A finite number of positive integer solutions 1/x+1/y=p/q

[lfdahl]

Given a rational number, $\frac{p}{q}$, show that there are only a finite number of positive integer solutions to the equation:

$$\frac{1}{x}+\frac{1}{y}=\frac{p}{q}$$

 

[kaliprasad]

Given a rational number, $\frac{p}{q}$, show that there are only a finite number of positive integer solutions to the equation:

$$\frac{1}{x}+\frac{1}{y}=\frac{p}{q}$$

Spoiler

without loss of generality we can assume x <= y so $\frac{1}{x} >=\frac{1}{y}$ Even if we require ordered pair number number of solutions cannot be > double of the case x <= y

further there exists n such that $\frac{1}{n+1} < \frac{p}{q} < \frac{1}{n}$ so $\frac{1}{x} > \frac{1}{2n}$

there are a finite values x < 2n and x >n) that satisfy the criteria hence finite solutions ( I have provided the estimates with +/- 1 and not exact bound)

 

[lfdahl]

Spoiler

without loss of generality we can assume x <= y so $\frac{1}{x} >=\frac{1}{y}$ Even if we require ordered pair number number of solutions cannot be > double of the case x <= y

further there exists n such that $\frac{1}{n+1} < \frac{p}{q} < \frac{1}{n}$ so $\frac{1}{x} > \frac{1}{2n}$

there are a finite values x < 2n and x >n) that satisfy the criteria hence finite solutions ( I have provided the estimates with +/- 1 and not exact bound)

Spoiler

Hi, kaliprasad!
Thankyou very much for your solution! Please explain how you conclude, that: $\frac{1}{x} >\frac{1}{2n}.$

 

[kaliprasad]

Spoiler

Hi, kaliprasad!
Thankyou very much for your solution! Please explain how you conclude, that: $\frac{1}{x} >\frac{1}{2n}.$

Spoiler

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

if ( a + b = 2c) and a > b then a > c else a + b < c + c or a +b < 2c

based on above $\frac{1}{x} >\frac{1}{2n}$

 

[lfdahl]

Spoiler

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

if ( a + b = 2c) and a > b then a > c else a + b < c + c or a +b < 2c

based on above $\frac{1}{x} >\frac{1}{2n}$

Thankyou very much, kaliprasad for your participation and a nice solution!