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

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Answer to a math question Prove that if p is a perfect number then no multiple o f p is a perfect number.

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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} $

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

Answer to a math question Prove that if p is a perfect number then no multiple o f p is a perfect number.

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