m
2 + 105 = n
2 ⇒ n
2 – m
2 = 105
⇒ (n – m) (n + m) = 105
Since m an d n are positive integers (n – m) < (n + m), then by splitting 105 in two factors, we get
⇒ (n – m) (n + m) = 1 × 105
For (n – m) = 1 and (n + m) = 105, (m, n) = (52, 53)
⇒ (n – m) (n + m) = 3 × 35
For (n – m) = 3 and (n + m) = 35 (m, n) = (16, 19)
⇒ (n – m) (n + m) = 5 × 21
For (n – m) = 5 and (n + m) = 21 (m, n) = (8, 13)
⇒ (n – m) (n + m) = 7 × 15
For (n – m) = 7 and (n + m) = 15, (m, n) = (4, 11)
Hence, there are four required pairs.
Shortcut approach :Number of pairs =
105 = 3 × 5 × 7
Factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105
Number of factors of 105 = 8
Hence, required number of pairs = 8 / 2 = 4