Let $m \geq 3$ be an odd integer. Prove that for any $a \in \mathbb{F}_{2^m}$, the equation $x^6+x+a=0$ has a zero or two distinct solutions in $\mathbb{F}_{2^m}$.  I have proven the following facts:  Let $a&gt;1$ be an integer. For any positive $n,d\in \mathbb{Z}$, $\,$ $d$ divides $n$ if and only if $a^d-1$ divides $a^n-1$. In particular, $\mathbb{F}_{p^d}\subseteq \mathbb{F}_{p^n}$.   Let $\phi$ denote the Frobenius map $x\mapsto x^p$ on the finite field $\mathbb{F}_{p^n}$. Then $\phi$ gives an isomorphism of $\mathbb{F}_{p^n}$ to itself. Also, $\phi^n$ is the identity map and no lower power of $\phi$ is the identity.   For any prime $p$ and any nonzero $a \in \mathbb{F}_p$ prove that $x^p-x+a$ is irreducible and separable over $\mathbb{F}_p$.   I don't know if any of these facts will help me prove this. I am currently studying Chapter 13 from Dummit and Foote. Any help on how to prove the statement is greatly appreciated, thank you.  ***My question is not exactly the same as the post in   because I found a counterexample for $x^2+ax+1=0$ having the same number of solutions as $x^6+ax+1=0$ in $\mathbb{F}_{2^m}$ whenever $m \geq 3$ is odd.
124300
I am reading a proof that narrows down to the following statement:      It is easy to see that the number of solutions of $x^6+x=a$ in $\mathbb{F}_{2^m}$, where $a\in \mathbb{F}_{2^m}$ and $m\geq 3$ is odd, is same as number of solutions of $x^2+ax+1=0$ in $\mathbb{F}_{2^m}$.   Why is this true? They say that $x^7=1$ for $x \neq 0$, which I also don't see. Thank you!
97895
Prove that every positive integer $n$ has a unique expression of the form: $2^{r}m$ where $r\ge 0$ and $m$ is an odd positive integer  if $n$ is odd then $n=2^{0}n$, but I dont know what to do when $n$ is even  and to prove that this expression is unique is it a good choice to  use the fundamental theorem of arithmetic?  I would really appreciate your help
58141
Given the congruence $x^3 \equiv a \pmod p$, where $p \geq 5$ is a prime and $\gcd(a,p)=1$, prove the following:   If $p \equiv1 \pmod 6$, then the congruence has either no solutions or three incongruent solutions modulo $p$ If $p \equiv5 \pmod 6$, then the congruence has a unique solution modulo $p$   Help me...  I don't know what I learn because Prof teaches so quickly.  So I need your help to organize my thoughts.
25349
Let $\mathbb{F}$ be a finite field.  Let $a_1,\dots,a_n \in \mathbb{F}$ be given.  I want to know whether there exists $x_1,\dots,x_n \in \mathbb{F}$ such that  $$a_1 x_1^2 + a_2 x_2^2 + \dots + a_n x_n^2 = 1.$$  My question: Can we characterize when this equation has a solution, i.e., by some criterion on $a_1,\dots,a_n$?  Is there a clean characterization, in terms of $a_1,\dots,a_n$ and the properties of $\mathbb{F}$?    Here's what I can see immediately.  First, if $n=1$, there is an easy answer (the equation has a solution iff $a_1$ is a square in $\mathbb{F}$).  Second, if any $a_i$ is a square in $\mathbb{F}$, the equation has a solution (set all other $x_j$'s to $0$), so we can assume that all the $a_i$'s are non-squares and that $n \ge 2$.  Third, if $-1$ is a square in $\mathbb{F}$, then the equation has a solution: we can take $x_2 =  \sqrt{-a_1/a_2} \cdot x_1$.  But what about the case where $-1$ is a non-square in $\mathbb{F}$?  Is it always guaranteed to have a solution in this case?
109631
So first of all, I used the following from my lecture notes:   If $f \in \mathbb{Z}[x]$ is primitive (gcd of all the coefficients is 1) - $f$ irreducible in $\mathbb{Z}[x] \Leftrightarrow$ f irreducible in $\mathbb{Q}[x]$.   So now I only have to show that $f_n = x^n + x + 3$ is irreducible in $\mathbb{Z}[x]$ for all $n \geq 2. $  Is there a pattern I should be appealing to in some way?
