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(Partial) Question in short: Let $p$ be a monic integer polynomial of degree $n$. Is there a natural number $k$ with $0 \leq k \leq n$ such that $p+k$ is irreducible over the integers?

Longer version:

Let $p$ be a monic polynomial over the integers. Define the irreducibility measure $d(p)$ of $p$ as the smallest integer $k \geq 0$ such that $p+k$ is irreducible over the integers. Define $M_n:=$ sup $\{ d(p) | deg(p)=n \}$ for $n \geq 2$. Here $deg(p)$ is the degree of $p$.

Question: Is it true that $M_n \leq n$? (Answer no, by Joachim König). Is there a good bound for $M_n$ ?

The question is based on some small computer experiments.

edit: Sorry I forgot the condition that the polynomials are monic (I did all computer experiments with that assumption. The answer by Joachim König gives a counterexample in the non-monic case).

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    $\begingroup$ It is perhaps better to define $d(p)$ as the least natural integer such that $p+i,p+i+1,p+i+k$ contains an irreducible polynomial for all $i\in\mathbb Z$. (This makes the definition invariant under translations $x\longmapsto x+\tau$ which seems natural for integral polynomials). $\endgroup$
    – Roland Bacher
    Feb 19 at 9:58
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    $\begingroup$ @RijulSaini It was posted by Joachim König that $6x^2+7x$ is a counterexample, but he remarked that for non-monic polynomials of degree 2 $d(p) \leq 3$. $\endgroup$
    – Mare
    Feb 19 at 15:10
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    $\begingroup$ Similar questions arise at mathoverflow.net/questions/59956/… See also mathoverflow.net/questions/149362/… $\endgroup$
    – Gerry Myerson
    Feb 19 at 21:36
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    $\begingroup$ @Mare At least $M_n\ge n$ is clear for all $n$, due to the polynomial $p(x)=x\cdot (x+1)\cdots (x+n-1) + x$, which is also given in one of the questions linked by Gerry Myerson. $\endgroup$
    – Joachim König
    Feb 20 at 1:40
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    $\begingroup$ It becomes true if we replace being reducible to having an integer root, that easily follows from $p(a)-p(b)$ being divisible by $a-b$. In particular, this yields $d=2,d=3$ cases. $\endgroup$
    – Fedor Petrov
    Feb 20 at 5:41

1 Answer 1

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For $f=x^6 - 3x^5 - 2x^4 + 10x^3 + x^2 - 8x - 5$, one has $$f=(x^3 - 2x^2 - 2x + 5)(x^3 - x^2 - 2x - 1),$$ $$f+1=(x-2)(x^5 - x^4 - 4x^3 + 2x^2 + 5x + 2),$$ $$f+2=(x^2-x-1)(x^4 - 2x^3 - 3x^2 + 5x + 3),$$ $$f+3=(x^2-2)(x^4 - 3x^3 + 4x + 1),$$ $$f+4=(x+1)(x^5 - 4x^4 + 2x^3 + 8x^2 - 7x - 1),$$ $$f+5=x(x^5 - 3x^4 - 2x^3 + 10x^2 + x - 8),$$ $$f+6=(x-1)(x^5 - 2x^4 - 4x^3 + 6x^2 + 7x - 1).$$

(Found more or less by brute force.)

PS: In order to not cause unnecessary confusion, the "answer" giving a counterexample for the ``non-monic case", currently mentioned in the OP, was $f(x)=6x^2+7x$; this was removed as a separate answer after the OP had been altered.

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  • $\begingroup$ Thanks. I leave the question open a bit before I accept the answer since Im interested to see what $M_n$ might be for small $n$ or whether there is a good bound. $\endgroup$
    – Mare
    Feb 21 at 4:52
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    $\begingroup$ @Mare I don't think you can hope for a really good general bound here, since any bound would be an effective result on the least admissible value in Hilbert's irreducibility theorem (for the special case $F(x,t) = f(x)+t$), see e.g. matwbn.icm.edu.pl/ksiazki/aa/aa69/aa6937.pdf (I could just about imagine $M_4=4$ or $M_5=5$ could be provable (in case true), since those translate to integer solutions to somewhat managable systems of equations given by certain resolvent polynomials.) $\endgroup$
    – Joachim König
    Feb 21 at 5:29
  • $\begingroup$ For any finite degree, in principle it's possible to enumerate the possible degrees of factors and for each such to set up a system of bilinear equations on their coefficients, so for small $n$ it's just a matter of a lot of computation. $\endgroup$
    – Peter Taylor
    Feb 21 at 7:37
  • $\begingroup$ @PeterTaylor But you will have more variables than equations (even after some w.l.o.g.'s), so the issue remains whether the system has integral solutions. $\endgroup$
    – Peter Mueller
    Feb 21 at 12:35

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