How to prove for any prime $$p \neq \{2,7\}, \exists n \text{ s.t. }p \mid n$$ where n is a number that consists only of 1s in octal base.  I started off but need some direction. I thought of using  $$ 8^{p-1} - 1 \text{ (mod p)} $$  and that  $$ 8^{p-1} - 1 $$ given $(p-1)$ number of 7s in octal form; and that this divided by 7 is all 1s in octal form.
110285
Prove that every number ending in a $3$ has a multiple which consists only of ones.      Eg. $3$ has $111$, $13$ has $111111$.   Also, is their any direct way (without repetitive multiplication and checking) of obtaining such  multiple of any given number( ending with $3$) ?
56575
how to prove $\forall p$ prime :   $\binom{2p}{p} \equiv 2 \pmod p$ we have:  $\binom{2p}{p} = \frac{2p (2p-1)(2p-3)...1}{p!p!}$ but how to continue?
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.
29486
I have to prove this : "$n^7-n$ is a multiple of 7". This is what I have done this so far :  $P(n):n^7-n$ On putting $n=1,$ $P(1):1^7-1=0$, which is a multiple of 7. So, $P(1)$ is true. Let $P(k)$ be a multiple of 7. So, $k^7-k$ is a multiple of 7. So, $k^7-k=7m$, where $m$ is any natural number. Let $P(k+1)$ be true, Hence, $(k+1)^7-(k+1)$ is a multiple of 7.  Now, how do I evaluate $(k+1)^7$ without using the binomial theorem ? Or is there some other way to do it ?
104354
I first noticed this is true for the integers of the sequence $9, 99, 999, 9999,\dots$, since for some term $a_n=10^n-1$ in the sequence and $p$ a prime other than $2$ or $5$, we have $a_n\equiv 0 (mod\ p)\Leftrightarrow 10^n\equiv 1 (mod\ p)$ and we can find infinitely many $n=k(p-1)$ so that $10^n=(10^{p-1})^k\equiv 1^k\equiv 1 (mod\ p)$ by Fermat's little theorem.   I'm curious, is there a similar way to show this is true for repunits $1, 11, 111,\dots$?
112475
I am trying to prove that ($\forall \ n\in\mathbb{N}$) there exists a prime number $q$ such that $n &lt; q \le 1 + n!$  I have made a graph with $n=0$ through $n=10$ and found solutions to all of them looking for a pattern and I see that $n!$ gets enormous fast and it becomes quite obvious that there is a prime number in between them.    I have considered trying to prove by contradiction that $q$ does not exist on that interval, but I don't know where to go from that statement.  Could anybody help me figure it out?  I have been staring at it for hours and I can't figure out where to go.  Thank you.
33291
If $p$ and $q$ are odd primes show that $$p|(9^q -1) \implies q|(p-1).$$ ....  I am having difficulty to prove this.  Yes that was the question I have misunderstood between 9 and q!!!
92895
I feel like there must exist a combinatoric proof of a theorem like: There is a prime between $n$ and $2n$, or $p$ and $p^2$ or anything similar to this stronger than there is a prime between $p$ and $(\prod_p p) + 1$ (Euclid's theorem).  I was trying to prove one by the Sieve on this grid      1     2     3     4 ... p   p+1   p+2   p+3   p+4 ... 2p  2p+1  2p+2  2p+3  2p+4 ... 3p ........................... ........................... ........................... p^2 p^2+1 p^2+2 p^2+3 p^2+4 ...   but it did not work.  Do any good arguments like this exist? I don't expect anything as strong as Prime Number Theorem or even Bertrand, but surely a direct combination proof can prove that there are lots of primes?
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
957
Show that    $$1^{p-1} + 2^{p-1} +\ldots + (p-1)^{p-1} \equiv -1 \mod p$$   So, I use Fermat's little theorem, that is if $p$ does not divide $a$, then $a^{p-1}$ is congruent to $1$ (mod $p$). But I'm still confused. If $1^{p-1}$ is congruent to $1$ (mod $p$), $2^{p-1}$ is congruent to $1$ (mod $p$), and $p$ does not divide $p-1$, so $(p-1)^{p-1}$ is congruent to $1$ (mod $p$). How does that equal $-1$ (mod $p$)?
68761
I found the following question and can't make much headway with it.  Show that there exist infinitely many positive integers $n$ such that $2^n \equiv n \mod p$ where $p$ is a odd prime...  I started by writing the number as $n=(p-1)k+c$ which implies $$2^c \equiv c-k \mod p$$ where I used Fermat's Little theorem. Now if $k$ is of the form $pm$ for some integer $m$ we get $n=(p-1)(p)m+c$   $$2^c \equiv c \mod p$$ and we have reduced $n$ to some smaller integer $c$ but I can't figure out how to find the smallest number $c$ which would then generate infinitely many $n$.  Any help in finding a possible approach to this would be highly appreciated.
103641
Given that $p$ is a prime and $p\mid a^n$, prove that $p^n\mid a^n$.   I know that the fundamental theorem of arithmetic states that any positive integer can be represented as a product of primes but how do I apply this to the proof?
96129
I know how to prove that, if $2^n-1$ is prime and $n&gt;1$, then $n$ is prime.  But how do we prove that, if $a^n-1$ is prime and $n&gt;1$, then $a$ must equal 2?
111147
Prove that there is no arithmetic progression that consists only of primes.  A question that I've been set; I'm guessing it makes use of primes being written in the form 4k+1 and 4k+3? Not sure where to start.  Thanks.
62238
If $2^p-1$ is a prime, (thus $p$ is a prime, too) then $p\mid 2^p-2=\phi(2^p-1).$  But I find $n\mid \phi(2^n-1)$ is always hold, no matter what $n$ is. Such as $4\mid \phi(2^4-1)=8.$  If we denote $a_n=\dfrac{\phi(2^n-1)}{n}$, then $a_n$ is , but how to prove it is always integer?  Thanks in advance!
37470
To prove the following fact: given any sequence of digits in any base, eg 314159265358979323 base 10, there are infinitely many primes that start with these digits,eg when expressed in decimal they start with 314159265358979323. I think using a prime number theorem will be helpful . But I am still not getting a point of how to start the proof.
20547
Prove if $p$ is a prime then $p \mid \binom pk$ for $k\in\{1,\ldots,p-1\}$  I don't really know where to begin with this one.   I can see that I have to use the fact that $p$ is prime somewhere - the same is not true for composite numbers, for example $4\nmid 6=\binom42$.  I have checked that this is true for the first few primes:    $3$ divides $\binom 31=\binom32=3$ $5$ divides both $\binom 51=\binom 54=5$ and $\binom 52=\binom 53=10$ $7$ divides $\binom71=\binom76=7$, $\binom72=\binom75=21$ and $\binom73=\binom74=35$.