Find two consecutive natural numbers such that the sum of their squares is 61.

Let the required numbers be = x , x + 1

Given, sum of squares of the numbers = 61

$\Rightarrow x^2 + (x + 1)^2 = 61 \\[1em] \Rightarrow x^2 + (x^2 + 1 + 2x) = 61 \\[1em] \Rightarrow x^2 + x^2 + 1 + 2x = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 - 61 = 0 \\[1em] \Rightarrow 2x^2 + 2x - 60 = 0 \\[1em] \Rightarrow 2(x^2 + x - 30) = 0 \\[1em] \Rightarrow x^2 + x - 30 = 0 \\[1em] \Rightarrow x^2 + 6x - 5x - 30 = 0 \\[1em] \Rightarrow x(x + 6) - 5(x + 6) = 0 \\[1em] \Rightarrow (x + 6)(x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 6 = 0 \text{ or } x - 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -6 \text { or } x = 5$

Since the numbers are natural number so x ≠ -6.

∴ x = 5 , x + 1 = 6.

Hence, the required natural numbers are 5 , 6.