Timeline for Integers not represented by $ 2 x^2 + x y + 3 y^2 + z^3 - z $
Current License: CC BY-SA 2.5
13 events
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| Apr 8, 2011 at 20:52 | comment | added | Will Jagy | I showed this problem to Henri Cohen, whom I met yesterday, including the class number 5 discriminant (-47) variant. I asked if he knew you, he said "Of course." Hendrik Lenstra was not in his office next door. I also emailed this Math Overthrow link, but I don't know that Prof. Cohen checks his msri.org email, many ignore it. | |
| Mar 31, 2011 at 17:58 | comment | added | Kevin Buzzard | He has to write it yet :-) Deadline some time in June I guess. | |
| Mar 30, 2011 at 19:56 | comment | added | Will Jagy | When the time comes, if you judge it appropriate, I would love to see a pdf of the thesis. Do you think the same result works for monic quintics and fifth powers in the class group? I remember that example of discriminant -47 from Appendix C in Henri Cohen, Advanced Topics in Computational Number Theory. For that one, I never did figure out what numbers ought to be ruled out by taking the binary quadratic form non-principal (also discriminant -47, so x^2 + x y + 12 y^2, 2 x^2 +- x y + 6 y^2, 3 x^2 +- x y + 4 y^2 ). | |
| Mar 30, 2011 at 18:43 | comment | added | Kevin Buzzard | I set a masters thesis for a student in this area, to see what they could do. They can prove things like "x^2+xy+6y^2+F(z) represents all integers" for F any monic cubic with integer coefficients, but with a non-principal form which is not a cube in the class group I don't think that either he, or I, have a good method for proving that most integers are represented. | |
| Mar 29, 2011 at 22:43 | history | edited | Gerry Myerson | CC BY-SA 2.5 |
improved formatting
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| Mar 29, 2011 at 20:59 | comment | added | Will Jagy | Kevin, thanks for letting me know about the edit (I had one new comment when I logged in). And thanks for putting in some effort to finalize your presentation. As when you first suggested the idea, I remain fascinated by the possibility of proving that all other numbers are integrally represented by the polynomial. Do you think you can do the difficult direction for the (much) easier polynomial in my Problem in the December 2010 M.A.A. Monthly? Of course I did not ask the difficult direction as part of the Problem, but have been hoping people might include that in their submitted solutions. | |
| Mar 29, 2011 at 12:26 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
I am less optimistic about being able to prove what I claim in this post, so edited the post to reflect this.
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| Feb 17, 2010 at 12:12 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
deleted 362 characters in body
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| Feb 16, 2010 at 16:53 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
Added Lenstra's proof.; added 6 characters in body
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| Jan 29, 2010 at 18:57 | vote | accept | Will Jagy | ||
| Jan 29, 2010 at 18:50 | comment | added | Anweshi | @buzzard. I suggest that you also make a community wiki post with your mail to NMBRTHRY, and possible responses coming in there. | |
| Jan 29, 2010 at 14:28 | comment | added | Franz Lemmermeyer | As for conjecture 2: the primes dividing D are the index divisors of the cubic polynomial defining the Hilbert class field. I know that common index divisors necessarily split; if this is true for all index divisors that are primes (except for finitely many divisors of the coefficients), Conj. 2 would follow since primes splitting in the Hilbert class field must be represented by the principal form. I guess there's some classical result on index divisors that takes care of Conj. 2. | |
| Jan 29, 2010 at 11:59 | history | answered | Kevin Buzzard | CC BY-SA 2.5 |