Question

find out two non continuous positive completenumbers where tge sum of this should be 290​

Answer 1

Let one of the odd positive integer be x 

then the other odd positive integer is x+2

their sum of squares = x² +(x+2)²

                               = x² + x² + 4x +4

                               = 2x² + 4x + 4

Given that their sum of squares = 290

⇒ 2x² +4x + 4 = 290

⇒ 2x² +4x = 290-4 = 286

⇒ 2x² + 4x -286 = 0 

⇒ 2(x² + 2x - 143) = 0 

⇒ x² + 2x - 143 = 0

⇒ x² + 13x - 11x -143 = 0 

⇒ x(x+13) - 11(x+13) = 0 

⇒ (x-11) = 0 , (x+13) = 0

Therfore , x = 11 or -13

We always take positive value of x 

So , x = 11 and (x+2) = 11 + 2 = 13 

Therefore , the odd positive integers are 11 and 13

Answer 2

You can use any combination..

290 = 200 + 90

Or

190 + 100