If `alpha` is complex fifth root of unity and `(1+alpha +alpha^(2)+ alpha^(3))^(2005) = p + qalpha + ralpha^(2) + salpha^(3)` (where p,q,r,s are real), then find the value of `p+ q+r+s`.

Explanation

To solve the problem, we need to analyze the expression \( (1 + \alpha + \alpha^2 + \alpha^3)^{2005} \), where \( \alpha \) is a complex fifth root of unity.

### Step-by-Step Solution:

1. **Understanding the Roots of Unity**:
- The complex fifth roots of unity are the solutions to the equation \( x^5 = 1 \). These roots are given by \( 1, \alpha, \alpha^2, \alpha^3, \alpha^4 \), where \( \alpha = e^{2\pi i / 5} \).
- The sum of all fifth roots of unity is zero:
\[ 1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 = 0 \]

2. **Rearranging the Sum**:
- From the above equation, we can express \( 1 + \alpha + \alpha^2 + \alpha^3 \) as:
\[ 1 + \alpha + \alpha^2 + \alpha^3 = -\alpha^4 \]

3. **Substituting into the Expression**:
- We substitute this into our original expression:
\[ (1 + \alpha + \alpha^2 + \alpha^3)^{2005} = (-\alpha^4)^{2005} \]

4. **Simplifying the Power**:
- This can be simplified as:
\[ (-1)^{2005} \cdot (\alpha^4)^{2005} = -1 \cdot \alpha^{8020} \]

5. **Finding \( \alpha^{8020} \)**:
- Since \( \alpha^5 = 1 \), we can reduce the exponent \( 8020 \) modulo \( 5 \):
\[ 8020 \mod 5 = 0 \quad \text{(since } 8020 = 5 \times 1604\text{)} \]
- Thus, \( \alpha^{8020} = (\alpha^5)^{1604} = 1^{1604} = 1 \).

6. **Final Result**:
- Therefore, we have:
\[ (1 + \alpha + \alpha^2 + \alpha^3)^{2005} = -1 \cdot 1 = -1 \]

7. **Comparing with the Given Form**:
- We need to express this as:
\[ -1 = P + Q\alpha + R\alpha^2 + S\alpha^3 \]
- From this, we can see:
- \( P = -1 \)
- \( Q = 0 \)
- \( R = 0 \)
- \( S = 0 \)

8. **Calculating \( P + Q + R + S \)**:
- Now, we compute:
\[ P + Q + R + S = -1 + 0 + 0 + 0 = -1 \]

### Final Answer:
Thus, the value of \( P + Q + R + S \) is \( \boxed{-1} \).