I am trying to solve the following exercise of Dummit and Foote Book(page # 551).   Let $a&gt;1$ be an integer. Prove for any positive integers $n,d$ that $d$ divides $n$ if and only if $a^d-1$ divides $a^n-1$. Conclude in particular that $\mathbb{F}_{p^d}\subseteq\mathbb{F}_{p^n}$ if and only if $d$ divides $n$.  I did the first part and I know that for all $\alpha\in \mathbb{F}_{p^d}$, $\alpha^{p^d}=\alpha$. How can I apply the first part for the second? Any help is greatly appreciated. Thank you.
123033
I am looking to prove that $GF(p^n)$ contains a unique subfield isomorphic to $GF(p^m)$ if and only if $m$ is a divisor of $n$. In the Wikipedia article:    They say that this statement has been proven by E. H. Moore in 1893.  How to go about proving the theorem? I know first I need to generate a subset, prove that it's a subfield, and then assume they are two and show it's unique.  But how to generate it in the first place? Consider the set of polynomials of the form $x^{p^m}-x=0$? Is there a way to show that it's a subfield without checking all the axioms?
1132
Show that $n$ does not divide $2^n - 1$ where $n$ is an integer greater than $1$?  Clearly $2^n - 1$ is an odd integer, therefore, let $n$ be an odd integer and it divides $2^n - 1$. We can write $2^n - 1 = (2-1)(2^{n-1} + \cdots \cdots + 1) = (2^{n-1} + \cdots \cdots + 1)$ From here on I don't know how to proceed?
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!
31274
This is from Pinter, A Book of Abstract Algebra, p.265.   Given $p(x) \in F[x]$ where $F$ is a field, I would like to show that $p(x)$ divided by $x-c$ has remainder $p(c)$.  This is easy if $c$ is a root of $p$, but I don't see how to prove it if $c$ is not a root.
37253
Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$. I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt and if it's the former part, please verify my solution-n=1.
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!
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?
107216
I think prime factorization is needed for this question:   Suppose $a$, $b$ and $n$ are positive integers. Prove that $a^n$ divides $b^n$ if and only if a divides $b$.
50999
How would I formally prove that for the integers $a$, $b$, and $d$, if $d\mid da+b$, then $d\mid b$? Would a direct proof be the best option? If I do a direct proof I seem to get stuck pretty quickly... in fact I don't even know how to start...   Assume $d\mid da+b$ ... ?  This might look like a HW dump but I honestly just don't know how to continue...
41202
Let $d=(a^n,b^n), d_1=(a,b)^n$; need show $d\mid d_1$ &amp; $d_1\mid d$ to prove.  For the general case of $(a,b)\ne 1, (a,b)= (d_1)^\frac{1}{n}$, else $(a,b)=d_1$.  Let, $d_2= (a,b)=d_1^\frac{1}{n}\implies a=a'd_2, b = b'd_2\implies a^n= (a')^nd_2^n, b^n= (b')^nd_2^n$  Unable to progress further.
12276
I'm asked to show that $a^{61} \equiv a\pmod{1001}$ for every $a \in \mathbb{N}$. I've tried to tackle this using Fermat's Little Theorem and Euler's Theorem, but I can't even get started. My main problem seems to be the "for every $a \in \mathbb{N}$" part, because if it restricted $\gcd(a, 1001) = 1$ I would be able to factor out 1001 and try to show that it's factors divide $a^{61} - a$. I'm kinda lost here. Any help is appreciated.
65408
Let $a,b,m,n$ be positive integers such that g.c.d.$(a,b)=1$ and $a^m+b^m|a^n+b^n$ , then is it true that $m$ divides $n$ ?
46790
So I've said $x^{2}+1$ is reducible over $\mathbb{F}_{p^{n}} \iff \mathbb{F}_{p^{n}}$ contains a root $\alpha$. Hence if $\alpha$ is such a root then $\alpha^2 = -1$ so that $\alpha^4 =1$ and hence $|\alpha|=4$ in $\mathbb{F}^{\times}_{p^{n}}$ and hence $4\mid p^n -1 \implies p^n \equiv 1\bmod 4$. Hence if $p^{n}$ is congruent to $0,2,3$ mod $4$ then $x^{2}+1$ is irreducible over $\mathbb{F}_{p^{n}}$.  My problem is with $p^{n}$ congruent to $1$ mod $4$ being sufficient to conclude that $x^{2}+1$ is reducible. How can I show this?
44709
I'm trying to prove using induction that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:   Base case: $P(1) = (1) + (8) + (27) = 36, 36$ can be divided by $9$ so the base case is valid Inductive step: let $P(n)$ be the statement $9$ divides $n^3 + (n+1)^3 + (n+2)^3$. Assume $P(k)$ is true, so $9$ divides $k^3 + (k+1)^3 + (k+2)^3$.   And this is where I'm stuck. I'm not sure how to demonstrate that $9$ divides $P(n)$ when $n = k+1$. If someone could step me in the right direction that would be awesome. Thanks!
108664
Let $\mathbb N^*$ be the set of positive integers and $n$ be a fixed positive integer. How many functions $ f : \mathbb N^*\to\mathbb N^*$ are there such that for all positive integers $x$ and $y$, $f(x) + f(y)$ divides $x^n + y^n$?
38542
The function $d(n)$ gives the number of positive divisors of $n$, including n itself. So for example, $d(25) = 3$, because $25$ has three divisors: $1$, $5$, and $25$.  So how do I prove that the product of all of the positive divisors of $n$ (including $n$ itself) is $n^{\frac{d(n)}{2}}$.  For example, the divisors of $12$ are $1$, $2$, $3$, $4$, $6$, and $12$. $d(12)$ is $6$, and $1 · 2 · 3 · 4 · 6 · 12 = 1728 = 12^3 = 12^{\frac{6}{2}} = 12^{\frac{d(n)}{2}}$
24127
I believe I have the solution to this problem but post it anyway to get feedback and alternate solutions/angles for it.  For all $n \in \mathrm {Z_+}$ prove $n$ is a perfect square if and only if $n$ has odd # of positive divisors. Thought to use induction but the perfect squares don't increase by 1.    $\Rightarrow$: If $n$ is a perfect square it must consist of 1+ prime factors each to an even power.  If $n$ consists of 1+ prime factors each to an even power it must have an odd # of positive divisors.  $\Leftarrow$: If $n$ has an odd number of positive divisors it must consist of 1+ prime factors each to an even power.  If $n$ consists of 1+ prime factors each to an even power it must be a perfect square.  Thanks.