To solve the problem of finding the number of positive integers (n \leq 100) such that the equation (x^2 - y^2 = n) has at least one solution in positive integers (x) and (y), we proceed as follows:
Key Observations:
- (x^2 - y^2 = (x - y)(x + y) = n).
- (x > y \geq 1), so (x + y > x - y \geq 1), and both factors ((x - y)) and ((x + y)) have the same parity (since their sum (2x) is even).
Valid Cases for (n):
-
(n) is odd:
Let (x - y = 1) and (x + y = n). Solving gives (x = \frac{n+1}{2}) and (y = \frac{n-1}{2}), which are integers if (n) is odd. -
(n) is divisible by 4:
Let (x - y = 2) and (x + y = \frac{n}{2}). Solving gives (x = \frac{n+4}{4}) and (y = \frac{n-4}{4}), which are integers if (n) is divisible by 4.
Counting Valid (n):
- Number of odd (n \leq 100): (\frac{100}{2} = 50).
- Number of multiples of 4 (\leq 100): (\frac{100}{4} = 25).
Total Valid (n):
(50 + 25 = 75).
Answer: (\boxed{75})
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