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6 cze 2014 · Prove that if $2^{p}-1$ is prime then $$n=2^{p-1}(2^p-1)$$ is a perfect number here is what i did: We need to prove the $\sigma(n)=n$ so $\sigma(n)=\sigma(2^{p-1})\sigma(2^p-1)$ since $2^{p}-1$ is a prime thus $\sigma(2^p-1)=2^p$
- $2^n + 1$ is prime $ \\implies n$ is a power of $2$
$$n \neq 2^k \quad \text{for all} \ k \in \mathbb{Z}...
- Disprove that "if $p$ is a prime number, then $2^p-1$ is also a prime ...
If prime numbers are randomly distributed, as is widely and...
- For any prime $p > 3$, why is $p^2-1$ always divisible by $24$?
$n^2 = 1 \pmod{24}$ for $n=1,5,7,11$, by checking each case...
- $2^n + 1$ is prime $ \\implies n$ is a power of $2$
15 sie 2017 · $$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.
If prime numbers are randomly distributed, as is widely and strongly believed, then $x=2^p-1=prime$ is false. Proof: Assume $p_s$ is the smallest possible prime. Then, $x=2^{p_{s}}- 1$ must be the next prime. This means that $x$ determines the next prime in the sequence and so forth.
$n^2 = 1 \pmod{24}$ for $n=1,5,7,11$, by checking each case individually. $(n+12)^2 = n^2 + 24n + 144 = n^2 \pmod{24}$. Therefore, $n^2 = 1 \pmod{24}$ when $n$ is odd and not divisible by $3$, and so $n^2-1$ is divisible by $24$ for these $n$. You don't need primality of $p$ here!
26 kwi 2023 · Theorem. Let a ∈ N a ∈ N be an even perfect number . Then a a is in the form: 2n−1(2n − 1) 2 n − 1 ( 2 n − 1) where 2n − 1 2 n − 1 is prime . Similarly, if 2n − 1 2 n − 1 is prime, then 2n−1(2n − 1) 2 n − 1 ( 2 n − 1) is perfect .
Since sqrt( 2 64) = 2 32 and sqrt(1) = 1 , I've factorized 2 64 - 1 into product of (sum of integers * subtraction of integers). Both of these are integers as well, therefore 2 64 - 1 can be factorized into integers (where neither of them is 1), which means it is not prime.
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 \(n.\)
