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.
Best Answer
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.