Question

The sum of squares of two consecutive natural numbers is $85$. Find the numbers

Answer 1

Let us assume the two consecutive natural numbers as $a$ and $a+1$.
Given, $a^2+(a+1{)}^2=85$
$\Rightarrow a^2+a^2+1+2a=85$
$\Rightarrow 2a^2+2a+1=85$
$\Rightarrow 2a^2+2a-84=0$
$\Rightarrow a^2+a-42=0$
$\Rightarrow a^2+7a-6a-42=0$
$\Rightarrow a(a+7)-6(a+7)=0$
$\Rightarrow (a+7)=0$ or $(a-6)=0$
$\Rightarrow a=-7$ or $a=6$
Considering positive number (because they are natural numbers), $a=6$
The two consecutive numbers are $a=6$ and $a+1=7$.