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Is every polynomial of the form $2x^{2n} - x^n +1$ irreducible for $n>0$?

Motivation: A few years ago a student asked if $29$ was the largest number which is prime and one more than a perfect number. The obvious sort of thing to do here is to try and prove this is to say that any even perfect number $n$ must be of the form $2^{p-1}(2^p -1)$ and then make a series of congruence restrictions for $p$ based on this. One can try to make a covering congruence this way for large possibilities. One can for any finite list of primes get that $n$ is not $0$ mod $p$ for any of those primes. But one can can get many other restrictions. For example, if $p \equiv 1 \pmod 3$, then  $2^{p-1}(2^p -1) +1 \equiv 0 \pmod 7$, and if $p \equiv 7 \pmod{10}$, then   $2^{p-1}(2^p -1) +1 \equiv 0 \pmod{11}$. However, any finite list of this sort is going to fail to solve the problem. That's because for any prime $q$ if $p \equiv 1 \pmod{q-1}$, then $$2^{p-1}(2^p -1) +1 \equiv 2 \pmod q.$$

So, I told the student that this problem looked not really attackable because the only obvious approach had a fundamental obstruction. But it then occurred to me just recently that there might be a way around this. In particular, if $2x^{2n} - x^n +1$ has a non-trivial factorization for some $n$, then this lets us rule out primes where $p \equiv 1 \pmod{n}$ and so we could possibly get a list of congruences which is sufficient. However, this seems to not happen. In particular, for $1 \leq n \leq 50$, $2x^{2n} - x^n +1$ is irreducible. I've also checked this for a few isolated larger $n$, such as $n=360$. Note that if $2x^{2n} - x^n +1$ is irreducible then so is $2x^{2d} - x^d +1$ for any divisor $d$ of $n$.

The obvious directions to try to prove this is irreducible don't seem to work. The reciprocal polynomial is $p(x)= x^{2n} -x^n +2$ but that doesn't seem to be very helpful here. There's no obvious substitution to use Eisenstein. Perron fails in general for both the polynomial and the reciprocal polynomial. The Cohn criterion fails because one of the coefficients is negative. There are some papers which prove specific rules for irreducibility of trinomial like Koley and Reddy's paper "Irreducibility criterion for certain trinomials" but that seems insufficient. One might hope for using Theorem 3 from their paper on the reciprocal polynomial, but that doesn't seem to be strong enough. Theorem 2 of their paper also fails since the reciprocal polynomial has $2$ and not $2^2$ for the constant.

More generally, it seems like every polynomial of the form $2x^{2n} - x^n +k$ is irreducible for $n>0$ and $k \geq 1$, but I'm much less confident that is the case, and for those I have checked a much smaller range.

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    $\begingroup$ The Mathematica function "IrreduciblePolynomialQ" says that the answer is yes for $1\leq n\leq 300$ after about three minutes. Another six minutes or so get it up to $n\leq 400$. I'm sure this could be pushed further, but this is already quite convincing. $\endgroup$
    – Pace Nielsen
    Commented Feb 15 at 19:45
  • $\begingroup$ I had a little trouble parsing the bit where you talked about ruling out $p$ being $0$ modulo $p$, but I'm pretty sure that it was about $n$ being $0$ modulo $p$, and edited accordingly. I hope that that was correct. $\endgroup$
    – LSpice
    Commented Feb 15 at 21:16
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    $\begingroup$ @LSpice, yes thank you. $\endgroup$
    – JoshuaZ
    Commented Feb 15 at 21:17
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    $\begingroup$ A friend of mine unable to comment on MO wants to point out that there are indeed other primes which are 1 more than a perfect number, for example 33550337 $\endgroup$
    – Achim Krause
    Commented Feb 16 at 13:30
  • $\begingroup$ @AchimKrause Thanks! So the answer is it is just false. I should not have assumed the student had checked a bit more. I guess then the same question still makes sense but with just "is the set of primes of this sort finite" and presumably is still a very hard problem. $\endgroup$
    – JoshuaZ
    Commented Feb 16 at 13:33

1 Answer 1

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Yes it is. The polynomial $f(x)=2x^{2n}-x^n+1$ has all $n$ roots inside the open unit disc: if $|x|>1$, then obviously $2|x^{2n}|>|x^n|+1$, and if $|x|=1$, the equality takes place, but if $|-x^n+1|=2$, then $x^n=-1$ and still $f(x)\ne 0$. But if $f(x)=g(x)h(x)$ with $g,h \in \mathbb{Z}[x]$ of positive degrees, then one of $g,h$ has both leading and constant terms $\pm 1$, thus the product of roots is $\pm 1$. A contradiction.

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    $\begingroup$ Thinking about your answer more, I'm surprised that a more general theorem is not listed somewhere in the literature about showing irreducibility. It seems like your argument shows the more general claim that if $P(x) \in \mathbb{Z}$, and $P(x)= a_n x^n + a_{n-1}x^{n-1} \cdots +a_1 x^1 +a_1$, and $a_1= \pm$ where there $a_i$ are integers, $a_1 = \pm 1$, $a_n$ is a prime, then under very general circumstances as long as $a_n$ is large compared to the sum of the other $a_i$, $P(x)$ will be irreducible. My guess is that this is already somewhere "officially" I've never seen it. $\endgroup$
    – JoshuaZ
    Commented Mar 13 at 19:02
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    $\begingroup$ I have seen some theorem of this type in Prasolov's book Polynomials $\endgroup$
    – Fedor Petrov
    Commented Mar 13 at 19:21

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