dfrac2x<\dfrac1x+\dfrac1y=\dfrac12\implies x>4$ as $x>0$

$$y=\dfrac{2(x-2)+4}{x-2}$$

$$\iff y-2=\dfrac4{x-2}$$

So, we need $x-2$ to divide $4$

$\implies x-2\in[\pm1,\pm2,\pm4]$

But keep in mind $x>4,y>0$

Answer 4

I looked at the previous answers, and I'd like to head a similar route, but try to be more specific.

Firstly, you solve $\frac 1 x +\frac 1 y = \frac 1 2$ by subtracting $\frac 1 x$, finding the LCD and subtracting the two fractions on the right, then take the reciprocal to get $y=\frac {2x} {x-2}$.

We want to write $x-2$ and not $2-x$ because if $x\gt2$ then y would be negative, so for $x-3$ using $y=\frac 6 {-1}$ give us $\frac 1 3 -\frac 1 6=\frac 1 2$ which would instead give us $\frac 1 6$ on the left. Obviously, $\frac 1 6 \ne \frac 1 2$, but using $y=\frac 6 1$ gives us $\frac 1 3 +\frac 1 6=\frac 1 2$ which is true. If you use the other form, then $y=-\frac {2x} {2-x}$ for all $-\infty\lt x \lt 0, 0\lt x \lt2, 2\lt x\lt\infty$. For example, using $x=1$ gives us $1-\frac 1 2= \frac 1 2$ which is true. On the other hand, $x=10$ gives $y=-\big(\frac {20} {-8}\big)=\frac 5 2$ which means $\frac 1 {10} +\frac 2 5=\frac 1 x +\frac 1 y=\frac 5 {10}=\frac 1 2$, and that's true as well. I'd rather ignore the whole negative thing and just say $y=\frac {2x} {x-2}$.

So, we have $\frac 1 x +\frac 1 y=\frac 1 2$ as long as $y=\frac {2x} {x-2}$ (or $y=-\frac {2x} {2-x}$), but what about x? Well, if $x=2$ then $y=\frac 4 0$ and we can't have that, it would destroy reality as we know it. So $x\ne2$, nor can $x$ be zero, but it can be negative (example: $x=-1$,$y=\frac 2 3$,$\frac 1 x +\frac 1 y=-1+\frac 3 2=\frac 1 2$) it can be $\gt2$ or $\lt2$ (example: $x=1$,$y=-2$,$\frac 1 x+\frac 1 y=1-\frac 1 2=\frac 1 2$.

Listing all possible pairs is impossible in terms of just single numbers, for arguments sake, lets use $x=20$. Then, according to our formula, $y=\frac {2x} {x-2}=\frac {40} 18=\frac {20} 9$ and $\frac 1 x +\frac 1 y=\frac 1 {20} +\frac 9 {20}= \frac {10} {20}=\frac 1 2$. So, as long as $-\infty \lt x \lt 0, 0\lt x \lt 2, 2\lt x\lt \infty$ and $y=\frac {2x} {x-2}$, then $\frac 1 x +\frac 1 y=\frac 1 2$. I would list the intervals along with the expression of $y$ in terms of $x$ as my answer, or list the expression for $y$ and say $x\ne2,x\ne0$. The number of possibilities are not finite, so the only way to list them by hand is by writing down an expression for $y$ in terms of $x$ and listing the intervals of $x$ for which the equation remains true.

Answer 5

You got

$$x+y=\frac{xy}2,$$

which is equivalent to

$$xy-2x-2y=0,$$ which is equivalent to

$$(x-2)(y-2)=4.$$

All such equtions that only contain the terms $xy,x,y$ and $1$ with rational coefficients can be brought in that or similar form with integer coefficients (just multiply the equation with the common denominator). If the coefficient before $xy$ is not $1$, multiply the whole equation with that coefficient, $2xy-3x+5y=7$ becomes $4xy-6x+10y=14$ or $4xy-6x+10y-14=0$, for example. The last eqution becomes

$$ 0 = 4xy-6x+10y-14 = (2x+5)(2y-3) + 1$$

or equivalently

$$(2x+5)(2y-3)=-1$$

Since all numbers are integers, you can now equate each factor on the left hand side to an integer divisor of the right hand side. In this case, this leads to the 2 solutions

$$2x+5=1, 2y-3=-1 \Longleftrightarrow x=-2, y=1$$

and

$$2x+5=-1, 2y-3=1 \Longleftrightarrow x=-3, y=2.$$

The question at hand $(x-2)(y-2)=4$ by lab bhattacharjee in their answer.