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Let $P\in \mathbb{Z}[x]$ be a polynomial of degree 2011. Can you show that there is an infinite amount of such polynomials, such that $P$ and $P+2$ remain irreducible?

$x^{2011}+15ax+3$ is a polynomial of degree 2011. By Schönemann-Eisenstein, it follows that it must be irreducible since: 3 doesn't divide $1$, 9 doesn't divide 3 but $3|15$. So this is irreducible over $\mathbb{Q}[x]$. It also follows that 25 respectively 5 doesn't divide $5$ respectively $1$, but does divide 15 for all $a\in \mathbb{N}$, so by using 5 as a prime it follows that P+2 is also irreducible.

The book asks for polynomials in $\mathbb{Z}[x]$, I have shown it for $\mathbb{Q}[x]$. Gauss lemma states that if it is for integers only over $\mathbb{Q}[x]$, it also counts for $\mathbb{Z}[x]$

Is my reasoning correct? Please do tell

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    (1) You should review the *proof* of the Eisenstein irreducibility criterion as far as the role of ${\mathbf Z}[x]$ and ${\mathbf Q}[x]$ are concerned, and then you'll be able to answer your own question. (2) You are using a bit of notation without having explained what it means (e.g., a_0 is the constant term of $P+2$), which makes it a little frustrating to read. (3) The integers are not a subfield of the rationals. Perhaps for that last part you should review the definitions of irreducibility in ${\mathbf Q}[x]$ and ${\mathbf Z}[x]$.2011-11-17
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    KCd, Thanks. ….2011-11-17

2 Answers 2

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Dear VVV your example is very nice, and your polynomials are irreducible over $\mathbb Z$.
There is a subtlety involved in the Schönemann-Eisenstein criterion:

1) The Schönemann-Eisenstein criterion only gives irreducibility in $\mathbb Q[X]$.
For example you can apply it to $f(X)=2X^2+6$ with $p=3$: the criterion will tell you that $f(X)$ is irreducible in $\mathbb Q[X]$.
However $f(X)$ is not irreducible in $\mathbb Z[X]$ because there you have a genuine factorization $2X^2+6=(2).(X^2+3)$.
The apparent paradox is explained by the fact that $2$ is not invertible in $\mathbb Z[X]$, but it is invertible in $\mathbb Q[X]$

2) However in your case, your polynomials are also irreducible over $\mathbb Z$ because of the following result:

Given a polynomial $f(X)\in A[X]$ where $A$ is a UFD with fraction field $K$, we have the equivalence:
$ f(X)$ is irreducible in $A[X]$
$ \iff $
$f(X)$ is irreducible in $K[X]$ and no prime element in $A $ divides all the coefficients of $f(X) $

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    I would have said instead that for monic polynomials in ${\mathbf Z}[x]$ there is no difference between irreducibility over ${\mathbf Z}$ and over ${\mathbf Q}$. That is of course just a special case of what is written in part (2) of this answer, but it does apply to the case under consideration in the original question.2011-11-17
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    @KCd: well, yes, of course. Obviously the result I quote is also more general in that $\mathbb Z$ is replaced by $A$. This general result is well worth remembering, and I believe that it may be instructive to show how a result is integrated in a broader perspective.2011-11-18
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To make things harder, we prove a result about polynomials that have shape very much like the one described by the OP.

We will show that there are infinitely many irreducible polynomials $x^{2011} +p(p+2)x+p$ with $p$ prime and $x^{2011} +p(p+2)x+p+2$ irreducible. For each of these two polynomials, the irreducibility proof uses the Eisenstein Criterion.

A possible approach is to show that $p+2$ is prime for infinitely many primes $p$. Then, as in the OP's example, $x^{2011} +p(p+2)x+p$ and
$x^{2011} +p(p+2)x+p+2$ are both irreducible by Eisenstein's Criterion. However, this requires proving the Twin Prime Conjecture, so it is not the easy way to go.

Instead, recall that a number $n$ is called powerful(l) if for every prime $q$ that divides $n$, its square $q^2$ divides $n$. Powerful numbers are relatively scarce. There is a constant $C$ such that the number of powerful numbers less than $x$ is less than $Cx^{1/2}$ (see for example the Wikipedia article).

Since the primes have much greater asymptotic density than the powerful numbers, there are infinitely many primes $p$ such that $p+2$ is not powerful. Indeed, for "most" primes $p$, the number $p+2$ is not powerful. Let $p$ be any prime such that $p+2$ is not a powerful number.

Then $x^{2011}+p(p+2)+p$ and $x^{2011}+p(p+2)+p+2$ are both irreducible. For the second polynomial, we use the Eisenstein Criterion with $q$ any prime which divides $p+2$ but whose square does not divide $p+2$. There is such a prime $q$ since $p+2$ is not powerful.

Comment: The proof in principle is of irreducibility over $\mathbb{Q}[x]$. But irreducibility over $\mathbb{Z}[X]$ is an immediate consequence, since our polynomial is monic, and consequently the $\gcd$ of the coefficients is $1$.

By using Chen primes, we can find infinitely many primes $p$ such that $p+2$ is prime or has at most two prime factors. That is the nearest we can come to imitating your $3$, $5$ example.

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    You don't even need $p$ to be prime, as long as $p$ and $p+2$ are not powerfull this trick works. If $p_1$ and $p_2$ are two distinct primes, you can use Chinese Remainder Theorem: $n =p_1 \mod p_1^2$ and $n+2 = p_2 \mod p_2^2$ has a solution $n$, and then $x^{2011}+n(n+1)x+n$ works....2011-11-17
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    @N.S.: I wanted to get as close as I could to the OP's example without mentioning Chen primes. Your family of examples uses much less machinery than mine.2011-11-17
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    Great answer! +12016-01-01