Timeline for A generalisation of the equation $n = ab + ac + bc$
Current License: CC BY-SA 4.0
27 events
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| Dec 16, 2020 at 4:55 | comment | added | individ | math.stackexchange.com/questions/419766/… | |
| Dec 16, 2020 at 4:54 | comment | added | individ | math.stackexchange.com/questions/872324/… | |
| Dec 15, 2020 at 22:48 | comment | added | Gerry Myerson | Numbers of the form $abc + abd + acd + bcd$, with $0 < a < b < c < d$ tabulated at oeis.org/A179796 with the observation, "All numbers greater than $181799$ appear to be in this sequence." That page also links back to this question. | |
| Dec 15, 2020 at 20:53 | history | edited | YCor | CC BY-SA 4.0 |
formatting (the question was bumped anyway)
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| Dec 15, 2020 at 20:23 | answer | added | Nikos Bagis | timeline score: -1 | |
| Jul 18, 2011 at 13:22 | history | bounty ended | Jernej | ||
| Jul 11, 2011 at 22:57 | comment | added | Will Jagy | GH, good. So my impression of usage is that the factorization trick involved in showing $x^2 + y^2 + z^9 \neq 216 p^3$ is not usually referred to as a local obstruction. Something similar happens with $x^2 + 27 y^2 + 7 z^3,$ another one by Kap with an even larger sum-of-exponent-reciprocals, but having coefficients this time. I sent a variant of that one to the M.A.A. Problems and Solutions, it appeared in December 2010 Monthly. The same trick gives most, or all, easy examples of spinor exceptional integers for positive ternary integral quadratic forms, but not the splitting integers. | |
| Jul 11, 2011 at 22:41 | comment | added | GH from MO | @Will: Thanks for the clarification! | |
| Jul 11, 2011 at 22:40 | comment | added | Will Jagy | GH, I must agree, I was just illustrating that the rough density argument may not be enough. I remember, though, R.C. Vaughan telling Kaplansky that the obstruction here could not be detected $p$-adically, and as such this defeated a conjecture in the first edition of his book on the Hardy-Littlewood method. So, in the second edition, on page 127 it says "There are some exceptions to this, see Exercise 5," then Exercise 5 on page 146 is about $x^2 + y^2 + z^9.$ | |
| Jul 11, 2011 at 21:23 | comment | added | GH from MO | @Will: I am not sure if I am following you, but the heuristic is that if there are no local obstructions (unlike in your example) and the number of variables is sufficiently large (more precisely the sum of reciprocals of the various degrees is large) then there are no global obstructions (i.e. the equation has a solution). There are ambitious conjectures along these lines like $G(k)\ll k$ in the Waring problem, but we are far from proving these. | |
| Jul 11, 2011 at 21:13 | comment | added | Will Jagy | On GH's note, I do not see how to take advantage of homogeneity. So this example may not have the correct taste, then, but $ x^2 + y^2 + z^9 \neq 216 p^3$ for integers $x,y,z$ and any (positive) prime $p \equiv 1 \pmod 4.$ The point is that the number of integer triples $(x,y,z)$ such that $x,y,z \geq 0, \; x^2 + y^2 + z^9 \leq N$ for some large positive $N$ is roughly $C N^{10/9}$ for a constant $C.$ So one might think large values of $N$ would be represented. | |
| Jul 11, 2011 at 20:53 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Jul 11, 2011 at 20:43 | history | edited | Will Jagy | CC BY-SA 3.0 |
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| Jul 11, 2011 at 20:37 | comment | added | GH from MO | This looks like a very difficult problem as the number of variables $m$ is almost the same as the degree $m-1$. It is of similar flavor to the question "Can we write every sufficiently large number as a sum of four cubes?" or "Is $G(k)<100k$ in the Waring problem?" I would be surprised if this problem were resolved in the next 20 years. | |
| Jul 11, 2011 at 20:20 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Jul 11, 2011 at 12:58 | history | edited | Jernej | CC BY-SA 3.0 |
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| Jul 11, 2011 at 12:51 | comment | added | Harry Altman | By "taken from a set", what you mean is that they're distinct, right? "Taken from a set" doesn't really mean anything. But it probably is not worth worrying about distinctness at first... | |
| Jul 11, 2011 at 12:33 | history | bounty started | Jernej | ||
| Jul 11, 2011 at 12:33 | history | edited | Jernej | CC BY-SA 3.0 |
changed representation of n to the one suggested in the comments
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| Jul 27, 2010 at 0:46 | comment | added | Wadim Zudilin | It looks like the special case $m=3$ has roots to the Erdős–Straus conjecture (en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Straus_conjecture) and the related one due to Wacław Sierpiński. The wikipedia article at least contains many useful links on this case but also on $m>3$ (including Erdős's paper in Hungarian!). | |
| Jul 26, 2010 at 23:46 | comment | added | Wadim Zudilin | Why don't you mention the case $m=2$? It's (almost) the abc conjecture! ;-) | |
| Jul 26, 2010 at 22:32 | comment | added | tdnoe | If we allow equality, the $n_0$ have been conjectured in oeis.org/classic/A027565, which appears to be growing exponentially. | |
| Jul 26, 2010 at 18:37 | comment | added | Will Jagy | I would write this as asking for a representation with all positive $x_j$ as $$ n = x_1 x_2 \ldots x_m \left( \frac{1}{x_1} + \frac{1}{x_2} + \ldots + \frac{1}{x_m} \right) $$ for any $ n > n_0.$ This agrees with your examples for $m=3$ and $m=4.$ Is this what you want? I find your way of writing this and speaking of a set $X$ as clouding the issue. | |
| Jul 26, 2010 at 17:54 | comment | added | Daniel Litt | Ah I see, sorry. | |
| Jul 26, 2010 at 16:03 | comment | added | Jernej | I think it is OK to omit the inequality as long as the numbers are taken from a set. | |
| Jul 26, 2010 at 15:48 | comment | added | Daniel Litt | To make this a true generalization, do you want to require $0<x_1<x_2< ...< x_m$? | |
| Jul 26, 2010 at 15:38 | history | asked | Jernej | CC BY-SA 2.5 |