Question

Find two consecutive positive odd numbers, the sum of whose squares is $802$

Answer 1

Let the two consecutive positive odd numbers be $a$ and $a+2$

Given:$a^2+{(a+2)}^2=802$

$\Rightarrow a^2+a^2+4a+4-802=0$

$\Rightarrow 2a^2+4a-798=0$

$\Rightarrow a^2+2a-399=0$

$\Rightarrow a^2+21a-19a-399=0$

$\Rightarrow (a+21)(a-19)=0$

$\Rightarrow a=-21$ or $a=19$ since $a$ cannot be negative

$\therefore a=19$

$\therefore$ the numbers are $19$ and $21$