Timeline for Distinct integer roots for a degree 7+ polynomial and its derivative
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 13, 2021 at 0:09 | vote | accept | Benjamin Dickman | ||
| S Sep 13, 2021 at 0:09 | history | bounty ended | Benjamin Dickman | ||
| S Sep 13, 2021 at 0:09 | history | notice removed | Benjamin Dickman | ||
| Sep 7, 2021 at 23:23 | comment | added | Benjamin Dickman | @PeterTaylor Thank you for trying! | |
| Sep 7, 2021 at 22:56 | comment | added | Peter Taylor | An approach which didn't go very well: I was inspired by mathoverflow.net/q/402225 to try lifting a solution mod 29 to the integers. By brute force find solution $P(x) = \prod_i (x - a_i) \pmod{29}$ with derivative $P'(x) = \prod_i (x - b_i) \pmod {29}$. Lift to $\tilde{P}(x) = \prod_i (x - a_i + 29v_i)$ with the $v_i$ as variables; lift $P'$ to $\tilde{P'}$ similarly and equate the coefficients of $\tilde{P'}$ with the coefficients of $\tfrac{\textrm{d}}{\textrm{d}x} \tilde{P}$ and ask a CAS to find the prime ideals of the ideal so generated. I aborted when Sage ran out of RAM. | |
| Sep 7, 2021 at 19:11 | answer | added | Terry Tao | timeline score: 19 | |
| Sep 7, 2021 at 17:36 | comment | added | Benjamin Dickman | @TerryTao I'm not sure what a reasonable expectation is for the bounty period; so, a clear explanation [as an answer, not comment] as to the potential barriers could be bountiful + helpful [at least for my purposes!]. Still holding hope but not breath that someone has a novel idea or wants to run a refined(ish) search and see whether a "major algebraic miracle" arises... | |
| Sep 7, 2021 at 17:21 | comment | added | Terry Tao | Looking again at Choudry's paper, all of their birational transformations preserve the degree deficit of $1$ (for symmetric $d=6$), ending up at an elliptic curve (45) in quartic form, which after clearing denominators is a degree 4 equation on 3 integer variables (so with deficit $4-3=1$). But one can of course place this elliptic curve in Weierstrass form by an appropriate Mobius transformation (sending a rational point to infinity), reducing the deficit to $3-3=0$. But I think it unlikely that such tricks can shave the deficit by much more than 1 without a major algebraic miracle. | |
| Sep 7, 2021 at 3:14 | comment | added | Terry Tao | (correction, I meant the (symmetrised) $d=8$ case, rather than $d=6$.) | |
| Sep 7, 2021 at 1:08 | comment | added | Terry Tao | In the paper of Choudhry linked to by Benjamin, some ad hoc birational transformations are used to locate an elliptic curve inside this particular projective variety that lets one overcome the naive surplus in degree. It's potentially possible that some similar ad hoc birational embedding may also overcome the surplus in higher degree but it gets quadratically harder as the degree increases - for instance for $d=6$ one has to overcome a degree deficit of $2+4+6-(4+3)=5$ instead of just $2+4-(3+2)=1$ which looks way too ambitious to work to me. | |
| Sep 7, 2021 at 0:36 | comment | added | Will Sawin | @TerryTao if the unknowns are the roots, then the equations are given by the equalities of the coefficients in $ \frac{d}{dx} \prod_{i=1}^d (x-a_i) = d \prod_{i=1}^{d-1} (x-b_i)$, which have degrees $1,\dots, d-1$. This explains why there are solutions with $d=1,2,3,4$. For $d=6$, symmetric, I see $3 +2 $ variables and equations of degrees $2,4$, so I still don't see an easy explanation of why there are solutions there... | |
| S Sep 6, 2021 at 23:23 | history | bounty started | Benjamin Dickman | ||
| S Sep 6, 2021 at 23:23 | history | notice added | Benjamin Dickman | Draw attention | |
| Sep 5, 2021 at 17:11 | comment | added | Benjamin Dickman | @TerryTao I wonder whether the sort of methods employed e.g. here suggest that "naive probabilistic heuristics" are not fine enough to draw conclusions about $d > 6$ examples. (Candidly, the content from (29) onward is beyond my ken – elliptic curves, APECS/Mordell-Weil basis, etc. Hoping someone gives this a shot!) | |
| Sep 4, 2021 at 16:38 | comment | added | Benjamin Dickman | @StevenStadnicki A degree eight example would answer my question in the affirmative! Not sure if anyone is up for implementing a search; some of the linked background materials in the OP used tactics much finer than brute force... Drawing from elliptic curves and techniques outside my bailiwick. | |
| Sep 3, 2021 at 20:27 | comment | added | Terry Tao | One is asking to solve a diophantine system of $d-1$ equations of degree $d-1$ in $2d-1$ integer unknowns. Naive probabilistic heuristics then suggest that this equation is likely to have no non-trivial solutions once $d$ gets large; it's already somewhat lucky that there are symmetry reductions and other degree-lowering transformations that allow solutions at $d=6$. So I would imagine that there are no non-trivial solutions for $d>6$, but that actually proving this would be beyond current technology. | |
| Sep 3, 2021 at 20:07 | comment | added | Steven Stadnicki | Thinking about it more, I believe a symmetric degree-5 example is impossible (and likewise degree-7); in degree 5, for instance, we have $x(x^2-a^2)(x^2-b^2)$ $= x^5-(a^2+b^2)x^3+a^2b^2x$ with derivative $5x^4-3(a^2+b^2)x^2+a^2b^2$ $=5(x^4-\frac35(a^2+b^2)x^2+\frac15a^2b^2)$ and of course the last term can't be a product of two integer squares itself. I wonder whether it might be easier to look for a symmetric degree-8 example than an asymmetric degree-7 one. | |
| Sep 3, 2021 at 19:20 | comment | added | Steven Stadnicki | I haven't done the relevant complex multiplications yet, but I suspect it's not entirely coincidence that all of the roots of the degree-6 polynomial and its derivative in the symmetric example are writable as sums of two squares (and therefore norms of complex numbers). It might be worth a short computer search for symmetric degree-5 examples; assuming their roots are of comparable size to the degree-6 ones, that's only a few million cases to check and they'll easily fit in 64-bit integer arithmetic. | |
| Sep 3, 2021 at 18:38 | comment | added | Benjamin Dickman | @StevenStadnicki I don't know of any quintics with that symmetry. The linked MSE question includes a quintic but it is not of that form; specifically, p. 23 in this paper identifies: $p(x) = x(x − 180)(x − 285)(x − 460)(x − 780)$ as having the desired property. | |
| Sep 3, 2021 at 18:34 | comment | added | Steven Stadnicki | Are there any quintics known with this 'special form' (roots symmetric about the origin?) Reducing the search space would certainly help. | |
| Sep 3, 2021 at 18:13 | history | asked | Benjamin Dickman | CC BY-SA 4.0 |