How Many Solutions Satisfy This Diophantine Equation?

[Albert1]

$x,y\in N$
$\dfrac {1}{x}+\dfrac{1}{y}=\dfrac{1}{1987}---(1)$
find :$max(x+y)$ and $min(x)$
How many solutions of (x,y) will satisfy (1) ?

 

[mathbalarka]

Spoiler

I will answer your second question first : If $$\frac1x + \frac1y = \frac1{n}$$ For some nonzero positive integer $n$ then it's clear by rearranging that $n(x + y) = xy$. If $(x, y) = k$, i.e., $x = a \cdot k$ and $y = b \cdot k$ where $(a, b) = 1$, then $n \cdot (a + b) = k \cdot ab$. Thus, $a | n$. But then $n/a \cdot ( a + b) = k \cdot b$, and $b$ must divide $n/a$ (which is an integer). Thus, the number of solutions $(a, b)$ are

$$\sum_{a | n} \sum_{b | n/a} 1 = \tau(n)$$

Where $\tau(n)$ is the number of divisors of $n$. In this case, however $1987$ is a prime, thus $\tau(1987) = 2$. This makes answering the former question a lot easier : one clear solution is $(3974, 3974)$ as $1/2 + 1/2 = 1$. Another solution is $(1988, 3950156)$, coming from a little nonobvious identity $(a + 1)^{-1} + \left ( a \cdot (a + 1) \right )^{-1} = a^{-1}$. By the calculations above, these are all the possible solutions. It is thus clear that $\text{max}(x + y) = 1988 + 3950156 = 3952144$ and $\text{min}(x) = 1988$.

 

[mathbalarka]

Spoiler

A terribly elementary solution inspired by http://mathhelpboards.com/challenge-questions-puzzles-28/find-ab-bc-ca-11916.html#post56296 :

$$\frac1{x} + \frac1{y} = \frac1{n} \Rightarrow n( x+ y) = xy \Rightarrow xy - n(x + y) + n^2 = n^2 \Rightarrow (x - n)(y - n) = n^2$$

In this case, $n = 1987$ is a prime, so WLOG either $(x - n, y- n) = (n^2, 1)$ or $(n, n)$. If the former, then $x = n^2 + n = 3950156$ and $y = n + 1 = 1988$ and if otherwise then $x = y = 2n = 3974$. Thus, $\max(x + y) = 3952144$ and $\min(x) = 1988$

 

[Albert1]

Spoiler

A terribly elementary solution inspired by http://mathhelpboards.com/challenge-questions-puzzles-28/find-ab-bc-ca-11916.html#post56296 :

$$\frac1{x} + \frac1{y} = \frac1{n} \Rightarrow n( x+ y) = xy \Rightarrow xy - n(x + y) + n^2 = n^2 \Rightarrow (x - n)(y - n) = n^2$$

In this case, $n = 1987$ is a prime, so WLOG either $(x - n, y- n) = (n^2, 1)$ or $(n, n)$. If the former, then $x = n^2 + n = 3950156$ and $y = n + 1 = 1988$ and if otherwise then $x = y = 2n = 3974$. Thus, $\max(x + y) = 3952144$ and $\min(x) = 1988$

perfect !

 

[CellCoree]

As a scientist, my response to this content would be as follows:

To find the maximum and minimum values of x+y and x, we need to first solve for the variables in equation (1). Multiplying both sides by xy, we get:

xy * (1/x) + xy * (1/y) = xy * (1/1987)
y + x = 1987

Now, to find the maximum value of x+y, we can use the fact that the sum of two numbers is maximum when they are equal. Therefore, the maximum value of x+y would be 1987.

To find the minimum value of x, we can use the fact that the minimum value of 1/x occurs when x is maximum. Since x can only take positive integer values, the minimum value of x would be 1.

To find the solutions of (x,y) that satisfy equation (1), we can use the fact that x and y must be positive integers. Therefore, the solutions would be all pairs of positive integers (x,y) that satisfy the equation y + x = 1987. There are multiple solutions for this, such as (1,1986), (2,1985), (3,1984), etc.

In conclusion, the maximum value of x+y is 1987 and the minimum value of x is 1. There are multiple solutions for (x,y) that satisfy equation (1).