Which of the following statements must be true about n in the expression for Mersenne prime 2^n-1

Question

Statement 1: it must be an even number
Statement 2: it must be an odd composite number
Statement 3: it must be a prime number

Answer 1

To understand which statements about \( n \) in the expression for Mersenne primes \( 2^n - 1 \) are true, let's analyze each statement.

Mersenne primes are primes of the form \( 2^n - 1 \) where \( n \) is a positive integer.

1. Statement 1: \( n \) must be an even number.
   This statement is not true. For \( 2^n - 1 \) to be a prime, \( n \) must be a prime number, but it does not have to be even. In fact, \( 2^3 - 1 = 7 \) is a Mersenne prime, and here \( n = 3 \) is an odd number. Thus, \( n \) can be odd or even, but for the Mersenne number \( 2^n - 1 \) to be prime, \( n \) needs to be a prime number.

2. Statement 2: \( n \) must be an odd composite number.
   This statement is also not true. A composite number is a positive integer that has at least one positive divisor other than one and itself. Mersenne primes occur when \( n \) is a prime number, so it cannot be an odd composite number since composite numbers are not prime.

3. Statement 3: \( n \) must be a prime number.
   This statement is true. For \( 2^n - 1 \) to be a Mersenne prime, \( n \) must be prime. If \( n \) is not prime, \( 2^n - 1 \) is not necessarily prime. For example, if \( n = 4 \) (which is composite), then \( 2^4 - 1 = 15 \), which is not prime.

In conclusion, the only statement that must be true about \( n \) in the expression for Mersenne primes \( 2^n - 1 \) is:

Statement 3: it must be a prime number.

Answer 2

Are you sure

Answer 3

Yes, I am sure about the analysis. Let's summarize the statements again with clarifications regarding \( n \) in the context of Mersenne primes \( 2^n - 1 \):

1. Statement 1: \( n \) must be an even number.

   • False. Mersenne primes can occur for odd \( n \). For example, \( n = 3 \) is odd, and \( 2^3 - 1 = 7 \), which is a prime. The only even prime number is 2, and \( 2^2 - 1 = 3\), which is also a prime, but \( n \) being even does not guarantee that \( 2^n - 1 \) will yield a prime for any other even \( n \).

2. Statement 2: \( n \) must be an odd composite number.

   • False. \( n \) must be a prime number for \( 2^n - 1 \) to potentially be a Mersenne prime. Odd composite numbers (e.g., 9, 15) yield \( 2^n - 1 \) that are not necessarily prime. For instance, \( n = 9 \) gives \( 2^9 - 1 = 511\), which is not prime.

3. Statement 3: \( n \) must be a prime number.

   • True. For \( 2^n - 1 \) to be a Mersenne prime, \( n \) must indeed be a prime number. This is a requirement because if \( n \) is composite, say \( n = ab \) with \( a, b > 1 \), then \( 2^n - 1 = (2^a - 1)(2^a + 1) \), which cannot be prime.

Thus, the conclusion is that only Statement 3 is true: \( n \) must be a prime number for \( 2^n - 1 \) to be a Mersenne prime. If you have any further questions or need clarification, feel free to ask!