Let's $v$ the smallest positive integer > 1 for which $2 ^ {v} - 1$ divisible by $p$. Prove that $2 ^ {m} - 1$ ($m &gt; v$) divisible by $p$ then and only then $v$ divide $m$?  There is a hint that enough to divide $m$ into $v$ with the remainder, but it gives me nothing.
123308
I figured we'd have to show $a=qb$ from $2^a-1=r(2^b-1)$ for $q,r\in\mathbb{Z}$ for the first implication and vice versa, how should I proceed from here?
45924
When the polynomial $P(x)$ is divided by $x^2 -1$, the remainder is $3x-1$. What is the remainder when $P(x)$ is divided by $x-1$?  I'm not sure what to do. I know that $P(1)=P(-1)=3x-1$, but I don't think that's how you're supposed to do it.  Any help would be appreciated, thanks.
107
Prove that for all natural number $x$ and $n$, $x^n - 1$ is divisible by $x-1$.  So here's my thoughts: it is true for $n=1$, then I want to prove that it is also true for $n-1$  then I use long division, I get:  $x^n -1 = x(x^{n-1} -1 ) + (x-1)$  so the left side is divisible by $x-1$ by hypothesis, what about the right side?
71134
I understand the idea of the proof. I just want to make sure I wrote my proof well.  Suppose $n$ is not prime. Then $\exists x,y \in \mathbb{Z}$ such that $n = xy$.  $2^{xy} - 1 = (2^x)^y - 1$  $ = (2^y - 1)(2^{y(x-1)} + 2^{y(x-2)} + ... + 2^{y} + 1)$  Since $2^{n} - 1$ is divisible by $2^y - 1$ it must be that $2^n - 1$ is not prime. Contradiction. Thus $n$ must be prime.  How does this look?
64025
What is the greatest integer that divides $p^4-1$ for every prime number $p &gt; 5$?  This was on a practice math GRE so it's probably really easy.
61123
I need to show the following:  For any integer $x$, $x^2 + 4$ is not divisible by $3$.    I was trying proof by contraposition, but I do not believe that is the most efficient way to go about this.  Can anybody point me in the right direction, please?
40340
Prove that for any positive integers $m$ and $n$ , there exists a set of $n$ consecutive positive integers each of which is divisible by a number of the form $d^m$, where $d$ is some positive integer not equal to $1$.  I don't know how to approach this question.
79260
We are given that $m$ and $n$ are positive integers such that $\operatorname{lcm}(m,n) + \operatorname{gcd}(m,n) = m + n$.  We are looking to prove that one of numbers (either $m$ or $n$) must be divisible by the other.
16355
$P,Q,R,S$ are polynomials such that: $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$   I thought a lot on this but no result!! By the way,one idea is to insert some values for $x$ and try to produce a system of equations for the given polynomials,but I'm not sure it works.
75836
I am reading a math book. It states the following rule:      For an integer $n$ greater than 1, let the prime factorization of $n$ be $$n=p^a_1p^b_2p^c_3...p^m_k$$    Here $a, b, c,..., m$ are nonnegative integers, and $p_1,p_1,...,p_k$ are prime numbers. Then the number of divisors of $n$ is: $$d(n)=(a+1)(b+1)(c+1)...(m+1)$$   I can use the rule to solve problems correctly, but I don't understand the rule. How to prove it?
70311
I need to show that $n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer without using concepts of abstract algebra  I know that $$a^n\equiv 1\pmod {a^n-1}$$  How do I proceed from there?
117611
Let $p$ be a prime such that $a^p-1 \equiv 0 \pmod{p}$. Prove that $\gcd(a,p) = 1$.   We know from Fermat's Little Theorem that $a^{p-1}-1 \equiv 0 \pmod{p}$ if and only if $\gcd(a,p) = 1$, but how do we use this to solve the question?
94023
I need your help with this: Let $f$, $p$, and $g$ be a polynomials in $F[x]$, assume that $f$, $g$, and $p$ are nononzero.  If $\gcd(f,p)=1$ and $fg$ is divisible by $p$ with no remainder, I need to prove that $g$ is  divisible by $p$ with no remainder  Thanks.
40231
Let $p$ be prime. Consider the sequence: $1, 2^2, 3^3, 4^4, 5^5, 6^6 \cdots \pmod p$. Prove the sequence is periodic with smallest period $p(p-1).$ This means that $p(p-1)$ is the least among all positive integers $l$ with the property that whenever $n ≡ m (mod ~l),$ we have $n^n ≡ m^m (mod ~p)$ Prove that: (a). $n \equiv m \pmod {p (p-1)}$ implies that $n^n ≡ m^m \pmod p$ $(b).$ If $l ≥ 2$ has the property that whenever $n ≡ m \pmod l,$ we have $n^n\equiv m^m \pmod p,$ then $p(p-1) \leq l.$
66574
Show that $13$ divides $2^{70} + 3^{70}$.   My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two?  Thanks!
73568
Let $A_{n \times n}$ be a $n$ by $n$ matrix with all entries equal to $\pm 1$. I want to show $\text{det}(A)$ is divisible by $2^{n-1}$. How can I appraoch this?
73489
I know how to prove the fact that $ p$ | ${ p \choose k } $ (when writing it as a fraction, $p$ cannot be divided by any of the $1\times2\times...\times k$ or $1\times2\times...\times(p-k)$ because $p$ is prime).  When I try to apply the same rationale for $ p$ | ${ p^n \choose k } $ I get stumped because $p^n$ is in fact divisible by one of the $k! = 1\times2\times...\times k$, where $0 &lt; k &lt; p^n$.  If we take the example of $2^3$ and $k = 7$, we can easily see that $k! = 1 \times 2 \times ... \times 6 \times 7$, and $2$ clearly divides $2^3$.  How else can I approach this?