If p is a prime number and w, x, y, z are four natural numbers whose sum is less than p, then `(w+x+y+z)^(p)-(w^(p)+x^(p)+y^(p)+z^(p))` is always divisible by

Options

A. `p-1`
B. `p^(2)`
C. p
D. `p+1`

Answer

To solve the problem, we need to show that the expression \( (w+x+y+z)^p - (w^p + x^p + y^p + z^p) \) is always divisible by \( p-1 \) when \( p \) is a prime number and \( w, x, y, z \) are natural numbers whose sum is less than \( p \).

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the expression:
\[ (w+x+y+z)^p - (w^p + x^p + y^p + z^p) \]
Here, \( w+x+y+z < p \) and \( p \) is a prime number.

2. **Applying the Binomial Theorem**:
According to the Binomial Theorem, we can expand \( (w+x+y+z)^p \):
\[ (w+x+y+z)^p = \sum_{k=0}^{p} \binom{p}{k} (w+x+y+z)^{p-k} \cdot (w+x+y+z)^k \]
However, since \( w+x+y+z < p \), we can also apply Fermat's Little Theorem which states that for any integer \( a \) and a prime \( p \):
\[ a^p \equiv a \mod p \]
This means:
\[ w^p \equiv w \mod p, \quad x^p \equiv x \mod p, \quad y^p \equiv y \mod p, \quad z^p \equiv z \mod p \]

3. **Rewriting the Expression**:
From Fermat's Little Theorem, we can rewrite \( w^p + x^p + y^p + z^p \) as:
\[ w^p + x^p + y^p + z^p \equiv w + x + y + z \mod p \]
Therefore, we can express our original equation as:
\[ (w+x+y+z)^p - (w+x+y+z) \equiv 0 \mod p \]

4. **Factoring the Expression**:
We can factor the expression:
\[ (w+x+y+z)^p - (w+x+y+z) = (w+x+y+z) \left( (w+x+y+z)^{p-1} - 1 \right) \]
Since \( w+x+y+z < p \), the term \( (w+x+y+z)^{p-1} - 1 \) is divisible by \( p-1 \) (by Fermat's theorem).

5. **Conclusion**:
Thus, we conclude that:
\[ (w+x+y+z)^p - (w^p + x^p + y^p + z^p) \text{ is divisible by } p-1. \]

### Final Result:
The expression \( (w+x+y+z)^p - (w^p + x^p + y^p + z^p) \) is always divisible by \( p-1 \).