Timeline for Is $x^{2k+1} - 7x^2 + 1$ irreducible?
Current License: CC BY-SA 3.0
26 events
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| Jan 16, 2017 at 14:24 | comment | added | Michael Stoll | @Lucian ... but the numerator is, and Ljunggren tells you that when it is reducible, it is the product of two irreducible factors. | |
| Jan 16, 2017 at 13:28 | comment | added | Lucian | @MichaelStoll: $~\dfrac{x^{6k+2}-x+1}{x^2-x+1}~$ is not a trinomial. | |
| Jan 16, 2017 at 12:57 | comment | added | Michael Stoll | @Lucian The first, no; the second is answered (positively) by Ljunggren's paper linked in my first comment here. | |
| Jan 16, 2017 at 9:19 | comment | added | Lucian | @MichaelStoll: Do you have any idea on how to tackle these two questions ? | |
| Jan 8, 2017 at 19:21 | comment | added | Michael Stoll | See also mathoverflow.net/questions/56579/about-irreducible-trinomials (linked from the MSE thread KConrad mentions). | |
| Jan 8, 2017 at 16:21 | comment | added | KConrad | This looks similar to the proof I posted earlier for irreducibility of $x^n-x-1$ at math.stackexchange.com/questions/393646/irreducibility-of-xn-x-1-over-mathbb-q/393779 | |
| Jan 8, 2017 at 13:34 | comment | added | Will Sawin | @MichaelStoll (1) Indeed I meant that. (2) Sure, but in any case it's totally explicit so we can compute in a finite amount of time an exact description of which polynomials are irreducible and which are not. (3) Thanks! | |
| Jan 8, 2017 at 13:31 | comment | added | Michael Stoll | @WillSawin (1) You probably want the $a_j$ to be irreducible, to exclude any further possibilities of factorization. (2) It is possible that some of the roots of $d(x^{-1})d(x) - c(x^{-1})c(x)$ are roots of $c(x^{-1})x^N + d(x)$ for infinitely many $N$, e.g., when $c = 1$, $d = 1+x$ and $N \equiv 2 \bmod 3$. (3) I like the compactness argument. | |
| Jan 8, 2017 at 13:23 | comment | added | Will Sawin | I guess the point is any common roots of $c(x^-1) x^N + d(x)$ and $d(x^{-1}) x^N + c(x)$ must also be roots of $d(x^{-1}) d(x) - c(x^{-1}) c(x)$ and we can eliminate those finitely many roots by some explicit calculation. | |
| Jan 8, 2017 at 13:18 | comment | added | Will Sawin | For the second part, given $G(x) = \pm x^a f(x)$, we observe that since $f(x)$ and $f(x^{-1})$ have no common roots, $h(x)$ and $h(x^{-1})$ have no common roots, so all common roots of $f(x)$ and $G(x)$ must be roots of $g(x)$, so $h(x)=1$. We can run the same argument if $G(x) = \pm x^a f(x^{-1})$, but it seems we also need to check that $f(x)$ and $f(x^{-1})$ have no common roots in general, so maybe I was a bit hasty. | |
| Jan 8, 2017 at 13:14 | comment | added | Will Sawin | The first part is the same, with the key point being that if there is an alternate factorization for all finite powers with a given sum of squares, then by compactness there is an alternate factorization in power series with a given sum of squares, which must be a factorization in polynomials because there are only finitely many nonzero coefficients. | |
| Jan 8, 2017 at 13:14 | comment | added | Will Sawin | Let $a_1,\dots,a_n$ be integer polynomials with constant term one. Consider all the ways to divide $a_1,\dots,a_n$ to two groups and multiply the elements of each group together, obtaining two polynomials. Let $c$ and $d$ be the pair of polynomials obtained where the sums of the squares of coefficients of both polynomials is minimal. Assume that furthermore they are unique, and that $c(x^{-1})$ and $d(x)$ relatively prime to each other and to $x$. It seems that your argument implies $f(x) = c(x^{-1}) x^N +d(x)$ is irreducible for $N$ large enough, where "large enough" is explicitly computable. | |
| Jan 8, 2017 at 11:36 | comment | added | Michael Stoll | @GHfromMO I have made this clearer in my answer. | |
| Jan 8, 2017 at 11:35 | history | edited | Michael Stoll | CC BY-SA 3.0 |
made the condition on the size of the solution more precise, prompted by GHfromMO's comment
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| Jan 8, 2017 at 11:32 | comment | added | Michael Stoll | @GHfromMO We consider initial segments of the coefficient vector of $G$ from both ends that do not overlap (this leads to the condition $n>40$), so the sum of squares of both together is bounded by the sum of squares of all coefficients of $G$. | |
| Jan 8, 2017 at 11:31 | comment | added | Michael Stoll | The same method gives the irrducibility of $x^n - 11x^2 + 1$ for $n \neq 4$. The computation is more involved and needs to consider terms up to $x^{37}$. Some more experimentation shows that it proves that $x^n - mx^2 + 1$ is irreducible for $n > 4$ and $m = 3,5,6,7,8,10,11,12,13$. | |
| Jan 8, 2017 at 11:30 | comment | added | GH from MO | I am slightly confused by the condition $a_1^2 + a_2^2 + \ldots + b_1^2 + b_2^2 + \ldots \le 49$. Why is it not $a_1^2 + a_2^2 + \ldots\le 50$ and $b_1^2 + b_2^2 + \ldots \le 50$? I mean, we only know that $\|G\| = 51$. | |
| Jan 8, 2017 at 11:11 | history | edited | Michael Stoll | CC BY-SA 3.0 |
fixed details of computation
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| Jan 7, 2017 at 18:35 | history | edited | Michael Stoll | CC BY-SA 3.0 |
fixed a grammatical mistake
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| Jan 7, 2017 at 10:00 | comment | added | Michael Stoll | Ljunggren treats the case $x^n \pm x^m \pm 1$ in his paper (Theorem 3). | |
| Jan 7, 2017 at 9:47 | comment | added | Michael Stoll | @Pablo I found it on page 97 of Zannier's "Lecture notes on Diophantine Analysis" (footnote to exercise 3.20): "Selmer proved that actually $f_n$ is irreducible for such $n$. Here is a sketch of an argument due to Ljiungrenn. ..." Alas, there is no "Ljiungrenn" in MathSciNet or ZMATH. This should be "W. Ljunggren", and the original source appears to be this: mscand.dk/article/view/10593/8614. | |
| Jan 7, 2017 at 9:43 | vote | accept | Pablo | ||
| Jan 7, 2017 at 9:33 | history | edited | Michael Stoll | CC BY-SA 3.0 |
simplified part of the argument
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| Jan 7, 2017 at 9:32 | comment | added | Will Sawin | This looks awesome and potentially quite general! The first part can be expressed without integration in a way that to me is a little more intuitive: The identity $f(x) f(x^{-1}) = G(x) G(x^{-1})$ implies that the sum of squares of coefficients of $f(x)$ equals the sum of squares of coefficients of $G(x)$ because the sum of the squares of coefficients of $p(x)$ is the coefficient of $0$ in $p(x) p(x^{-1})$ by the obvious expansion. | |
| Jan 7, 2017 at 9:25 | comment | added | Pablo | That's amazing! do you have a reference for the original argument? | |
| Jan 7, 2017 at 8:55 | history | answered | Michael Stoll | CC BY-SA 3.0 |