Prove that, for every natural number $n &gt; 2$, there is a prime number between $n$ and $n!$. [Hint: There is a prime number that divides $n! - 1$.]
13021
How to prove that there exists a prime number between $n$ and $ n!$, for all $ n&gt; 2$?    ( gives a much better bound, but this question is about obtaining a self-contained proof.)
60492
If $2^n+1$ is prime how to show that  $n$ must be a power of $2$  Being at elementary level I am at a loss what to begin with?
11650
Prove that if $2^n - 1$ is prime, then $n$ is prime for $n$ being a natural number   I've looked at      It is known that $2^n-1$ can only be prime if $n$ is prime.  This is because if $jk=n$, $2^n-1=\sum_{i=0}^{n-1}2^i=\sum_{i=0}^{j-1} 2^i \sum_{i=0}^{k-1} 2^{ij}$ So they will only continue to alternate at twin primes.  In particular, $2^{6k+2}-1, 2^{6k+3}-1$ and $2^{6k+4}-1$ will all be composite   What I dont understand is how do I get:   $$\sum_{i=0}^{n-1}2^i=\sum_{i=0}^{j-1} 2^i \sum_{i=0}^{k-1} 2^{ij}$$  If I am looking at the wrong answer, how can I proof the above?     Then again, maybe I am indeed looking at the wrong thing?      In particular, $2^{6k+2}-1, 2^{6k+3}-1$ and $2^{6k+4}-1$ will all be composite   This doesnt say why $2^{6k+2}-1$ is composite?
599
This an interesting problem my friend has been working on for a while now (I just saw it an hour ago, but could not come up with anything substantial besides some PMI attempts).  Here's the full problem:  Let $x_{1}, x_{2}, x_{3}, \cdots x_{y}$ be all the primes that are less than a given $n&gt;1,n \in \mathbb{N}$.   Prove that $$x_{1}x_{2}x_{3}\cdots x_{y} &lt; 4^n$$  Any ideas very much appreciated!
32987
Prove that:  If $n$ is greater than 2, then $(3n)!$ is divisible by $n! \times  (n + 1)! \times  (n + 2)!$  From Barnard &amp; Child's "Higher Algebra".  I know that the highest power of a prime $p$ contained in $N$! is  $$ \left\lfloor{\frac{N}{p}}\right\rfloor + \left\lfloor{\frac{N}{p^2}}\right\rfloor + \left\lfloor{\frac{N}{p^3}}\right\rfloor ... $$  I'm unable to show that the formula above gives a higher value for $N = 3n$ than for the sum of its values when $N$ = $n$ , $n + 1$ and $n + 2$, considering the condition that $n &gt; 2$.
26619
I was thinking of how to prove this -  Let $p_{k}$ be the $k$th prime.Then to show that there exists infinite number of $k$ such that $p_{k+1}-p_{k} &gt; 2$.  I was thinking of Twin Prime Conjecture stating that there are an infinite number of primes differing by 2, but how can it help with infinite number consecutive primes differing greater than 2?  I was thinking of going through the method of contradiction but could not proceed
28740
As in the title: Let $q$ be a positive integer, such that for all primes $p$, greater than a given natural number $N_q$, $p \equiv 1 \ \ (\text{mod } q)$. Then $q$ equals 2. I've thought about using some clever fact about the size of $(\mathbb{Z}/p\mathbb{Z})^*$ or $\phi(p)$ but alas nothing seems natural. Any help on how to proceed? Many thanks.
75422
I need help with this question.     Let $a$ and $b$ be positive integers such that $a^{n}+n$ divides $b^{n}+n$ for every natural number $n$. Show that $a=b$.   Any help would be appreciated! Thanks!
102177
I'm back again So there's another problem that I can't get to prove  If we take 21 numbers randomly from $1, 2, 3, ..., 40$ then between those $21$ numbers we will be able to find two numbers, of which the smaller one will divide the bigger one  I've been reading james hein "discrete structures, logic and computability" but still can't get to think logically myself. I would be very grateful if I could get some directions/tips/hints or anything, thanks in advance
84012
I need help with this question: There are infinitely many prime numbers $p$ for which $p + 2$ and $p + 4$ are also primes.  Where should I start? what proof techniques will be useful?  Any input will be very useful to me.
119566
Let x,n ∈ $\mathbb{N}$  such that :  1+$x^1$+$x^2$+$x^3$+....+$x^{n-1}$ is prime .  Proof that n is prime .  I actually suppose that n isn't prime absurdly so n=p×k such as p is prime ,I also use the $a^n$-$b^n$ formula  and find that $x^k$-1/$x^1$-1 so that means that k=1 and then  n=p which is absurd. Does Anyone has another answer ?
56603
I need help proving that for every positive integer $n$, there exist $n$ consecutive positive integers, each of which is composite. The hint that came with the problem is: Consider the numbers $2+(n+1)!,3+(n+1)!,...,n+(n+1)!,n+1+(n+1)!$.  I honestly have no idea how to begin here. I am specifically confused about what it means when it says "there exist $n$ consecutive positive integers". Does this mean that if $n = 5$, for example, then somewhere in the positive integers there are 5 consecutive composite integers, and that we want to prove that? I get that composite means that they are not prime, so they have more factors than 1 and themselves.  Any help would be greatly appreciated.  Thank you!
42691
Exactly what the name entails. The function $2^n-1$ I see largely tends to generate primes when $n$ is prime. However, a week ago I heard that this was horribly false. Please show me a disproof.
12122
Prove that for all $n&gt;2$ there exists at least one prime $p$ such that $\sqrt{n}&lt;p&lt;n$ using elementary methods.  My try: If not, $\sum_{p&lt;\sqrt{n}} (\lfloor n/p \rfloor-\lfloor \sqrt{n}/p \rfloor) \geq n-\sqrt{n}$  then...?
59309
Let $f(n)$ be the first $n$ digits of $\pi$ . How many times is $f(n)$ prime on the interval $[1, k]$ ? Are there infinitely many prime $f(n)$'s on $[1, \infty)$?  I know this is probably be a very hard question and most likely an open problem, but I am just curious.
57235
Let $p_n$ be the $n$th prime number in its natural order. We want to show that  $$p_{n+1} \leq p_1\dotsc p_n + 1 \tag{1}$$  I know $p_1\dotsc p_n + 1 &gt; 1$, so it has at least one prime divisor, but the text says that $p_{n+1}$ cannot exceed the smallest of these divisors, but I cannot understand why, especially when I imagine $p_{n+1}$ being very large. Although some examples confirm that this statement is true, I cannot make a connection as to why the $n+1$ prime number cannot exceed the smallest prime divisor. Once this has been shown to be true, $(1)$ immediately follows.
4565
I'm trying to figure out how to prove that if $p$ and $p^2+2$ are prime numbers then $p^3+2$ is a prime number too. Can someone help me please?