How can we prove that if : $n+1|n!+1$ then $n$ is a prime number. My try: I try to do contrapositive: If $n$ is composite then $n+1 \nmid n!+1$ and then to get that if $n$ is composite so $n+1=ab$ when $2 \le a \le n $ so that $a|n+1$ but here I don't have idea how to continue, is this the good way?
122670
... namely that if $n &gt; 1$ and $(n − 1)!\equiv−1\pmod{n}$, then $n$ is prime.  This is for a number theory class I'm in at Penn State. My idea is to follow accordingly, but I can't get it concretely..  Reason by contradiction, suppose that $n$ is not prime.   Then $n = ab$, where $1 &lt; a, b &lt; n$  In particular $(n-1)! = ai$ for some $i\in\Bbb{Z}$. $\\$ If $(n-1)! = nj - 1$ for some $j\in\Bbb{Z}$ then it follows that a probably doesn't divide $-1$?
29324
We know 11...11 (n times) is a prime number prove n is also prime.  it is easy to understand n is an odd number and 3 does not divide it but I could not do some thing more
105585
$n &gt; 2$ and $n$ is a prime number then $n$ is odd.  Prove by contradiction assume $n$ is even then there is some $k\in\Bbb N, n = 2k$ then $n &gt;2, 2k &gt; 2 , k &gt; 1$  Is this  a sufficient proof?
104786
Problem: If $p\geq 5$ is a prime number, show that $p^2+2$ is composite.  Remarks: Now if one observes that $p$ takes the forms $6k+1$ and $6k+5$, the problem is resolved quite easily. However, if one were to choose other forms say $4k+1$ and $4k+3$ then $p^2$ would be of the form $4p+1$ which would yeild the general form $4p+3$ on addition with $2$. Obviously this does not lead to any concrete conclusion. I was wondering whether there is any specific procedure involved in finding the right quotients (For eg. $6$ and $4$) or is this just a random problem solving trick. Moreover, if anyone could provide some intuition as to why $6$ works would be much appreciated.
75474
In Euclid's proof that there are infinitely many primes, the number  $p_1 p_2 ... p_n + 1$ is constructed and proved to be either a prime, or a product of primes greater than $p_n$.  Trivially, we could also use the number $R_n=p_1 p_2 ... p_n - 1$ to prove the theorem, for n>2.  Intuitively, as $n$ grows, the probability that $R_n$ is prime gets smaller. Is there a proof that $R_n$ is not prime for any $n$ greater than some integer $M$ ? Or conversely, that there is an infinite number of prime $R_n$ numbers?  A possibly equivalent question: Is there a prime number greater than $p_n$ and smaller than $R_n$ for any $n$ greater than some integer $M$ ?
66390
I am stuck on this problem, and was wondering if anyone could help me out with this. The question is as follows:  Let $n$ be an integer such that $n ≥ 1$. Prove that $6$ divides $n(n + 1)(n + 2)$. Note: An integer $a$ divides an integer $b$, written $a|b$, if there exists $q ∈ Z$ such that $b = qa$. Alternatively, $a|b$ if dividing $b$ by $a$, $b ÷ a$, results in an integer.  Should I do a proof by induction? All help/input is appreciated!
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?
107011
Given two numbers $a$ and $b$, where $a$ is co-prime to $b$,     Show that for any number $c$, if $a|c$ and $b | c$ then $ab| c$. Is the reverse also true? In other words, if $ab |c$ and $a$ is co-prime to $b$, then do we have $a | c$ as well as $b|c$?
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Are there any theorems that give us any information about the prime factorization of some integer $n+1$, if we already know the factorization of $n$?  Recalling Euclid's famous proof for the infinity of the set of prime numbers, I guess we know that if $n = p_1 p_2 p_3$, then $n+1$ cannot have $p_1$, $p_2$, or $p_3$ as factors. But is there any way we could use the information about $n$'s factorization to determine something more precise about the factorization of $n+1$?
