query: $2^n-1$ is prime or composite as $n$ is prime or composite [without using induction method]
positive_passages.docid: 26514
positive_passages.text: Prove that if $n$ is a composite, then $2^n-1$ is composite.
negative_passages.docid: 105854
negative_passages.text: Prove $n^2(n^4-1)$ is divisible by 60 using Mathematical Induction.
negative_passages.docid: 42885
negative_passages.text: Fermat primes relation to $2^n+1$
negative_passages.docid: 11867
negative_passages.text: Prove by induction that $n < 2^n$ for all $n \ge 1$
negative_passages.docid: 44161
negative_passages.text: Using induction to prove $\sum_{k=0}^{N-1} 2^k = 2^N-1$
negative_passages.docid: 102738
negative_passages.text: Show that $e^n>\frac{(n+1)^n}{n!}$ without using induction.
negative_passages.docid: 34954
negative_passages.text: If $11..11$($n$ times 1) is prime, prove that $n$ is also prime
negative_passages.docid: 43939
negative_passages.text: Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1.
negative_passages.docid: 45700
negative_passages.text: How can I know if $2^{2^{2^{2^{2}}}}+1=?$ is prime?
negative_passages.docid: 72026
negative_passages.text: Induction proof of $\sum_{k=1}^{n} \binom n k = 2^n -1 $
negative_passages.docid: 23544
negative_passages.text: Complete induction proof that every $n > 1$ can be written as a product of primes
negative_passages.docid: 102889
negative_passages.text: If $2^n - 1$ is prime from some integer $n$, prove that n must also be prime.
negative_passages.docid: 100305
negative_passages.text: Prove $2^n-1$ is divisible by $(2^m-1)^2$ iff...
negative_passages.docid: 63356
negative_passages.text: Can $x^{n}-1$ be prime if $x$ is not a power of $2$ and $n$ is odd?
negative_passages.docid: 11642
negative_passages.text: Proving by induction that $2^n \le 2^{n+1}-2^{n-1} - 1$ . Does my proof make sense?
negative_passages.docid: 130093
negative_passages.text: Prove by Induction : $n < 2^n$
negative_passages.docid: 134963
negative_passages.text: Proving a non-prime (composite) number $> 1$ has a divisor $d \leq\sqrt q$