Let's represent the two natural numbers as x and (x + 3) since they differ by 3. The sum of their squares can be expressed as follows:
x^2 + (x + 3)^2 = 117
Now, we can solve the equation:
x^2 + (x^2 + 6x + 9) = 117
Combine like terms:
2x^2 + 6x + 9 = 117
Subtract 117 from both sides:
2x^2 + 6x - 108 = 0
Now, we need to factor the quadratic equation:
2(x^2 + 3x - 54) = 0
Now, factor the quadratic expression inside the parenthesis:
2(x + 9)(x - 6) = 0
Set each factor to zero and solve for x:
x + 9 = 0 --> x = -9 (This solution is not a natural number, as it's negative)
x - 6 = 0 --> x = 6
So, the two natural numbers that differ by 3 and whose sum of squares is 117 are 6 and 9.