Find two consecutive odd integers such that sum of their squares is 394.

Let the required two consecutive odd integers be x , x + 2

Given, sum of squares of two consecutive odd integers = 394

$\Rightarrow x^2 + (x + 2)^2 = 394 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x = 394 \\[1em] \Rightarrow 2x^2 + 4x + 4 -394 = 0 \\[1em] \Rightarrow 2x^2 + 4x - 390 = 0 \\[1em] \Rightarrow 2(x^2 + 2x - 195) = 0 \\[1em] \Rightarrow x^2 + 2x - 195 = 0 \\[1em] \Rightarrow x^2 + 15x - 13x - 195 = 0 \\[1em] \Rightarrow x(x + 15) - 13(x + 15) = 0 \\[1em] \Rightarrow (x + 15)(x - 13) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 15 = 0 \text{ or } x - 13 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -15 \text{ or } x = 13$

∴ When x = -15 , x + 2 = -13 and when x = 13 , x + 2 = 15.

Hence required integers are -15 , -13 or 13, 15 .