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

In summary, the equation 1/x + 1/y = p/q represents a mathematical problem where we are looking for values of x and y that, when plugged into the equation, will result in the fraction p/q. The phrase "finite number of positive integer solutions" means that there are a limited number of possible pairs of positive integers that satisfy the equation. To determine the solutions for 1/x + 1/y = p/q, we can use algebraic manipulation and number theory principles. No, there cannot be an infinite number of positive integer solutions for 1/x + 1/y = p/q. This is because the equation has a finite number of possible solutions, which can be determined using mathematical methods. Additionally, the phrase "
  • #1
lfdahl
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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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  • #2
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)
 
  • #3
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: $\frac{1}{x} >\frac{1}{2n}.$
 
  • #4
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 $\frac{1}{x} >\frac{1}{2n}$
 
  • #5
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!
 

1. What is the meaning of "A finite number of positive integer solutions 1/x+1/y=p/q"?

The equation 1/x + 1/y = p/q represents a mathematical problem where we are looking for values of x and y that, when plugged into the equation, will result in the fraction p/q. The phrase "finite number of positive integer solutions" means that there are a limited number of possible pairs of positive integers that satisfy the equation.

2. How do we determine the solutions for 1/x+1/y=p/q?

To determine the solutions for 1/x + 1/y = p/q, we can use algebraic manipulation and number theory principles. We can rearrange the equation to solve for one variable in terms of the other, and then use divisibility rules and prime factorization to determine which values of the other variable will result in a positive integer solution.

3. Can there be an infinite number of positive integer solutions for 1/x+1/y=p/q?

No, there cannot be an infinite number of positive integer solutions for 1/x + 1/y = p/q. This is because the equation has a finite number of possible solutions, which can be determined using mathematical methods. Additionally, the phrase "finite number of solutions" means that there is a limit to the number of solutions, and infinite is not a finite number.

4. Are there any specific restrictions on the values of p and q in 1/x+1/y=p/q?

Yes, there are some restrictions on the values of p and q in 1/x + 1/y = p/q. The values of p and q must be positive integers, and they cannot be equal to 1 simultaneously. This is because if p and q are both equal to 1, the equation would become 1/x + 1/y = 1, which has an infinite number of solutions.

5. Can we use this equation to solve real-world problems?

Yes, the equation 1/x + 1/y = p/q can be used to solve real-world problems in various fields such as physics, economics, and engineering. It can be used to find the possible combinations of two quantities that result in a given fraction, which can be helpful in solving various practical problems.

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