Question

Find two consecutive positive integers, the sum of whose squares is $313$.

Answer 1

Let the numbers be, $a,a+1$

Given,

$a^2+(a+1{)}^2=313$

$a^2+a^2+1+2a=313$

$2a^2+2a-312=0$

$a^2+a-156=0$

$a=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=\frac{-1\pm \sqrt{1^2-4\left(1\right)(-156)}}{2\left(1\right)}$

$a=\frac{-1\pm \sqrt{625}}{2}$

$a=\frac{-1\pm 25}{2}$

$\therefore a=12,13$

Thus the numbers are; $12,13$

as, ${12}^2+{13}^2=313$