111907
Question Statement:- Let $a,b,c$ be positive integers and consider all the quadratic equations of the form $ax^2-bx+c=0$ which have two distinct real roots in $(0,1)$. Find the least positive integers $a$ and $b$ for which such a quadratic equation exist.   Attempt at a solution:- Let $f(x)=ax^2-bx+c$. Now as the roots of $f(x)=0$ lie in $(0,1)$ hence $f(0)\cdot f(1)\gt0$. $$D\gt0\implies b^2-4ac\gt0$$ $$f(0)=c \\ f(1)=a-b+c$$ As, $a,b,c\in\mathbb{Z}^+$, so $$f(0)\ge1 \implies c\ge1\tag{1}$$ $$f(1)\ge1\implies a-b+c\ge1\tag{2}$$ $$f(0)\cdot f(1)\ge1\implies c(a-b+c)\ge1\implies c^2+(a-b)c-1\ge0\tag{3}$$ Now, for the quadratic inequality in $c$ that we obtained in $(3)$, for it to hold we get $$D\le0\implies (a-b)^2+4\le0$$ The above inequality doesn't hold $\forall a,b \in \mathbb{Z}^+$. After arriving at this conclusion I am kind of stuck on what to conclude from the above result or rather am I going in the right direction. A little push in the right direction in the form of a hint or rather what should be the line of thought for attempting the question. Please don't write the solution, just the line of thought to arrive at the solution would be enough.
72465
My question is, can we prove that the number of solutions to $x^n=e$ in group $G$ is divisible by $n$, for $n | |G|$   Apparently the answer is it is indeed true, infact for the equation $x^n = a$ for any $a \in G $ also, however I have had many ideas each of which having a small flaw with them.  For example I attempted induction on order of $G$ as follows:  Assume true for all groups of order $ \leq k$. Consider $|G| = k+1$.  If $G$ is cyclic, we show fairly easily the result holds. (In fact if I recall correctly there are $n$ solutions for each divisor $n$).  If $G$ is not cyclic, we can look at the subgroups of $G$. It'd be nice to say the number of solutions in each group is divisible by $n$ by the induction hypothesis and get the desired answer by inclusion-exlusion for solutions in all the subgroups, however we could potentially have a subgroup of order  less than $n$ and still contain a solution.   So if we were able to show it works for all the subgroups of orders $ &lt; n$ collectively we would be done. I have no idea whether this can work though.  I did see this question  however I couldn't find much help on it. From the hint given I managed to show $no. solutions = \sum_{ d|n} \lambda_{i} \phi(d)$ for some $\lambda_{i}$s, however they could be different so I can't just use the well know summing of euler-phi formula.  Any solution/ideas would be very much appreciated.
107169
Question: Let $p$ and $q$ be distinct odd prime numbers.  By considering primitive roots, we need to show $\exists c\in\mathbb{Z}$ with the property that:  $\bullet$ $c^n\equiv 1\pmod{p}$ whenever $n$ is a multiple of $p-1$, and   $\bullet$ whenever $n$ is not divisible by $q-1$, then $c^n\not\equiv 1\pmod{q}$  I am only interested with the beginning of this answer.  Answer: We let $c$ be a primitive root modulo $q$.  We are going to use Fermat's little theorem so we need $(c,p)=1$  If $(c,p)\neq 1$, then we have $p\mid c$, but then we have $(c+q, p)=1$  and since $c+q$ is a primitive root modulo $q$, we can replace $c$ with $c+q$ for $(c,p)=1$ for using Fermat's little theorem.  I understand the general method and argument, but...  Contention: Why is $c+q$ still a primitive root modulo $q$?  When is this generally true? Is there an identity for when things remain a primitive root?  Why is $(c+q)^{\phi(q)}\equiv 1\pmod{q}$?
80080
For any $g$ in $\mathbb{Z}/p\mathbb{Z}[x]$ prove that the degree of $f = \gcd(x^p - x, g(x))$ is exactly the number of distinct roots of $g$ in $\mathbb{Z}/p\mathbb{Z}$.   My main problem is that I do not really understand how to approach this problem at all. I am very new to solving proofs and any type of help would be appreciated thank you.
102996
Let there be integers $a,m,n$ such that $a=m^2=n^3$. Show that $a\equiv 0,1,28,36\pmod{63}$. I have established that $a$ must be a sixth power of an integer: $$a=\frac{a^3}{a^2} = \frac{m^6}{n^6} =\left (\frac{m}{n}\right )^6$$ $n^6\mid m^6$ implies $n\mid m$ so $\frac{m}{n}=:b\in\mathbb{Z}$. My objective is to show that $b\equiv 0,1,28,36\pmod{63}$. Then for every $f\in\mathbb{Z}[X]$: $$b\equiv r\pmod{n}\Longrightarrow f(b)\equiv f(r)\pmod{n} $$ (just pick $f(x)=x^6$). Trouble is, I have no idea how to show $b$'s congruences. How to proceed?    Alternatively, I could also state that $a = 63t+r, 0\leq r&lt;63$, but I don't know how to tie in the fact that $a$ is a perfect cube/square at the same time. Advice?
