Answer:
4 pairs
Explanation:
You want the number of pairs of integers (m, n) such that m > n and m² -n² = 96.
Difference of squares
The difference of squares is factored as ...
m² -n² = (m -n)(m +n)
In order for the difference to be 96, these factors must be factors of 96.
Factors of 96
The prime factorization of 96 is ...
96 = 2^5 · 3
The number of divisors of 96 is the product of the exponents of these factors, each increased by 1:
(5+1)(1+1) = 12
That is, there are 12/2 = 6 factor pairs ab such that a<b. They are ...
96 = 1·96 = 2·48 = 3·32 = 4·24 = 6·16 = 8·12
Solution
The values of m and n relate to the values of 'a' and 'b' by ...
m = (a+b)/2
n = (b-a)/2
In order for m and n to be integers, the values of 'a' and 'b' must both have the same parity. That is, they must both be even, or both be odd. This eliminates the factor pairs 1·96 and 3·32, leaving 4 (a, b) pairs, hence 4 (m, n) pairs. Possible ordered pairs are ...
(m, n) ∈ {(25, 23), (14, 10), (11, 5), (10, 2)} . . . . . 4 pairs