Find two consecutive odd positive integers,

Solution:

We have to find two consecutive integers sum of whose squares is 290.

Let the two consecutive integers be x and x+2

According to the question

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

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

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

Dividing both sides by 2

$x^{2}+2 x+2=145$

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

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

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

 

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

$x=-13,11$

Therefore two consecutive integers are 11,13