# Irreducibility measure of integer polynomials

(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).

• 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). Feb 19 at 9:58
• @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 $d(p) \leq 3$.
– Mare
Feb 19 at 15:10
• Similar questions arise at mathoverflow.net/questions/59956/… See also mathoverflow.net/questions/149362/… Feb 19 at 21:36
• @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. Feb 20 at 1:40
• 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. Feb 20 at 5:41

For $$f=x^6 - 3x^5 - 2x^4 + 10x^3 + x^2 - 8x - 5$$, one has $$f=(x^3 - 2x^2 - 2x + 5)(x^3 - x^2 - 2x - 1),$$ $$f+1=(x-2)(x^5 - x^4 - 4x^3 + 2x^2 + 5x + 2),$$ $$f+2=(x^2-x-1)(x^4 - 2x^3 - 3x^2 + 5x + 3),$$ $$f+3=(x^2-2)(x^4 - 3x^3 + 4x + 1),$$ $$f+4=(x+1)(x^5 - 4x^4 + 2x^3 + 8x^2 - 7x - 1),$$ $$f+5=x(x^5 - 3x^4 - 2x^3 + 10x^2 + x - 8),$$ $$f+6=(x-1)(x^5 - 2x^4 - 4x^3 + 6x^2 + 7x - 1).$$
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.
• 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.
• @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.) Feb 21 at 5:29
• 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. Feb 21 at 7:37