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Prove that if 2n − 1 is prime, then n is prime for n being a natural number. I've looked at https://math.stackexchange.com/a/19998. It is known that 2n − 1 can only be prime if n is prime. This is because if jk = n, 2n − 1 =∑n−1 i=0 2i = ∑j−1 i=0 2i∑k−1 i=0 2ij. So they will only continue to alternate at twin primes.
17 Δεκ 2020 · If 2^n - 1 is prime for some positive integer n, prove that n is also prime. Numbers in this format are called Mersenne primes.Question submitted through www...
6 Ιουν 2014 · For the given $n$, the sum of the factors is $\sum_{r=0}^{p-1}\: ^nC_r \cdot 2^r \cdot (2^p - 1) = 2^p \cdot (2^p -1) = 2n$. Thus, proved that n is a perfect number. Share
Theorem. If for some positive integer n, 2 n -1 is prime, then so is n. Proof. Let r and s be positive integers, then the polynomial xrs -1 is xs -1 times xs(r-1) + xs(r-2) + ... + xs + 1. So if n is composite (say r.s with 1 < s < n ), then 2 n -1 is also composite (because it is divisible by 2s -1). ∎. Notice that we can say more: suppose n > 1.
10 Οκτ 2015 · It is a (still open) conjecture that the number of primes of this type less than $N$ grows with bound as $1.32\, N/\log^2 N$, see Sieving for large twin primes and Cunningham chains of length 2 of the second kind (2012).
Let p be an integer other than 0, $\pm$1. Prove that p is prime if whenever r and s are integers such that p=rs, then r=$\pm$1 or s=$\pm$1
Introduction. How do we go about finding primes? And once we have found them, how do we prove they are truly prime? The answer depends on the size of the primes and how sure we need to be of their primality. In these pages we present the appropriate answers in several sections. Let us preview these chapters one at a time.
