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If $2^n-1$ is prime, then $n$ cannot be $1$. We show that $n$ cannot be composite. Suppose to the contrary that $n=ab$, where $a$ and $b$ are greater than $1$. Then $$2^n-1=2^ {ab}-1= (2^a)^b-1.$$. Let $x=2^a$. Then $2^n-1=x^b-1$.
15 sie 2017 · $$n \neq 2^k \quad \text{for all} \ k \in \mathbb{Z} \implies 2^n + 1 \quad \text{is composite.}$$ Then I want to prove by contradiction: Suppose $n \neq 2^k \quad \forall k\in \mathbb{Z}$ and $2^n + 1$ is prime.
31 mar 2015 · Proof. Suppose that 2n − 1 is prime, and write n = st where s, t are positive integers. Since xs − 1 = (x − 1)(xs−1 +xs−2 + ⋯ + 1) , we can substitute x = 2t to see that 2t − 1 is a factor of 2n − 1. Since 2n − 1 is prime there are only two possibilities, 2t − 1 = 1 or 2t − 1 =2n − 1 . Therefore t = 1 or t = n.
26 kwi 2023 · Proof. Sufficient Condition. Suppose 2n − 1 is prime . Let a = 2n − 1(2n − 1) . Then n ≥ 2 which means 2n − 1 is even and hence so is a = 2n − 1(2n − 1) . Note that 2n − 1 is odd . Since all divisors (except 1) of 2n − 1 are even it follows that 2n − 1 and 2n − 1 are coprime .
20 mar 2023 · Prime Numbers - integers greater than \(1\) with exactly \(2\) positive divisors: \(1\) and itself. Let \(n\) be a positive integer greater than \(1\). Then \(n\) is called a prime number if \(n\) has exactly two positive divisors, \(1\) and \(n.\)
PROOFS THAT THERE ARE INFINITELY MANY PRIMES. Introduction. The fundamental theorem of arithmetic states that every positive integer may be fac-tored into a product of primes in a unique way. Moreover any nite product of prime numbers equals some positive integer.
22 sty 2022 · Prove or disprove that \[p_1p_2\dotsm p_n+1\nonumber \] is prime for all \(n\ge 1\). (Hint: If \(n=1\) we have \(2+1=3\) is prime. If \(n=2\) we have \(2\cdot 3+1=7\) is prime.
