I tried this problem for a while, but didn't see the application of Eisenstein's irreducibility criterion here. All the coefficients, including the leading coefficient, are equal to 1.   p is a prime number.  The polynomial is $f(x) = x^{p-1} + x^{p-2} + \cdots + x + 1$  How do I proceed? Many thanks.
37307
As opposed to the generic polynomial form for utilizing the Eisenstein Criterion ($a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\in\mathbb{Z}[x]$ is irreducible in $\mathbb{Q}$) how do we prove that if $p$ is a prime, $x^{p-1}+x^{p-2}+\dots+x+1$ is irreducible over $\mathbb{Q}$?
67525
Show that   $f_p (x^{p^{e-1}})$ is irreducible over $\mathbb{Q}$. I know it is obviously true for $ p=2$. Please suggest me any hints
33610
Let $P(x)$ be a nonconstant polynomial whose coefficients are positive integers. If $P(n)$ divides $P(P(n) - 2015 )$ for every natural number n, then prove that $$P (-2015) = 0$$  I've observed that the proof depends on  $ P(n)$ divides $ P(P(n)-2015)-P(-2015)\,$ but I cannot see how to prove that. Could someone please elaborate on how to proceed.
41057
Find all primes $p&gt;2$ for which $x^2+x+1$ is irreducible in $\mathbb{F}_p[x]$   Attempt. Since $x^2+x+1$ is of degree 2, it is reducible iff it has a root in $\mathbb{F}_p$. It has a root in $\mathbb{F}_p$ iff $x^3-1=(x-1)(x^2+x+1)$ has a root other than 1. The latter has a root other than $1$ iff $\mathbb{F}_p$ has elements of order $3$, which happens iff $3\mid |\mathbb{F}_p^{\times}|=p-1$, which happens iff $p\equiv 1\bmod 3$. So $x^2+x+1$ is irreducible iff $p\not \equiv 1\bmod3 $.   I am getting a different answer than my professor, who has $p\equiv 1\bmod 3$. If I am incorrect, were am I going wrong?  Update. Now I am unsure about my own reasoning. What if $1$ is a root of $x^2+x+1$? I don't know why I am considering roots of $x^2+x+1$ other than $1$ when $1$ can be a root.
67146
Let $F$ be a field of arbitrary characteristic, $a\in F$, and $p$ a prime number. Show that $$f(X)=X^p-a$$ is irreducible in $F[X]$ if it has no root in $F$.   This  to a related question mentions the result is due to Capelli.  I can prove the result if $F$ has characteristic $p$ as follows.  Suppose $f$ is reducible: $f(X)=g(X)h(X)$ with $g(X)$ an irreducible factor of degree $m$, $1\le m&lt;p$. Then if $\alpha$ is a root of $g$ in some extension field $K$ of $F$, we have $$f(X)=X^p-\alpha^p=(X-\alpha)^p$$ so its divisor $g(X)$ must be of the form $(X-\alpha)^m$. Since the coefficient of $X^{m-1}$ in $g$ is in $F$, we have $m\alpha\in F$.  So $\alpha\in F$ because $m$ is invertible modulo $p$.  How would you show the result in other characteristics?
87223
Assume that $F$ is a field and $\operatorname{char}(F)=p$. Let $a$ be an element in $F$ without $p$th root, then the polynomial $$x^{p^n}-a$$ is irreducible and inseparable over $F$ for all $n$.    I have proved the inseparable part by considering the derivative of the polynomial, but I'm having trouble with the irreducible part. Any help?
113998
I'm having a hard time showing this:     If $K$ is an extension of $\mathbb{Q}$ with degree $m$ and $f(x)$ an irreducible polynomial over the rationals with degree $n$, such that $\gcd(m, n)=1$, then $f(x)$ is irreducible over $K$.   I have tried it by writing $f(x)=a(x)b(x)$ and then looking at the coefficients of those polynomials (some of them must belong to $K-\mathbb{Q}$ which could possibly result in a contradiction) to no success. I have no idea where to use the hypothesis of m and n being relatively prime.  Any help would be appreciated.
106784
Determine whether $X^4-16X^2+4$ is irreducible in $\mathbb{Q}[X]$.  To solve this problem, I reasoned that since $X^4-16X^2+4$ has no rational roots hence irreducible.  But there is a hint to this question that uses different approach: Try supposing it is reducible, then it must factor into a product of two monic quadratic polynomials with integer coefficients. Then show that it is impossible, then conclude the original polynomial is irreducible.  My questions regarding the hints: 1. Why can't we just show that since $X^4-16X^2+4$ has no rational roots hence irreducible? 2. Why do we have to factorise into a product of two monic quadratic polynomials with integer coefficients? Why monic? And why can't we factorise into polynomial with degree 1 and 3? Also, lastly why the coefficients have to be integers?  Thank you for any explanations!    Edit: Thanks for all the answers.  I saw that Second Gauss Lemma is used. But I only learned the first one in class. Which is: Let $R$ be a Unique Facorisation Domain. If $f,g\in\mathbb{R}[X]$ are primitive, then so too is their product $fg$.  Is it inevitable to use Second Gauss Lemma? Is there any other way around?
