Question

find two consecutive odd positive integers sum of whose square is 290.

Answer 1

Let the two consecutive odd number be $x$ and $x+2$.

Given: Sum of the squares of two consecutive odd positive integers is 290.

According to question,

$x^2+(x+2{)}^2=290$

$x^2+x^2+4x+4=290$

$2x^2+4x+4=290$

$2x^2+4x=290-4$

$2x^2+4x=286$

$x^2+2x=143$

$x^2+2x-143=0$

$x^2+13x-11x-143=0$

$x(x+13)-11(x+13)=0$

$x=11$ (since they are positive integers)

$x+2=11+2=13$

Thus, the two consecutive odd positive integers are 11 and 13.