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

Number of pairs = \(\frac{\text{number of factors\ 105}}{2}\)
\(105= 3 \times5 \times7\)
Number of factors = \(2 \times2\times 2 \times8\)
Therefore, the necessary quantity of pairs is = \(\frac{8}{2} =4\)

Detailed Explanation:
\(m^2+105=n^2\)
\(⇒ n^2-m^2=105\)
\(⇒(n−m)(n+m)=105\)

Given that both m and n are positive integers, \((n−m)<(n+m)\)
Splitting 105 in two factors, we get

\(⇒(n−m)(n+m)=1×105\)
When \((n−m)=1 and (n+m)=105,\) the pair \((m,n)=(52,53)\) yields \((n−m)(n+m)=3×35.\)
Similarly, when
\((n−m)=3\) and \((n+m)=35,\) the pair \((m,n)=(16,19)\) yields \((n−m)(n+m)=5×21.\)
For \((n−m)=5\) and \((n+m)=21,\) the pair \((m,n)=(8,13)\) yields \((n−m)(n+m)=7×21.\)

And when \((n−m)=7\) and \((n+m)=21,\) the pair \((m,n)=(4,11).\)

Thus, there are four pairs that satisfy the given conditions.