97520
I am trying to prove or disprove following statment:     $x^2-31x+257$ evaluates to a prime number whenever $x$ is a natural number.   First of all, I realized that  we can't factorize this  polynomial  using its square root like  $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ because discriminant is negative, also I used excel for numbers from 1 to 1530 (just for checking), and yes it gives me prime numbers, unfortunately I dont know  what the formula is, which evaluates  every natural number for given input, maybe we may write $x_{k+1}=k+1$     for $k=0,\ldots\infty$, but how can I use  this  recursive statment? I have tried  instead of  $x$, put $k+1$ in this polynomial, and so I got  $$(k+1)^2-31k+257=k^2+2k+1-31k+257=k^2-29k+258$$ but now this last  polynomial for $k=1$ evaluates to $259-29=230$, which is not prime, but are my steps correct? Please help me.
9624
The polynomial $f(x) = x^3 -3x + 1$  is irreducible, and I'm trying to find the splitting field of the polynomial.  We've been given the hint by our lecturer to allow an arbitrary $\alpha$ to be a root and then to check that $\alpha$ is a root, then $2-\alpha^2$ is also root. I've done this with no problem.  My issue is since this is a polynomial of degree $3$, there must be another root that is somehow related to $\alpha$ so that I can create the splitting field and in turn the Galois group of my polynomial but I have no idea what this last root could be.
7994
Q. Let $P(x)$ be a non-constant polynomial whose coefficients are   positive integers. If $P(n)$ divides $P(P(n)-2015)$ for every natural   number $n$, prove that $P(-2015)=0$.   In one of the sources, the solution given is as follows: Note that $P(n)-2015-(-2015)=P(n)$ divides $P(P(n)-2015)-P(-2015)$ for every positive integer $n$. But $P(n)$ divides $P(P(n)-2015)$ for every positive integer n.   Therefore, $P(n)$ divides $P(-2015)$ for every positive integer $n$. Hence $P(-2015)=0$.  I am not able to understand that how $P(n)-2015-(-2015)=P(n)$ divides $P(P(n)-2015)-P(-2015)$. Please help me out.
73197
Question is to prove :   Irreducibility of $(x-1)(x-2)\cdots (x-n)- 1$ over  $\mathbb{Z}$ for all $n\geq 1$ Irreducibility of $(x-1)(x-2)\cdots (x-n)+ 1$ over  $\mathbb{Z}$ for all $n\geq 1$ and $n\neq 4$   Hint for first bullet is      If the polynomial factors consider the value of the factors at $x=1,2,\dots,n$   For second bullet :  Suppose $p(x)=(x-1)(x-2)\cdots (x-n)-1$ is reducible we have:  $p(x)=q(x)r(x)$ with $\text {Max {degree of p(x), degree of r(x)}}&lt;n$  Hint is suggesting me to use that $p(i)=-1$ for all $1\leq i\leq n$   i.e., $q(i)r(i)=-1$ for all $1\leq i\leq n$  i.e., $q(i)=-1; r(i)=1$ or $q(i)=1;r(i)=-1$  for all $1\leq i\leq n$  For second bullet :  Suppose $p(x)=(x-1)(x-2)\cdots (x-n)+1$ is reducible we have:  $p(x)=q(x)r(x)$ with $\text {Max {degree of p(x), degree of r(x)}}&lt;n$  Hint is suggesting me to use that $p(i)=1$ for all $1\leq i\leq n$   i.e., $q(i)r(i)=1$ for all $1\leq i\leq n$  i.e., $q(i)=r(i)=1$ or $q(i)=r(i)=-1$  for all $1\leq i\leq n$  I am getting some vague ideas but could not bind them to prove this.  I would be thankful if some one can help me to clear this.  Thank you.  P.S : Please give "just hints". Do not write whole answer at once. This is a "request". Thank you :)  Edit : I have changed the title from   Irreducibility of $(x-1)(x-2)\cdots (x-n)\pm 1$ over $\mathbb{Z}$  to   Existence of Irreducible polynomials over $\mathbb{Z}$ of any given degree  for two reasons :   Irreducibility of $(x-1)(x-2)\cdots (x-n)\pm 1$ over $\mathbb{Z}$ for any $n\geq 1$ implies Existence of Irreducible polynomials over $\mathbb{Z}$ of any given degree The title looks atractive
26962
Let $A$ be a field and let $p(X)$ be a polynomial over $A$. Let $a\in A$.   Want to show:     $p(X)$ is irreducible if and only if $p(X+a)$ is irreducible.    I suspect that I should use the substitution principle somehow, but that's as far as I've come. Completely stumped.
