Question

Find two consecutive natural numbers such that the sum of their squares is $365$.

Answer 1

Step 1: Use the given condition to form the equation:

Let the two consecutive natural numbers be $x$ and $x+1$.

According to the given question,

$\begin{array}{rcl}{x}^2+{\left(x+1\right)}^2& =& 365\end{array}$

Simplify the above equation as follows:

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & x^2+x^2+1+2x=365\end{array} \\ \Rightarrow 2x^2+2x+1=365 \\ \Rightarrow 2x^2+2x-364=0 \\ \Rightarrow x^2+x-182=0 \end{gathered}$$

Step 2: Factorise the obtained equation.

$$\begin{gathered} x^2+x-182=0 \\ \Rightarrow x^2+14x-13x-182=0 \\ \Rightarrow x\left(x+14\right)-13\left(x+14\right)=0 \\ \Rightarrow \left(x+14\right)\left(x-13\right)=0 \\ \Rightarrow x=-14,13 \end{gathered}$$

The value of $x$ cannot be negative because $x$ is a natural number. So, take $x=13$ and neglect $x=-14$.

Hence, the two consecutive natural numbers are $13$ and $13+1=14$.