The sum of the squares of two consecutive odd positive integers is 394. Find then.
Let the consecutive odd integers be a, a + 2
⇒ a2 + (a + 2)2 = 394
⇒ a2 + a2 + 4a + 4 = 394
⇒ a2 + 2a – 195 = 0
⇒ a2 + 15a – 13a – 195 = 0
⇒ a(a + 15) – 13(a + 15) = 0
⇒ (a – 13)(a + 15) = 0
⇒ a = 13, -15
The numbers are 13 and 15