Question

Find two consecutive natural numbers whose squares have the sum 313.

Answer 1

Let the first number be a
Second number = a+1

According to question

=> a^2 +(a+1)^2 = 313

=> a^2 +a^2 +2a +1 = 313

=> 2a^2 +2a +1 - 313 = 0

=> 2a^2 +2a - 312 = 0

=> a^2 +a - 156 = 0

=> a^2 +13a - 12a - 156 = 0

=> a(a+13) - 12(a+13) = 0

=> (a-12)(a+13) = 0


a= 12 and - 13

But it is a natural number
So, a = 12

First number = 12

Second number = 13

Answer 2

Let the two number be x, x+1
Therefore , x² + (x+1)² = 313
x² + x² + 2x + 1 = 313
2x² +2x = 312
x² + x -156 = 0
x² +13x - 12x - 156 = 0
x( x + 13 ) - 12( x + 13) = 0
(x-12)(x+13) =0
Therefore , x=12,-13(Neglected)
Therefore the two number are 12 and 13