Find two natural numbers which differ by 3 and whose squares have the sum 117.

Let first number be x

Since difference between two numbers is 3 hence, the other number is (x + 3).

Given , sum of the squares of number = 117

∴ x² + (x + 3)² = 117

$\Rightarrow x^2 + x^2 + 9 + 6x = 117 \\[1em] \Rightarrow 2x^2 + 6x + 9 - 117 = 0 \\[1em] \Rightarrow 2x^2 + 6x - 108 = 0 \\[1em] \Rightarrow 2(x^2 + 3x - 54) = 0 \\[1em] \Rightarrow x^2 + 3x - 54 = 0 \\[1em] \Rightarrow x^2 + 9x - 6x - 54 = 0 \\[1em] \Rightarrow x(x + 9) - 6(x + 9) = 0 \\[1em] \Rightarrow (x - 6)(x + 9) = 0 \\[1em] \Rightarrow x - 6 = 0 \text{ or } x + 9 = 0 \\[1em] x = 6 \text{ or } x = -9$

Since, numbers are natural hence, x ≠ -9.

∴ x = 6 , x + 3 = 9.

Hence, the required numbers are 6, 9.