After doing some numerical test, I find $10^{k} + 1$ is not prime unless k = 1, 2. I know that If k is odd, then $11 \mid 10^{k} + 1 = (11 - 1)^k - 1$, and if k is even and not a power of 2, then using the same method, $10^{k} + 1$ is not prime. So the only case is to check whether there exist some k &gt; 1, such that $10^{2^k} + 1$ is prime?
85270
I have observed that all numbers of the form $10^{k} + 1$ are composite for $k{\gt}2$, and am trying to prove the same.   Progress  For odd $k$ , it is fairly obvious, as we get 11 as a divisor.  I am having difficulty with $k$ even. Have no clue how to approach. Would prefer a non-inductive approach, if possible  Developments  Thanks to crostul, we can say it is composite for $k$ having an odd divisor. Hence, it only remains to prove it for $k = 2^n$
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?
81964
Why is $1$ not considered a prime number?  Or, why is the definition of prime numbers given for integers greater than $1$?
32497
So basically we have a number $10101...10101$ that contains $2016$ zeros and can be written as$ \sum _{ k=0 }^{ 2016 }{ 100^{ k } }$ . I want to prove that this number is not a prime without using anything besides a piece of paper and a pen. I'm stuck on this for quite a few days now.
36993
Prove that if $p$ and $q$ are odd primes and $q$ can divide $a^p-1$, then either $q$ divides $a-1$ or else $q=2kp+1$ for some integer $k$.  I thought the supposition is $a\equiv 1 (mod\ q)$. So $$(a-1)(a^{p-1}+a^{p-2}+...+1) \equiv 0 (mod\ q)$$ Then I proved $q$ may divide $a-1$. But I don't know how to check that $q=2kp+1$. thks
118685
I remember coming across this fact a while ago in a pdf somewhere, but I haven't been able to find it again. Can someone show me how to prove it? I would appreciate easier proofs.  EDIT: I'm very sorry, I forgot to actually state it.  Prove that $-1$ is not a quadratic residue modulo primes of the form $4k+3$.  Thanks!
36335
I need to show that if $(n-2)! \equiv 1 \pmod n$ then $n$ is prime. I think that if $n$ is composite, then we'll have every factor of $n$ in $(n-2)!$, and it would yield that $(n-2)! \equiv 0 \pmod n$. However, I didn't use the fact that it specifically congruent to $1 \bmod n$, so I think I'm getting something fundamental wrong. Is my solution correct? Why do we demand congruence to $1 \bmod n$?
86566
Fix $k \in \mathbb{Z}_+$. Prove that we can find infinitely many primes of the form $a2^k +1,$ where $a$ is a positive integer. We can use the result that: If $p \ne 2$ is a prime, and if $p$ divides $s^{2^t}+1$, for $s &gt; 1$ and $t \ge 1$, then $p \equiv 1 \pmod {2^{t+1}}$.  Ive trying to get something inductively: For $k = 1$, there are infinitely many primes of the form $2a + 1$. Suppose there are infinitely many primes of the form $a2^k + 1$, and then show that there are infinitely many primes of the form $a2^{k+1} + 1$. If there are infinitely primes of the form $a2^k + 1$, where $a$ is even, then we have that, $a2^k + 1 = (2q)2^k + 1 = q2^{k+1} + 1$. Hence we are done. Therefore, suppose there are only infinitely many primes of form $a2^k + 1$, where $a$ is odd. - but I can't get a contradiction out of this.
53347
Prove that if $m\not =n,a$ are positive integers then $(a^{2^n}+1,a^{2^m}+1)$ is $1$ if $a$ is even and $2$ if $a$ is odd.    I solve the problem in the following way: I assume that $m&gt;n$ then  $$a^{2^m}+1=a^{2^m}-1+2$$ With this expression if I take a divisor prime $p$ of $a^{2^n}+1$ then if $p|a^{2^m}+1$ this prime must divide 2 and the result follows.   But, how can I prove this statement without use of primes? This exercise appears before the concept of prime is introduced then I prefer do not use it.    Any idea?   Thanks!
855
If $n&gt;1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.  If you know , then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.  Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.
39607
Prove that there are infinitely many primes and non-primes in the numbers $10^n+1$, where $n$ is a natural number. So numbers are 101, 1001, 10001 etc.
19878
Let $p, q$ be prime numbers and $n ∈ \mathbb{N}$ such that $p \not\mid n- 1$.  If $p \mid n^q -1$ then show that $q \mid p-1$.   My ideas: I was thinking about the formula $$n^q -1 = (n-1)(n^{q-1} + n^{q-2} +\cdots+1),$$ but now here I do not know how to show that $q \mid p-1$.  Any hints or solution can be appreciated.
25735
for $N \ge 4$. Show for prime numbers,  $p \equiv 1$ mod $(N!)$ that none of the numbers $1,2,...,N$ are primitive roots modulo $p$   I can't figure out where to start with this question, all I can think to use is the Legendre symbol and Euler's Criterion but I haven't been able to do it. Any help would be much appreciated.
122999
Is there an easy way of showing that $2^n \pm 1$ is never a perfect power, except for $2^3 + 1 = 3^2 $?  I know that   gives the result directly, but I'm hopefully for a more elementary method.    I can show that it is never a square, except for $2^3 + 1$.  Proof: Cases $n=1, 2, 3$ are easily dealt with. Henceforth, let $n\geq4$.  $2^n -1 \equiv 3 \pmod{4}$ hence is never a square.     If $2^n +1 =x^2$, then $2^n = (x-1)(x+1)$ and both of these are powers of 2. Thus we must have $(x-1) = 2, (x+1) = 4$. This gives the solution of $2^3 + 1 = 3^2$.    Let's now do an odd prime.  Say $2^n +1 = x^p$. Then $2^n = x^p - 1 = (x-1)(x^{p-1}+x^{p-2} + \ldots +1)$, so both terms are powers of 2. We have $ x = 2^m+1$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since this is clearly greater than 1, we get a contradiction.  Say $2^n -1 = x^p$. Then $2^n = x^p + 1 = (x+1)(x^{p-1} - x^{p-2} + \ldots -x +1 )$, so both terms are powers of 2. We have $x = 2^m -1$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since $x^p + 1 \neq x+1$ except for $p=1$, this means that the term is greater than 1. Hence we get a contradiction.
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
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?
38510
Prove: If a prime number $p\in \mathbb N$ is from the form $p=4k+3,k\in \mathbb N$, then its also a prime number in $\mathbb Z[i]$,i.e. if $p|(z_1\cdot z_2)$ then $p|z_1$ or $p|z_2$.  I dont have any idea how to solve it, so I am looking for some hints.  Thanks in advance :)