Question

The sum of squares of two consecutive natural numbers is 244; find the numbers.

Answer 1

Disclaimer: There is an error in the given question. Here it should be two consecutive even natural numbers.

Let the two consecutive natural numbers be $x\mathrm{and}x+2$.
Sum of squares of these two consecutive natural numbers is 244
$$\begin{gathered} x^2+{\left(x+2\right)}^2=244 \\ \Rightarrow x^2+x^2+4+4x=244 \\ \Rightarrow 2x^2+4x-240=0 \\ \Rightarrow x^2+2x-120=0 \\ \Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \Rightarrow x=\frac{-2\pm \sqrt{2^2-4\times 1\times \left(-120\right)}}{2} \\ \Rightarrow x=\frac{-2\pm \sqrt{4+480}}{2} \end{gathered}$$
$$\begin{gathered} \Rightarrow x=\frac{-2\pm \sqrt{484}}{2} \\ \Rightarrow x=\frac{-2\pm 22}{2} \\ \Rightarrow x=\frac{-2+22}{2},\frac{-2-22}{2} \\ \Rightarrow x=\frac{20}{2},\frac{-24}{2} \\ \Rightarrow x=10,-12 \end{gathered}$$
But the natural number cannot be negative so,
The two numbers are 10 and 10 + 2 = 12.