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

Let the first integer be x.

The next consecutive positive integer will be x + 1.

According to the given question,

x² + ( x + 1)² = 365

Using, (a + b)² = a² + 2ab + b²

x² + (x² + 2x + 1) = 365 

2x² + 2x + 1 = 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 and x + 14 = 0

x = 13 and x = - 14

The value of x cannot be negative.

Hence x = 13 and x + 1 = 14