104310
I need to prove that for every natural n that happens and the prove needs to be by induction, I wont prove for n=1 because its obvious,so lets move to n and n+1 $$ \frac{(2n!)}{(n!)^2} ≥ \frac{(4^n)}{(2n+1)} $$ the assumption for n+1 $$ \frac{(2n+2)!}{(n+1)!^2} ≥ \frac{(4^{n+1})}{2n+3} $$ now I dont now how to progress,I tried this for n+1 $$ \frac{(2n+2)(2n+1)(2n!)}{(n+1)!^2} ≥ \frac{(4^{n}4^1)}{(2n+1)+2)} $$ and had a suggestion to use $ 2n! = (n!) \prod _{n=n}^{2n}\:n\cdot )! $ but cant see how it helps,ty.
92631
The question is:      Prove that if $p \in \mathbb{Z}$ is irreducible, then $p$ is also prime.    Irreducible is defined as follows: $n \in \mathbb{Z} $ is irreducible iff the only way to write $n=ab$ for some integers $a$ and $b$ is if $a = \pm1$ or $b = \pm1$. Prime is defined as follows: an integer $n$ is prime iff whenever $a,b \in \mathbb{Z} $ and $p\mid ab$ then it must be true that $p\mid a$ or $p\mid b$ (or both).    Hint: Prove that if $n\mid ab$ but $n$ does not divide $a$, then $\gcd(a,n) = \pm1$. Then use Bézout's theorem and prove $n\mid b$.
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.
13594
Can $n!$ be a perfect square when $n$ is an integer greater than $1$?  Clearly, when $n$ is prime, $n!$ is not a perfect square because the exponent of $n$ in $n!$ is $1$.  The same goes when $n-1$ is prime, by considering the exponent of $n-1$. What is the answer for a general value of $n$?  (And is it possible, to prove without Bertrand's postulate. Because Bertrands postulate is quite a strong result.)
41006
Recall that if $n$ is a positive integer, $n! = n(n-1)\cdots 3\cdot2\cdot1$.  a) Let $n$, $k$ be positive integers with $1&lt; k \le n$. Prove that $n! + k$ is composite.  b) Using part a, find $100$ consecutive integers all of which are composite.  c) In part b you gave a sequence of $100$ consecutive integers all of which are composite. You should be able to use the same idea to find $\ell$ consecutive integers which are composite for any positive integer $\ell$. In other words, you can find sequences of any length which consist of consecutive integers and contain no primes. Does this contradict the fact that there are an infinite number of prime numbers?
24138
I was doing some basic Number Theory problems and came across this problem :     Show that if $a$ and $n$ are positive integers with $n\gt 1$ and $a^{n} - 1$ is prime, then $a = 2$ and $n$ is prime   My Solution : (Sloppy)        $a^{n}-1$ = $(a-1)$ . $(a^{n-1} + a^{n-2} + ... + a + 1)$   This means that $(a-1)$ | $(a^{n}-1)$    But $(a^{n}-1)$ is prime   So , $(a-1)$ = 1 $\Rightarrow$ $a = 2$   Now , let $n$ be composite   $n = kl$ , where $1 \lt k \lt n$ and $1 \lt l \lt n$   Now , $a^{kl}-1$ = $(a^{k}-1)$ . $(a^{k.{(l-1)}} + a^{k.{(l-2)}} + ... + a^k + 1)$   This means that $2^{n} -1$ is composite    Hence , we have achieved a contradiction      My Question : Am I correct ?
114649
I'm looking prove using an indirect proof that:  If an integer $q &lt; 1$ is composite, then $q$ will have some divisor $d$ such that $d &gt; \sqrt{q}$.  I really just don't know how to even start this thing, most examples i've seen or done have been much more trivial than this.  My best guess is that I must assume $q &gt; 1$ is prime and that it has a divisor d such that $d &gt; \sqrt q$?  What then can I say about this statement that would prove, im assuming by contraposition that the about statement is true?  Any help with this would be greatly appreciated, I know it's a lot to ask.
74289
Prove that if a number $n &gt; 1$ is not prime, then it has a prime factor  $\le \sqrt{n}$.  My answer is that this is not always true, because you can pick a non-prime number that is greater than $1$ and its factor is equal to its square root.