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[math-fun] Information about "Primes are P"?
by Ki Song 27 Mar '03

27 Mar '03

Hello, Has anyone heard of a research paper titled "Primes are P" that was published last year by an Indian mathematican? Ki Song _________________________________________________________________ Protect your PC - get McAfee.com VirusScan Online http://clinic.mcafee.com/clinic/ibuy/campaign.asp?cid=3963

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[math-fun] Wanna be hung like a horse?
by Christina Long 27 Mar '03

27 Mar '03

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[math-fun] Re: [EMHL] Square determined by distances to 3 vertices
by Richard Guy 25 Mar '03

25 Mar '03

Certainly not off topic for this member! Of interest in connexion with The Inquisitive Problem Solver (Vaderlind, Guy & Larson) Problem 234. There, we're concerned with the perimeters of the triangles formed by a side and two of the distances, and the relevant equation is a sextic, I believe I recall, though we don't bother our readers with that. When you move into higher dimensions you must make sure that all dimensions are represented. As you observe, in three dimensions, the four corners of one face won't do, because the point doesn't know that it's outside that plane. Of course, in general, four distances won't be compatible. There's also a connexion with an unsolved problem, now possibly describable as clasical (see D19 in Unsolved Problems in Number Theory): Is there a point, in the plane of a rational sided square, at rational distances from all four corners of the square? Let me know if any Hyacinthian or Math Funster solves this! R. On Tue, 25 Mar 2003, mjc_w wrote: > Recently, inspired by a thread in sci.math, I worked out the > solution to determining the size of a square given the > distances from a point to 3 of its vertices. I know that this > has been done many times before, but I found some of my results > interesting. > > To summarize, if the distances to 3 consecituve vertices from a > point are a, b, and c, with b in the middle, then the side s > has its square t=s^2 satisfying the quadratic > > 0 = 2*t^2 - 2*t*(a^2+c^2) + (a^2-b^2)^2 + (c^2-b^2)^2 > > So, the side is clearly constructable. > > I noticed that the discriminant of this quadratic can be factored as > > -( (a-c)^2 - 2*b^2 )*( (a+c)^2 - 2*b^2 ) > > so that, since this must be positive to have a real root, > the condition for the square to exist is > > |a-c| <= b*sqrt(2) <= a+c. > > I have extended this to the case of determining the side > of an n-dimensional cube from the distances of a point to > n+1 of the vertices. It turns out that this can be done > if the ordinates of the vertices do not omit any of the > ordinates of the sides of the cube (e.g., for a 3-d cube, having > all 4 vertices on one face does not work). In particular, > having n vertices all adjacent to the n+1-st vertex works. > > The resulting equation is, again, a quadratic with roots, > when they exist, being the square of the side of the > n-dimensional cube. > > If anyone is interested, I would be glad to show details. > > BTW, if this if off topic for this group, my apologies. > > Martin Cohen > > > ------------------------ Yahoo! Groups Sponsor ---------------------~--> > Your own Online Store Selling our Overstock. > http://us.click.yahoo.com/rZll0B/4ftFAA/46VHAA/6HYolB/TM > ---------------------------------------------------------------------~-> > > > > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/ > >

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[math-fun] geometric algebra & blinding galois fields
by Mike Stay 24 Mar '03

24 Mar '03

I think I found two ways of approaching that "galois field in the exponent" idea I had: geometric algebra and blinding. Blinding is the easiest to explain: you have an element of GF(p^n) = a_0 + a_1 x + ... + a_{n-1} x^{n-1} that you wants to blind. You pick q such that p | q-1. You pick a generator g of a subgoup with order p. Then you blind each of the a_i by calculating g^{a_i}. You can add two blinded elements by multiplying each blinded term and then dividing by the corresponding terms of the primitive polynomial as many times as is necessary. You can multiply a blinded element by a non-blineded one by shifting & dividing in the obvious way. This isn't very cryptographically useful because p is usually small. But it illustrates the idea of a "logarithm" that behaves like a galois field in that when you multiply them, you add the field elements. In geometric algebra, one has various grades of elements: scalars, vectors, bivectors, etc. In 3-d geometric algebra, the unit blades are 1, e_1, e_2, e_3, e_1 e_2, e_2 e_3, e_3 e_1, and e_1 e_2 e_3. Since any vector squared is 1, we can do the same sort of thing with blades as with the blinded galois field: (e_1 e_2)(e_2 e_3) = e_1 e_3, 110+011 = 101. Of course, there's a lot more structure in geometric algebra, and it's not clear in either case how to do the modular reduction of the polynomial when one has the sum of two differently-graded elements like 1+e_1, but I thought the relationship between them was worth noting. It also begs the question of whether you can come up something like geometric algebra to deal with larger primes in the field: Say e_i cubed is 1 rather than squared. Looking at the "phase factor" we get when exchanging elements, e_1 e_2 e_1^2 = x e_2 e_1 e_1^2 = x e_2 but e_1 e_2 e_1^2 = x^2 e_1 e_1^2 e_2 = x^2 e_2 so x=x^2, x=0 or 1. The same sort of thing happens for all odd numbers, so it looks like whatever algebra would satisfy that relation would be symmetric and commutative. Can anyons come into play here? It seems like we can't get higher roots of 1 because multiplication is a binary operation. Is there any way to define a ternary "multiplication" that would give a phase factor of 2pi/3? Thanks, -- Mike Stay staym(a)clear.net.nz http://www.xaim.com/staym

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[math-fun] Cool Lego math sculptures...
by Henry Baker 23 Mar '03

23 Mar '03

at http://www.lipsons.pwp.blueyonder.co.uk/lego.htm >From MAA's "Focus" mag., 23,3 (Mar 2003).

