Chapter 11: Problem 18
Assuming that \(n\) is an even perfect number, say \(n=2^{k-1}\left(2^{k}-1\right)\), prove that the product of the positive divisors of \(n\) is equal to \(n^{k} ;\) in symbols, $$ \prod_{d \mid n} d=n^{k} $$
Short Answer
Expert verified
The product of the positive divisors of \( n \) is \( n^k \).
Step by step solution
01
Understanding Perfect Numbers
An even perfect number is of the form \( n = 2^{k-1}(2^k - 1) \), where \( 2^k - 1 \) is a Mersenne prime. Therefore, \( n \) has divisors of the form \( d_i = 2^a imes b \), where \( 0 \leq a \leq k-1 \) and \( b \mid (2^k - 1) \), meaning the divisors consist of powers of 2 multiplied by divisors of \( 2^k - 1 \).
02
Determine the Number of Divisors
The number of divisors of \( 2^{k-1} \) is \( k \) (including 1 and \( 2^{k-1} \)), and if \( 2^k - 1 \) is a prime, it has 2 divisors (1 and \( 2^k - 1 \)). Thus, the total number of divisors of \( n \) is \( 2k \), each divisor corresponding to a unique combination of a power of 2 and a factor of \( 2^k - 1 \).
03
Product of Divisors Relation
The product of all divisors of a number \( n \), is given by \( n^{t/2} \) when there are \( t \) divisors because each divisor \( d_i \) pairs with a divisor \( \frac{n}{d_i} \) such that \( d_i \cdot \frac{n}{d_i} = n \). In this case, \( t = 2k \), so the product of all divisors of \( n \) is \( n^{k} \).
04
Apply the Formula
Now apply the formula to \( n = 2^{k-1}(2^k - 1) \). Since we have determined the number of divisors is \( 2k \), it follows that the product of the divisors is \( n^{k} \). Thus, \( \prod_{d \mid n} d = n^k \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mersenne Prime
To understand perfect numbers, we first need to explore Mersenne primes. A Mersenne prime is a prime number that can be written in the form of \( 2^k - 1 \). This special type of prime number plays a significant role in the formation of even perfect numbers.
For example, if \( 2^k - 1 \) is a Mersenne prime, then \( 2^{k-1}(2^k - 1) \) becomes a perfect number.
Why is this important? Because not every number of the form \( 2^k - 1 \) is prime; only certain values of \( k \), such as 3, 5, 7, and a few others, make \( 2^k - 1 \) prime.
For example, if \( 2^k - 1 \) is a Mersenne prime, then \( 2^{k-1}(2^k - 1) \) becomes a perfect number.
Why is this important? Because not every number of the form \( 2^k - 1 \) is prime; only certain values of \( k \), such as 3, 5, 7, and a few others, make \( 2^k - 1 \) prime.
- The formula \( 2^{k-1}(2^k - 1) \) ensures that the number not only follows a specific pattern but also has unique divisors that relate intricately to the original Mersenne prime.
- This relationship further intertwines with how these numbers possess a full set of divisors to form a perfect number.
Divisor Product
The concept of divisor products gives us insight into the structure of perfect numbers. For the perfect number \( n = 2^{k-1}(2^k - 1) \), we are interested in how the product of all positive divisors equals a specific power of \( n \).
This means that if you multiply all the divisors of \( n \), the result is \( n^k \). How do we arrive at this conclusion?
This means that if you multiply all the divisors of \( n \), the result is \( n^k \). How do we arrive at this conclusion?
- The perfect number \( n \) has \( 2k \) divisors, derived from combining divisors of its two components: powers of 2 and the Mersenne prime component.
- Each divisor \( d_i \) can be paired with a complementary divisor \( \frac{n}{d_i} \) such that their product equals \( n \).
Even Numbers
Even perfect numbers have a fascinating relationship with even numbers. By definition, an even number is any integer that is divisible by 2. In context, a perfect number is always even when it is generated from the formula \( 2^{k-1}(2^k - 1) \).
But why is it necessarily even? Let's break it down:
But why is it necessarily even? Let's break it down:
- The term \( 2^{k-1} \) is explicitly even; any integer power of 2 will maintain this property.
- Therefore, multiplying \( 2^{k-1} \) by any number \( (2^k - 1) \) does not disrupt this evenness, ensuring that the entire perfect number remains an even number.