Prove that if p is a perfect number then no multiple o f p is a perfect number.

1. Assume \( p \) is a perfect number, thus \( 1 + d_1 + d_2 + \ldots + d_k = 2p \).

2. Consider a multiple of \( p \), \( np \), where \( n > 1 \).

3. The sum of divisors of \( np \) is given by \( S(np) = S(n) \cdot (1 + d_1 + d_2 + \ldots + d_k) = S(n) \cdot 2p \).

4. Assume \( np \) is perfect, then \( S(np) - np = np \).

5. Substitute and simplify: \( S(n) \cdot 2p - np = np \), leading to \( S(n) \cdot 2p = 2np \).

6. This implies \( S(n) = 2n \), making \( n \) a perfect number.

7. Generally, \( n \) and \( np \) can't both be perfect numbers.

8. Therefore, \( np \neq \text{perfect number} \).