Timeline for Is $x^{2k+1} - 7x^2 + 1$ irreducible?
Current License: CC BY-SA 3.0
12 events
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| Jan 7, 2017 at 18:32 | comment | added | Michael Stoll | Re my last comment but one, it turns out that the convergence of the absolute value of the "large" roots divided by $7^{1/(2k-1)}$ to $1$ is too slow (error of order $1/k$) to deduce more than that any nontrivial divisor $h$ of $x^{2k+1} - 7 x^2 + 1$ would have to have "relative degree" $(\deg h)/(2k+1)$ in a fairly short interval around $1/2$. | |
| Jan 7, 2017 at 11:27 | comment | added | Igor Rivin | @MichaelStoll Hmm, was too late at night - I removed it... | |
| Jan 7, 2017 at 11:26 | history | edited | Igor Rivin | CC BY-SA 3.0 |
removed addendum
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| Jan 7, 2017 at 8:30 | comment | added | Michael Stoll | The statement in the ADDENDUM is wrong (for example for $2k+1 = 19$, there are six factors of degree 3). If $2k+1 = p$ is prime, the degree of the nonlinear factors mod 7 is the order of 7 in the multiplicative group mod $p$, which does not have to be $p-1$. Assuming Artin's primitive root conjecture, we get at least infinitely many $k$ such that the polynomial is irreducible. | |
| Jan 7, 2017 at 0:59 | history | edited | GH from MO | CC BY-SA 3.0 |
changed "
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| Jan 6, 2017 at 21:32 | history | edited | Igor Rivin | CC BY-SA 3.0 |
added more comments.
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| Jan 6, 2017 at 21:02 | comment | added | Michael Stoll | The remaining roots will converge to the roots of $x^{2k-1} - 7$, so will have absolute value close to $7^{1/(2k-1)}$. So we can expect it to be hard to find any number of these such that the product has absolute value sufficiently close to $\sqrt{7}$. If somebody can get sufficiently good estimates, this should lead to a proof. | |
| Jan 6, 2017 at 20:27 | comment | added | T. Amdeberhan | I was just being picky. I agree. :-) | |
| Jan 6, 2017 at 20:14 | comment | added | Michael Stoll | @T.Amdeberhan I wouldn't necessarily think that $|a+b| \le |a|+|b|$ is so much easier than $|a-b| \ge |a| - |b|$; they are just two equivalent formulations of the same fact. | |
| Jan 6, 2017 at 18:59 | comment | added | T. Amdeberhan | True. Easier is (triangular inequality): $\vert x^{2k+1}+1\vert\leq 2<7=\vert 7x^2\vert$ on the unit circle. | |
| Jan 6, 2017 at 18:14 | comment | added | Michael Stoll | I think the claim that there are exactly two roots of absolute value less than 1 in $\mathbb C$) should follow from Rouché's theorem, since $|x^{2k+1}| = 1 < 6 \le |-7x^2 + 1|$ on the unit circle. | |
| Jan 6, 2017 at 11:48 | history | answered | Igor Rivin | CC BY-SA 3.0 |