91247
Let $p$ be a prime number, and let $\mathbb{F}_p$ be the field with $p$ elements. How many elements of $\mathbb{F}_p$ have cube roots in $\mathbb{F}_p$?   I had this question on an exam and after reviewing I am still not sure. Any help would be appreciated.
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 $p$.
99129
Let $F$ be a field of characteristic $p$, and let $\alpha \in F$. Let $f \in \mathbb{Z}_p[x]$ be such that $f(\alpha) = 0$. Apparently, we have $f(\alpha^p) = 0$. This was mentioned but not proved in my lectures. Can someone explain why please?  Thanks.
33506
I am having trouble with Silverman's exercise 1.10(b). The converse of (a) is easy because there is no integer solution to the equation when $p \equiv 3$ mod $4$. However, this method does not work for (b) because we always get empty set when intersecting whatever $V_p (p \equiv 3 $ mod $4)$ with $\mathbb{Q}$. I tried to intersect two varieties $V_p, V_q$ with something else, for example $\mathbb{Q}({\sqrt{p}})$ and try to tell the difference, but this is hard since it is very possible for $V_q$ to have nonempty intersection with $\mathbb{Q}({\sqrt{p}})$ and I think it is true that if so then $V_p, V_q$ are isomorphic over $\mathbb{Q}({\sqrt{p}})$. Are there some invariants I can use in this circumstance to show that two varieties are not isomorphic over $\mathbb{Q}$?
10635
Let $P \in \mathbb{Z}[x], P(x) = \displaystyle\sum\limits_{j=0}^n a_j x^j, a_n \neq 0$ and $a_0 \neq 0$; if $p/q$ is a root of P (with p and q coprimes) then $p|a_0$ and $q|a_n$  I've managed to prove the first part ($p|a_0$) and I suppose I'm not far from proving the second, though I'd really like some feedback since I'm just starting with making proofs of my own.  Proof:  $P(x) = a_n(x-p/q)\displaystyle\prod\limits_{j=2}^n (x-r_j)$, with $r_j$ being the other n-1 roots of P(x).  It follows that $a_0 = a_n(-p/q)\displaystyle\prod\limits_{j=2}^n (-r_j)$  Then, $-p/q|a_0$ and obviously $p/q|a_0$. Rephrasing, $a_0 = l\frac{p}{q} = \frac{l}{q} p$ with $l \in \mathbb{Z}$.  This implies $p|a_0$ if $l/q \in \mathbb{Z}$, but this is trivial since $q|lp$ and q and p are coprimes, so $q|l$.  Therefore, $p|a_0$.  As for the second part, we want to see that $q_i|a_n \forall i \leq n$.   We define $d$ as the least common multiple of $\{q_1, q_2,...,q_n\}$. Then, $q_i|a_n \forall i \leq n \iff d|a_n$.  Also, it follows that $d|\displaystyle\prod\limits_{j=1}^n q_j$, so we want to see that $a_n = l \displaystyle\prod\limits_{j=1}^n q_j$ with $l \in \mathbb{Z}$. Here's where I have my doubts with the proof as I have no way to show that l is indeed an integer.  Rearranging the previously given equation for $a_0$:  $a_n = a_0 \displaystyle\prod\limits_{j=1}^n \frac{-q_j}{p_j}$  Using the previous reasoning, as $p_i|a_0 \forall i \leq n$, then $a_0 = k \displaystyle\prod\limits_{j=1}^n p_j$.  Replacing $a_0$:  $a_n = k \displaystyle\prod\limits_{j=1}^n -q_j$, which is equivalent to $q|a_n$ as shown earlier.
26908
Let $p, q$ be distinct odd primes, I would like to compute the number of solution $\mod p$ to the congruence $x^q \equiv 1 \mod p$. (Show that it's $gcd(q, p - 1))$  If $x^q \equiv 1 \mod p$, since $q$ is prime, this must mean that $x$ is either $1$ or has order $q$.   Hence, the problem reduces to finding the number of elements of $U_p$ which is either $1$ or has order $q$.  If $ q \nmid  p - 1$, then there is only one element, namely $1$.  If $q \mid p - 1$, then the number of elements of order $q$ is $\phi(q) = q - 1$, and so the number of solutions to the congruence $\mod p$ is $q - 1 + 1 = q$, and in either case the number of solutions is $\gcd(q, p - 1)$.  Does this look fine?
71932
I am stuck with this exercise:     Show that $\forall s &gt;0 \quad \exists N &gt;0$ such that $$x^2 \equiv 1\pmod N$$ has more than $s$ solutions.   I think I cannot use the theory if quadratic residues, because the modulus is not prime, so maybe I must prove it with square roots or roots of unity, but I don't know how to approach it at all.  Thanks for any help.