Find three consecutive odd integers , the sum of whose squares is 83.

Let the required numbers be x , x + 2 , x + 4.

Given, the sum of squares of these numbers = 308

$\Rightarrow x^2 + (x + 2)^2 + (x + 4)^2 = 83 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x + x^2 + 16 + 8x = 83 \\[1em] \Rightarrow 3x^2 + 12x + 20 = 83 \\[1em] \Rightarrow 3x^2 + 12x - 63 = 0 \\[1em] \Rightarrow 3(x^2 + 4x - 21) = 0 \\[1em] \Rightarrow x^2 + 4x - 21 = 0 \\[1em] \Rightarrow x^2 + 7x - 3x - 21 = 0 \\[1em] \Rightarrow x(x + 7) - 3(x + 7) = 0 \\[1em] \Rightarrow (x + 7)(x - 3) = 0 \\[1em] \Rightarrow x + 7 = 0 \text{ or } x - 3 = 0 \\[1em] x = -7 \text{ or } x = 3$

∴ When x = -7 , x + 2 = -5 , x + 4 = -3 and when x = 3 , x + 2 = 5 , x + 4 = 7.

Hence , the required numbers are -7, -5, -3 and 3, 5, 7.