Question

Find two consecutive positive integers, sum of whose squares is 365.

Answer 1

In that numbers, let one of the numbers be ’x’. It’s consecutive positive integer is (x + 1) 

Sum of their squares is 365. 

∴ (x)² + (x + 1)² = 365 

x² + x² + 2x + 1 = 365 

2x² + 2x + 1 = 365 

2x² + 2x + 1 – 365 = 0 

2x² + 2x – 365 = 0 

x² + x – 182 = 0

x² + 14x – 13x – 182 = 0 

x(x + 14) – 13(x + 14) = 0 

(x + 14) (x – 13) = 0 

If x + 14 = 0, then x = -14 

If x – 13 = 0, then x = 13 

In these positive integer is 13. 

∴ One number, x = 13 

It consecutive number is, x + 1 = 13 + 1 = 14 

∴ The Numbers are 14 and 13.