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Let $f(x)=1+x+x^2+ \ldots + x^n$. There is a theorem saying

"For 'most' $n$, $f'(x)$ is irreducible". (Ref: matwbn.icm.edu.pl/ksiazki/aa/aa90/aa9023.pdf)

$f'(x)$ has the property that its coefficients form an arithmetic progression. So I wonder if any generalizations applies for such polynomials. For the easiest case, can we say anything about $g(x)=1+3x+5x^2+ \ldots + (2n+1)x^n$ ?

Thanks!

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    $\begingroup$ One can probably get the statement "for each $n$, most arithmetic progressions give irreducible polynomials" from Hilbert irreducibility. $\endgroup$
    – Will Sawin
    Commented Mar 29, 2013 at 16:31
  • $\begingroup$ @Will How? Directly, Hilbert irreducibility doesn't let you vary the degree of the polynomial. $\endgroup$
    – David E Speyer
    Commented Mar 29, 2013 at 16:34
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    $\begingroup$ @David: I think that Will fixes $n$. And then it is certainly a consequence of Hilbert's irreducibility theorem, since $\sum_{k=0}^n(a+bk)x^k$ is irreducible over $\mathbb Q(a,b)$, for variables $a,b$. $\endgroup$
    – Peter Mueller
    Commented Mar 29, 2013 at 17:35
  • $\begingroup$ This seems like a promising avenue of research - to generalize the proof that you indicated to arbitrary arithmetic progressions. Note that $g(x) = cf(x) + df'(x)$ for suitable constants $c,d$, if it helps. $\endgroup$
    – Greg Martin
    Commented Mar 29, 2013 at 18:27
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    $\begingroup$ @Aaron: Of course, there are many obvious examples like this: If $x-x_0$ is to be a factor of $f_{a,b}(x)=\sum_{k=0}^n(a+bk)x^k$, then there is a linear equation for $a,b$ from $f_{a,b}(x_0)=0$. $\endgroup$
    – Peter Mueller
    Commented Mar 29, 2013 at 20:19

2 Answers 2

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I've heard from Zhi-Wei Sun that he recently considered this question. In a post a few days ago to OEIS Least integer b>2n+1 such that the numbers written as [1,3,...,2n-1,2n+1] and [2n+1,2n-1,...,3,1] in base b are both prime. He gives the first of what he conjectures are infinitely many bases (for each fixed $n$) with the named property. Other fairly specific conjectures concerning Galois groups, reducibility over $\mathbb{Z}_p$ and the like can be found on that page. Any one of the conjectures would imply that $1+3x+5x^2+\cdots+(2n+1)x^{n}$ is always irreducible over the integers. A similar post a few days earlier than that concerns $1+2x+\cdots+(n+1)x^{n}$ which he would also conjecture is always irreducible over the integers.

Of course there are integer arithmetic progressions such that $f(a,b,n)=\sum_0^n(a+bk)x^k$ does factor (with $a \ne 0$ of course). At least there is $1+x+x^2+\cdots+x^n$ which is irreducible when and only when $n+1$ is prime. A fairly simple minded search over small parameters turns up

  • $-n+(2-n)x+(4-n)x^2+\cdots+nx^n$ which has $(x-1)$ as a factor (and $(x+1)$ for even $n$) but no other factors up to $n=42$
  • After some manipulation, the integer quadratic examples can written $s(t-2s)+(t^2-s^2)x+t(2t-s)x^2$ with linear factor $(s+tx)$
  • One can first work over $\mathbb{Q}$ , stipulate a factor $x-r=x-t/s$ , set $c_{n-1}=1$ and then solve for $c_0,\cdots,c_{n-2}$ such that $(x-t/s)(c_0+c_1x+\cdots+c_{n-1}x^{n-1})=f(a,b,n)$ for $a=c_0-rc_1$ and $b=r+1-c_{n-2}$. The solutions will have $c_i$ rational functions in $r$ with denominator $n+(n-1)r+\cdots+r^{n-1}$. Then one can scale to integer examples.
  • Perhaps there are nice solutions which are reducible but without a linear factor.

later Here is the solution for degree $5$ from which the pattern becomes clear. Thanks to Joro and Peter for seeing what I did not. The coefficients below are in arithmetic progression with difference $b=-(s^5+s^4t+s^3t^2+s^2t^3+st^4+t^5).$ It is not immediate, but also is not too hard to check that $(s-tx)$ is a factor as $x=\frac{s}{t}$ is a root.

$$\left(5{s}^{5}+4{s}^{4}t+3{s}^{3}{t}^{2}+2{s}^{2}{t}^{3}+s{t}^{4}+0t^5\right)+\left( 4{s}^{5}+3{s}^{4}t+2{s}^{3}{t}^{2}+{s}^{2}{t}^{3}+0st^4-{t}^{5} \right) x$$ $$\ \ +\left( 3{s}^{5}+2{s}^{4}t+{s}^{3}{t}^{2}+0s^2t^3-s{t}^{4}-2{t}^{5}\right){x}^{2} +\left( 2\{s}^{5}+{s}^{4}t+0s^3t^2-{s}^{2}{t}^{3}-2s{t}^{4}-3{t}^{5} \right) {x}^{3}$$ $$\ \ \ \ +\left( {s}^{5}+0s^4t-{s}^{3}{t}^{2}-2{s}^{2}{t}^{3}-3s{t}^{4}-4{t}^{5 } \right) {x}^{4}+\left(0s^5-{s}^{4}t-2{s}^{3}{t}^{2}-3{s}^{2}{t}^{3}-4s{t}^{4}-5{ t}^{5} \right) {x}^{5} $$

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  • $\begingroup$ @Aaron: Thanks for all the information! Looks like the question regarding my $g(x)$ is still open. $\endgroup$
    – Ted Mao
    Commented Mar 31, 2013 at 3:29
  • $\begingroup$ I guess so and evidently, even if "most" is a theorem for $f'$ it is still open if the actual answer is "all" $\endgroup$
    – Aaron Meyerowitz
    Commented Mar 31, 2013 at 3:38
  • $\begingroup$ By equating coefficients found degree 3 identity in my answer. $\endgroup$
    – joro
    Commented Mar 31, 2013 at 8:54
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Extending Aaron's idea about identities.

Working over $\mathbb{Q}[x]$ this appears degree 3 identity for $c5,c6 \in \mathbb{Z}$: $a=-c5 c6, b= c6 (c5^3+c5^2+c5+1)/(3 c5^2+2 c5+1)$.

$$P=a+(a+b) x+(a+2 b) x^2+(a+3 b)x^3= (-x + c5) \cdot \text{quadratic}$$.

Scale to get integral values.

Found by equating coefficients and suppose higher degree identities exist.

Added Degree 4

a= -c5*c6
b= c6*(c5^4+c5^3+c5^2+c5+1)/(4*c5^3+3*c5^2+2*c5+1)
P=a+(a+b)*x+(a+2*b)*x^2+(a+3*b)*x^3+(a+4*b)*x^4 == (c5-x)*O(x^3)
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