Solving the Equation (X^2 - Y^2 100) for Positive Integer Pairs

The problem at hand is to find pairs of positive integers (X, Y) that satisfy the equation (X^2 - Y^2 100).

Factorization and Initial Setup

The given equation can be factorized using the difference of squares:

(X^2 - Y^2 (X Y)(X - Y) 100)

Let:

(a X - Y)

(b X Y)

Then we have:

(ab 100)

Conditions for X and Y to be Positive Integers

For (X) and (Y) to be positive integers, both (X Y) and (X - Y) must be of the same parity (both even or both odd).

Given that (a X - Y) and (b X Y), we need to check which pairs of (a) and (b) satisfy these conditions.

Positive Divisors and Their Pairs

The prime factorization of 100 is:

(100 2^2 times 5^2)

The positive divisors of 100 are:

(1, 2, 4, 5, 10, 20, 25, 50, 100)

We list the pairs ((a, b)) such that (ab 100):

(1, 100) (2, 50) (4, 25) (5, 20) (10, 10)

Next, we identify which pairs yield both (a) and (b) as either both even or both odd:

(1, 100): odd and even (not valid) (2, 50): even and even (valid) (4, 25): even and odd (not valid) (5, 20): odd and even (not valid) (10, 10): even and even (valid)

The valid pairs are (2, 50) and (10, 10).

Calculating X and Y for Valid Pairs

For the pair (2, 50):

(X frac{(2 50)}{2} 26)

(Y frac{(50 - 2)}{2} 24)

Thus, the pair (26, 24) satisfies the equation (X^2 - Y^2 100).

For the pair (10, 10):

(X frac{(10 10)}{2} 10)

(Y frac{(10 - 10)}{2} 0) (not valid since (Y) must be positive)

Hence, the only valid pair of positive integers (X, Y) that satisfy the equation is (26, 24).

Therefore, there is 1 pair of positive integers (X, Y) that satisfies the equation (X^2 - Y^2 100).