71194
So, this is Hungerford problem 9 on page 166. Here's the problem in full:     Let $f(x) = \sum_{i=0}^n a_i x^i \in \mathbb{Z}[x]$. Suppose that for some $k$, $0 &lt; k &lt; n$, and some prime $p$ such that $p \nmid a_n$, $p \nmid a_k$, and $p \mid a_i$ for all $i = 0, \dots, k-1$, but $p^2 \nmid a_0$. Show that $f$ has a factor of degree at least $k$ that is irreducible in $\mathbb{Z}[x]$.   Here's my progress:  If we construct a new polynomial $g$ where $g$ is $\sum_{i=0}^k a_i x^i$ , we know by Eisenstein's criterion that $g$ is irreducible in $\mathbb{Q}[x]$. If we somehow knew that $g$ was irreducible in $\mathbb{Z}[x]$ too (that would follow if $g$ were primitive), then we could say that the smallest thing we could possibly factor out of $f$ that isn't a unit has to be of degree $k$ and that $g$ must be a factor of it. I tried writing $f$ as $C(f) \cdot f_1$ where $C(f)$ is the content of $f$, so that we'd be dealing with a primitive $f_1$ (and of course, the nice things about $g$ don't go away since $p$ doesn't divide every term), but this doesn't necessarily result in $g$ being primitive (right?)!  So, here's where I'm stuck: how can I show I can even factor $f$? How can I show that $g$ is irreducible in $\mathbb{Z}[x]$, whether by showing it's primitive or by some other means?  Thanks so much, everyone!
54804
I want to prove that for any set of distinct integers $a_1,\ldots,a_n$, the polynomial $$h = (x-a_1)\cdots(x-a_n) + 1$$ is irreducible over the field $\mathbb{Q}$, except for the following special cases which are reducible:  $$\left.\begin{cases} a_1 = a\\ a_2 = a+2 \end{cases}\right\} \implies h = (x-a-1)^2$$  and $$\left.\begin{cases} a_1 = a\\ a_2 = a+1\\ a_3 = a+2\\ a_4 = a+3 \end{cases}\right\} \implies h = ((x-a-1)(x-a-2)-1)^2$$
80506
Every irreducible polynomial $f$ over a perfect field $F$ is separable.   Can you check my proof? Assume that $f$ is irreducible but inseparable. So we have $f=\sum_i h_ix^i$ and $f^p=\sum_i h_i^px^{ip}$. Now I use Frobenius mapping $\phi (f)=f^p=(\sum_i h_ix^i)^p=\sum_i h_i^px^{ip}$ so it is reducible. Contradiction. Should I also show for char(F)=0?
1406
Let $K$ be an arbitrary field, $p$ a prime and $a\in K$. Show $f=x^p-a$ is either irreducible in   $K[x]$ or has a root in $K$.   My strategy was to split this up into a case for each characteristic.   The characteristic $p$ case is easy. Assume $f$ is reducible $=gh$ both smaller degree $g$ irreducible, with $\alpha$ a root of $g$. Then in $K(\alpha)[x]$, $f=(x-\alpha)^p$. So $g=(x-\alpha)^k$, and all the other factors of $f$ are of the same form. Hence they are all equal to the $minpoly(\alpha)$ and the degree of $minpoly(\alpha)|p$ so it is $1$ or $p$ and we are done.  The characteristic $0$ case  and characteristic $q\ne p$ I get stuck on. If $K$ contains a primitive $p$th root of unity $\zeta$, then over $K(\alpha)$, $f$ splits into linear factors. So the degree of the field extension for any root of $f$ is the same, so the degree of such a root is either $1$ or $p$. If $K$ does not have a primitive $p$th root of unity, then simply adjoin one. We know $[K(\zeta):K]|p-1$ and by the previous step either $f$ is irreducible over $K(\zeta)$ or splits into linear factors. If it is irreducible we are done, if not we have it splits into linear factors of the form $x-\zeta^ia^{1/p}$, but I am unsure what to do from here. The characteristic $q$ case seems similar to the characteristic $0$ case but again I am stuck trying to finish the proof.   Some googling has lead me to believe that this problem is generally approached using the field norm and if an accessible online reference exists I would love to see it, but I feel that in the context I have seen it should be entirely possible that this is done without needing the norm. As I have not seen the field norm before, I would like to know if there is a possible way to prove this without using it and hopefully that follows in the same vein my attempt at a proof did.