Find two consecutive odd positive integers, sum of whose squares is 290.

Let the two consecutive odd integers be a and a + 2.

Given:

Sum of squares of numbers = 290

(a)² + (a + 2)² = 290

a² + 4a + 4 + a² = 290

2a² + 4a + 4 = 290

2a² + 4a – 290 + 4= 0

2a² + 4a – 286 =0

2(a² + 2a – 143) =0

a² + 2a – 143 =0

On factorizing the above equation,

a² + 2a – 143 =0

Sum = 2

Product = 143

Therefore the two numbers satisfying the above conditions are -11 and 13.

a² + 13a – 11a – 143 =0

a(a + 13) -11(a + 13) = 0

(a – 11) (a + 13) = 0

Solving first part,

a – 11 = 0

a = 11

Solving second part,

a + 13 = 0

a = -13

Given that the numbers are positive. So a = -13 is not possible.

So First number is 11 and consecutive positive odd number is 13.