21.
Find two consecutive positive integers, sum of whose square is 365.
Let first positive integer be x
So, other positive integer = x + 1
According to question,
x2 + (x + 1)2 = 365 x2 + x2 + 1 + 2x = 365
⇒ 2x2 + 2x – 364 = 0
⇒ x2 + x – 182 = 0
⇒ x2 + 14x – 13x – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x – 13) (x + 14) = 0
⇒ x – 13 = 0
or x + 14 = 0
⇒ x = 13
or x = –14
Here x = – 14 does not exist.
Therefore, required positive integers are : x = 13 and x + 1 = 13 + 1 = 14.
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