Question

How many pairs (m, n) of positive integers satisfy the equation \(m^2+105=n^2 \) ?

• A. 8
• B. 7
• C. 9
• D. 4

Answer 1

The Correct Option is D

We are given the equation: 

\[ m^2 + 105 = n^2 \Rightarrow n^2 - m^2 = 105 \Rightarrow (n - m)(n + m) = 105 \]

This means we must find the number of ways to express 105 as a product of two positive integers \( (n - m)(n + m) \), where \( n > m \).

Step 1: Prime Factorization of 105

\[ 105 = 3 \times 5 \times 7 \]

So, the number of positive divisors of 105 is: \[ (1+1)(1+1)(1+1) = 8 \]

Each pair of factors \((a, b)\) such that \( a \cdot b = 105 \) and \( a < b \), gives a unique solution.

Hence, number of such pairs: \[ \frac{8}{2} = 4 \]

Step 2: Listing Valid Factor Pairs (n - m, n + m)

• \((1, 105)\): \( n = \frac{1 + 105}{2} = 53,\ m = \frac{105 - 1}{2} = 52 \)
• \((3, 35)\): \( n = 19,\ m = 16 \)
• \((5, 21)\): \( n = 13,\ m = 8 \)
• \((7, 15)\): \( n = 11,\ m = 4 \)

Final Answer:

There are exactly 4 integer pairs (m, n) satisfying the equation \( m^2 + 105 = n^2 \).