Find two consecutive odd positive integers, sum of whose squares is 970.
Let the consecutive odd positive integers be ‘a’ and a + 2
⇒ a2 + (a + 2)2 = 970
⇒ 2a2 + 4a – 966 = 0
⇒ a2 + 2a – 483 = 0
⇒ a2 + 23a – 21a – 483 = 0
⇒ a(a + 23) – 21(a + 23) = 0
⇒ (a – 21)(a + 23) = 0
Thus, a = 21
Consecutive odd positive integers are 21, 23