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[math-fun] imagining numbers
by Helger Lipmaa 18 Mar '03

18 Mar '03

Has anybody read Mazur's book "Imagining numbers"? Any comments? Helger

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[math-fun] (No) Math-Fun rule for posting news articles
by Richard Schroeppel 15 Mar '03

15 Mar '03

We rely on the judgment of list members as to what is acceptable. The Math-Fun list has no particular rules about posting news articles. We do have formatting rules (stick to plain text), and the list-server will hold anything over 40KB for approval. If the subset of people who want to argue about copyright will get in touch with me, I'll see if we can set up another list. I'll kick off the discussion with a funny story about my attempts to subscribe to the NYT in rural Utah. Rich rcs(a)cs.arizona.edu

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Re: [math-fun] Extension of a theorem on n-gons
by Richard Schroeppel 15 Mar '03

15 Mar '03

I've edited Steve Gray's message to kill the PDF attachment. You can get the paper directly from him (stevebg(a)adelphia.net) Rich ----- Gene Salamin asked: > Could you provide a statement of the 100 year old result, and of your > extention to it? Steve Gray replied: The 98-year-old result has been called the Douglas-Neumann theorem, but was first found by K. Petr in 1905, so I call it the Petr-Douglas-Neumann theorem, after Jesse Douglas and B.H. Neumann. It concerns layers of isosceles triangles constructed on the sides of an n-gon. After n-2 layers have been constructed, the resulting n-gon is regular, and after one more layer, a single point results. The first layer of triangles have apex angle 360/n, the second twice that, the third 3x360/n, etc. In my generalization the triangles used are not isosceles but are derived from an arbitrary-shaped "model" n-gon which is in general not similar to the one the triangles are placed on. After n-2 layers one gets an n-gon similar to the model, and after one more layer, a single point results. This is the first of several generalizations I have found and proved. I am currently working on another generalization which, if I ever find and prove it, will be the basis of another paper. However I already have enough material for at least one more paper, and probably two or three. One interesting thing about all these theorems is that they are "angle-only" theorems, that is, the constructions and results can be stated in natural fashion without reference to distances or lengths, which appear only secondarily. They are among the few theorems involving n-gons, where n is arbitrary. In some of my variations n must be prime, and in some others, certain of the construction parameters must have (simple) number-theoretic relationships. I'm enclosing a PDF of the final paper. If you have any more specific questions, please write back, although I do want to keep certain details of the other theorems private until I send in the next paper. Coincidentally, the same issue of the Monthly contains a geometry problem I submitted in 9/2001 that I had forgotten about. Steve Gray

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[math-fun] small prime gaps
by Richard Schroeppel 14 Mar '03

14 Mar '03

Date: Thu, 13 Mar 2003 23:11:22 -0500 Reply-To: Hugh Montgomery <hlm(a)UMICH.EDU> Sender: Number Theory List <NMBRTHRY(a)LISTSERV.NODAK.EDU> From: Hugh Montgomery <hlm(a)UMICH.EDU> Subject: small gaps between primes To: NMBRTHRY(a)LISTSERV.NODAK.EDU Oberwolfach 13 March, 2003 This week at Oberwolfach, a meeting on Elementary and Analytic Number Theory is taking place, organized by Bruedern, Montgomery, and Vaughan. At this meeting, today, Dan Goldston spoke about his new joint work with Yildirim concerning primes in short intervals. They seem to have proved that liminf (p_{n+1} - p_n)/log p_n = 0. Indeed, it looks like they will be able to show that p_{n+1} - p_n is infinitely often as small as (log p_n)^{4/5}. It may be that the method can be applied also to long gaps between primes, with the prospect of possibly breaking the Erdos--Rankin order of magnitude. I would say that this is the biggest excietement that prime number theory has seen since the Bombieri--Vinogradov theorem was proved in 1966. Hugh Montgomery

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Re: [math-fun] Ulam(1,2)
by Dan Hoey 14 Mar '03

14 Mar '03

"David Wilson" <davidwwilson(a)attbi.com> wrote: > I took a look at Ulam(1,2), the Ulam sequence starting with (1, 2) and > including every subsequent number which is a unique sum of distinct > earlier terms. This is Sloane's A002858. > ... Can anyone confirm my data? Did you intend to use zero-based indexing? If u(0)=1, u(1)=2, I concur that u(1000)=12336, u(1999)=25511. Otherwise one of us has a bug. I also get u(10000)=132790, u(50000)=676043, u(74084)=1000002. But there's a pretty definite bug in the histogram. My histogram of f(n)=(u(n) mod 21.6)/21.6 agrees with yours in its general trend, but differs in particulars. For f(1000...1999) in 20 buckets I get 113 101 119 98 87 70 58 40 31 8 5 0 0 2 4 14 22 64 66 98 rather than your 110 101 120 97 90 67 58 40 31 9 4 0 0 2 7 11 22 64 66 101. As a test, I find that the five numbers in the [.5,.55) bucket arise from u(1091)=13533 (f=0.527771) u(1210)=14894 (f=0.53704834) u(1730)=21870 (f=0.5) u(1748)=22151 (f=0.50927734) u(1947)=24851 (f=0.50927734). What four numbers do you have? I agree exactly with Klaus Brockhaus's data for f(1000...1259), so I think I'm slightly likelier to be correct, but only slightly, and I'm pretty suspicious about the zero-based indexing. My data suggests that 21.6 should be about 21.60157 . Dan

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