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Quantitative Aptitude Question on Number Systems

How many pairs (m, n) of positive integers satisfy the equation m2+105=n2m^2+105=n^2 ?

A

8

B

7

C

9

D

4

Answer

4

Explanation

Solution

Number of pairs = number of factors 1052\frac{\text{number of factors\ 105}}{2}
105=3×5×7105= 3 \times5 \times7
Number of factors = 2×2×2×82 \times2\times 2 \times8
Therefore, the necessary quantity of pairs is = 82=4\frac{8}{2} =4

Detailed Explanation:
m2+105=n2m^2+105=n^2
⇒n2−m2=105⇒ n^2-m^2=105
⇒(n−m)(n+m)=105⇒(n−m)(n+m)=105

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

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

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

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