The fourth-degree equation $x^4-x-504=0$ has $4$ roots $r_1$, $r_2$, $r_3$, $r_4$. If $S_x$ denotes the value of ${r_1}^4+{r_2}^4+{r_3}^4+{r_4}^4$, compute $S_4$.

[i]2015 CCA Math Bonanza Individual Round #10[/i]
1. Given the polynomial equation \(x^4 - x - 504 = 0\), we know it has four roots \(r_1, r_2, r_3, r_4\).
2. For each root \(r_i\), we have \(r_i^4 = r_i + 504\). This follows directly from substituting \(r_i\) into the polynomial equation.
3. We need to find the value of \(S_4 = r_1^4 + r_2^4 + r_3^4 + r_4^4\).
4. Using the equation \(r_i^4 = r_i + 504\), we can express \(S_4\) as:
   \[
   S_4 = r_1^4 + r_2^4 + r_3^4 + r_4^4 = (r_1 + 504) + (r_2 + 504) + (r_3 + 504) + (r_4 + 504)
   \]
5. Simplifying the above expression, we get:
   \[
   S_4 = (r_1 + r_2 + r_3 + r_4) + 4 \cdot 504
   \]
6. By Vieta's formulas, the sum of the roots of the polynomial \(x^4 - x - 504 = 0\) is zero (since the coefficient of \(x^3\) is zero). Therefore:
   \[
   r_1 + r_2 + r_3 + r_4 = 0
   \]
7. Substituting this into the expression for \(S_4\), we get:
   \[
   S_4 = 0 + 4 \cdot 504 = 2016
   \]

The final answer is \(\boxed{2016}\).