Question

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

Answer 1

Let one of the odd positive integer be $x$

then the other odd positive integer is $x+2$

their sum of squares

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

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

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

Given that their sum of squares = 290

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

$2x^2+4x=286$

$2x^2+4x-286=0$

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

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

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

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

Therfore ,$x=11or-13$

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