The sum of the squares of two consecutive natural numbers is 421 find the numbers

Let two consecutive numbers
x, (x + 1)

As per given conditionx^2 + (x + 1)^2 = 313x^2 + x^2 + 1 + 2x = 3132x^2 + 2x - 312 = 0x^2 + x - 156 = 0x^2 + 13x - 12x - 156 = 0x(x + 13) - 12(x + 13) = 0(x - 12)(x + 13) = 0

x = 12, - 13

Negative value not possible

Therefore, Numbers are 12 and 13

Let the two consecutive natural numbers be x and (x + 1) 

Given x2 + (x + 1)2 = 421 x2 + x2 + 2x + 1 – 421 = 0 2x2 + 2x – 420 = 0 x2 + x – 210 = 0 x2 + 15x  – 14x – 210 = 0 (x + 15)(x – 14) = 0 Hence x = -15 or 14Since x is a natural number Thus x = 14 x + 1 = 15

Therefore the two consecutive numbers are 14 and 15.