Let $p$ and $q$ are two prime numbers. Also, let us assume $q|(2^p-1)$. Then show that $(2^p-1,2^{q-1}-1)=2^{(p,q-1)}-1$. Note- $(p,q)$ denotes HCF of $p$ and $q$.
2481
For all $a, m, n \in \mathbb{Z}^+$,   $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
12560
If $p$ and $q$ are distinct primes such that $p|c$ and $q|c$, prove that $pq|c$ ..  I start with this:   if $c=pk$  and $c=qt$ then $q|pk$ and $p|qt$  but a couldn't get the solution.  I need your help
61754
Problem:   If $x^{n-1} \equiv 1 \pmod{n}$, and for all divisors $q$ of $n - 1$, $a^{q} \not\equiv 1 \pmod{n}$, then $n$ is prime. $(n &gt; 1)$   I read the proof in the book and it was very straightforward; however, I wonder is there a way to prove it by just using congruence property?    And another related question about power residue: If we have $a^{n - 1} \equiv 1 \pmod{n}$. Is there any relation between $n - 1$ and $\phi(n)$?  Because I thought of $a^{\phi(n)} \equiv 1 \pmod{n}$, when $(a, n) = 1$. I try to think of away to connect these two ideas, but I still cannot see it. Any idea?    Thanks,
6856
Here is a problem which i have not been able to do for quite sometime.   For $n \geq 2$, show that $n \nmid 2^{n}-1$.   I have thought of proving this in two ways: One by using induction which didn't actually work. Next by Fermat's little theorem we have $2^{p-1} \equiv 1 \ (\text{mod} \ p)$, which actually says that for $p \mid 2^{p}-2$. But i couldn't proceed more than this. Any ideas by which i can actually solve the problem?
22046
I got a question to show that :      If $p$ is prime number, then   $$x^p - x \equiv x(x-1)(x-2)(x-3)\cdots (x -(p-1))\,\,\text{(mod }\,p\text{)}$$    Now I got 2 steps to show that the two polynomials are congruent:   Show they have same roots. Show the leading term's coefficients on both sides are the same.    For step 2, I know it is obvious $x^p$ has coefficient 1 on both sides.  For step 1, but I am not sure how to find all the solutions for $x^p -x$ modulo $p$.
27033
Let $p$ be prime number ($p\gt2$) and $a,b\in\mathbb Z$ ,$a+b\neq0$ ,$\gcd(a,b)=1$ how to  prove that $$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1~~\text{or}~~ p$$  Thanks in advance .
72644
Prove that for $ n \geq 2$, n has at least one prime factor.  I'm trying to use induction. For n = 2, 2 = 1 x 2. For n > 2, n = n x 1, where 1 is a prime factor. Is this sufficient to prove the result? I feel like I may be mistaken here.
30113
Take $p_1, p_2, \ldots, p_n, p_{n+1}$ be $n+1$ prime numbers in $\mathbb{P} \subseteq \mathbb{N}$. $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ seems to be quite clear, but still need a proof. I know some proofs are involved with Galois theory, which is not I want.
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?
43718
Suppose that $5\leq q\leq p$ are both prime. Prove that $24|(p^2-q^2)$.   This is what I got so far. I figured that since $p,q$ are bigger than $5$, there are only odd primes for this conjecture. Then I can rename $p=2m+1,q=2n+1,m,n∈Z$.  We want to show $(p^2-q^2 )=24k,k∈Z$. By substituting $2m+1$ for $p$ and $2n+1$ for $q$, we get  $((2m+1)^2-(2n+1)^2)$. If we expand it, we get $4(m^2+n^2+m-n)$ Q. I think I can make $4(m^2+n^2+m-n)$ look like $24k$. How should I approach ?  And my prof. commented that "I would take this proof in small steps.  Rather than trying initially to prove that $24 | p^2-q^2$, look at the progress you already have.  You proved $4 | p^2-q^2$.  Can you prove 8 is a factor?  What about 3?  Then see if you can put together what you have to get 24 as a factor."  I am stuck here. Any helps?
68719
If $p$ and $q$ are distinct primes and $a$ be any integer then $a^{pq} -a^q -a^p +a$ is divisible by $pq$.  Factorising we get $a^{pq} -a^q -a^p +a =a^p(a^q -1) - a(a^{q-1}-1)$ and we know $p \mid a^{p-1}-1$ and $q \mid a^{q-1}-1$.  I can't proceed further with the proof. Please Help!
95065
let $p$ is odd prime numbers, and there exsit postive ineteger $a$ such $$p!|a^p+1$$  show that $$p!|(a+1)$$  I think maybe can use $$a^p+1=(a+1)(a^{p-1}-a^{p-2}+\cdots-a+1)$$ then I can't Continue
33518
If $p$ and $p^2 +8$ are both prime number, prove that $p^3 +4$ is also prime.   I was thinking using two cases for even and odd. So $p$ is even thus there is no prime, the statement does not hold. So, $p$ is odd, $p$ has to be of the form of $2k+1$.   let $p=2k+1$ which is prime  $(2k+1)^2 + 8 = 4k^2+4k+1+8 = 2(2k^2+2k+4)+1$ which is prime   so $(2k+1)^3 +4 = 8k^3+12k^2+6k+1 +4= 2(4k^3+6k^2+3k+2)+1$ which is also prime.   is that all that i need to show
48933
Find all pairs of primes $p,q$ such that $pq \mid 2^p +2^q$.  My attempt  : When either one of them is $2$ then easy case checking gives me set of solutions. But what happens when neither of them is $2$?
107652
Find a prime number $p$ and an integer $b&lt;p$ such that $p$ divides $b^{p−1}−1$.  First I think of long divisions but it didn't work out. Now I'm stuck..
37176
Deduce that there is a prime gap of length $\geq n$ for all $n \in \mathbb{N}$   Show that if $2^n - 1$ is prime, then $n$ is prime.   Show that if $n$ is prime, then $2^n - 1$ is not divisible by $7$ for any $n &gt; 3$.       I'm not really sure how to do the first bit. For the second one, I'm not sure if this is correct but I have this: If $n$ is not prime, then $n = a b, a, b &gt; 1$. Also $2^n - 1 = (2 - 1)(1 + 2 + 2^2 + .... + 2^{n-1})$. So if we substitute for $n = ab$, we get $(2^a)^b - 1 = (2^a - 1)(1 + 2^a + (2^a)^2 + .... + (2^a)^{b - 1})$ which is now composite and so if $n = ab$, then $2^n - 1$. This means that $2^n - 1$ is prime if $n$ is prime. For the last one, we have a hint "Follow the example in lectures to show that $2^n - 1$ is not divisible by $3$", but I missed this lecture and haven't had a chance to get the note, so can someone help me with the divisible by $3$ bit and then I can try the divisible by $7$ bit myself.   Thank you  EDIT: I found a simpler proof for the second bit online:     Supposed $n$ is composite, it can be then written in the form $n = ab, a, b &gt; 1 \implies (2^a)^b - 1$ is prime with $b&gt;1$ and $2^a&gt;2$, contradicting statement $1$.   What is statement $1$? Maybe then I can understand why it contradicts it.
81606
I have come across this question, which reminded me of Fermats little theorem, i dont know if the Fermats theorem is actually in use in the following mathematical statements   an integer a is a coprime with p and a coprime with q (p and q are different prime numbers )then prove that  $$a^{(p-1)(q-1)} \equiv 1(\operatorname{mod} pq) $$  $$p^{q-1}+q^{p-1} \equiv 1(\operatorname{mod} pq)$$  any help would be appreciated, thanks in advance.