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

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The equation 1/x + 1/y = p/q has been analyzed, demonstrating that for a given rational number p/q, there exists only a finite number of positive integer solutions. The discussion highlights the mathematical reasoning behind this conclusion, focusing on the constraints imposed by the rational nature of p/q. Participants engage in providing insights and solutions, with kaliprasad contributing a notable solution. The finite nature of the solutions is attributed to the properties of rational numbers and their relationship with integer values. Overall, the discussion emphasizes the limited scope of solutions in this mathematical context.
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Given a rational number, $\frac{p}{q}$, show that there are only a finite number of positive integer solutions to the equation:

$$\frac{1}{x}+\frac{1}{y}=\frac{p}{q}$$
 
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lfdahl said:
Given a rational number, $\frac{p}{q}$, show that there are only a finite number of positive integer solutions to the equation:

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

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)
 
kaliprasad said:
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)

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

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

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

based on above $\frac{1}{x} >\frac{1}{2n}$
 
kaliprasad said:
$\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$

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!
 
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