May 2003 - mathfun@mailman.xmission.com

[math-fun] Re: Quadratic question
by Jpr2718＠aol.com 04 May '03

04 May '03

David Wilson asked: > For which (a, b) do all solutions of > a x^2 - b y^2 = [1] > have the form (r k, s j) for some fixed r, s > 1? D. T. Walker, ``On the Diophantine Equation $mX^2 - nY^2 = \pm 1$, American Mathematical Monthly, Volume 74, Issue 5, (May 1967), pp. 504-513, has a theorem that answers Professor Wilson's query. Consider (1) $aX^2 - bY^2 = \pm 1$. Theorem 9 in the above is (a and b are positive integers, not squares (and whose product is not a square)): If (1) is solvable for both of a, b > 1 and has r \sqrt{a} + s \sqrt{b} as its smallest solution; and if i is a nonnegative integer, then the formula r_i \sqrt{a} + s_i \sqrt{b} = (r \sqrt{a} + s \sqrt{b} )^{2i + 1} gives all possible solutions of (1). So r will always divide r_i and s will always divide s_i. Any a, b missing from the list in Professor Clark's reply apparently are not there because either the equation has no solution for those values, or because at least one of r, s is 1. John John Robertson 973-331-9978 jpr2718(a)aol.com

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[math-fun] Re: Quadratic question
by Jpr2718＠aol.com 02 May '03

02 May '03

David Wilson asked: > For which (a, b) do all solutions of > a x^2 - b y^2 = [1] > have the form (r k, s j) for some fixed r, s > 1? Consider the binary quadratic form equation Ax^2 + Bxy + Cy^2 = N, with discriminant D = B^2 - 4AC. Any two solutions x, y and x', y' are equivalent if one can be derived from the other by x = ((u-Bv)/2)x' - Cvy', y = Avx' + ((u+Bv)/2)y', where u, v is any solution to X^2 - DY^2 = 4. When N = 1, any two solutions are equivalent. (See the section ``2 The structure of solutions'' in Keith Matthews, The Diophantine equation ax^2 + bxy + cy^2 = N, D = b^2 - 4ac > 0, Journal de Theorie des Nombres de Bordeaux, 14 (2002) 257-270.) Let r, s be the minimal positive solution to ax^2 - by^2 = 1. Then (4ar^2 - 2)^2 - 4ab(2rs)^2 = 4. Assume, for the moment, that 4ar^2-2, 2rs is the minimal positive solution to X^2 - 4abY^2 = 4. If x', y' is a solution to ax^2 - by^2 = 1, and if x' is a multiple of r and y' is a multiple of s (as they are for the minimal solution), then x and y from the above formulas are also multiples of r and s respectively (because v = 2rs). So, by induction, all solutions are multiples of r and s. This says that solutions to ax^2 - by^2 = 1 can fail to have x and y be multiples of r, s only when 4ar^2-2, 2rs is not the minimal positive solution to X^2 - 4abY^2 = 4. I do not know whether this can occur. But, instead of Professor Wilson's question, we have the question as to whether there are any equations ax^2 - by^2 = 1 where x and y for any solution are _not_ multiples of those for the minimal solution. Richard Mollin probably discusses this, and it seems to me there is a paper in the Monthly from the 1950s that may address this issue. John John Robertson 973-331-9978 jpr2718(a)aol.com

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[math-fun] Quadratic question
by David Wilson 02 May '03

02 May '03

When I solved [1] 8 x^2 - 343 y^2 = 1, all solutions (x, y) were of the form (26041 j, 3397 k). An similar situation appears to be true of 22 x^2 - 29 y^2 = 1, where all solutions (x, y) seem to be of the form (31 j, 27 k). For which (a, b) do all solutions of a x^2 - b y^2 = 0 have the form (r k, s j) for some fixed r, s > 1?

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Re: [math-fun] hex on manifolds
by Dan Hoey 01 May '03

01 May '03

Bill Thurston writes: > Does anyone have some thoughts about generalizations of hex to > various manifolds? > ... For CP^2, a good rule seems to be to declare the winner to be > anyone such that H^2(their colors) is non-zero. Forgive me if this is exactly what you said, but I suggested playing hex on the projective plane a few years ago. Bill Taylor pointed out the natural winning configuration: a path that is not contractible to a point. (If that isn't what you said, what's CP^2?) You can play on a "Y" board, which is graph-isomorphic to a hemi-icosahedron with the sides triangulated. Edge moves are made by playing two antipodal stones. On the commercial board (available from Kadon), you play on the vertices of the triangulation. If you prefer ASCII art, I wrote a program that produces drawings where you play in the spaces. Here are three sizes of board. For the smallest, there's only one starting move, only one response, and then two of the four third moves are winners. The next size up has two openers--are they both winners? ____ ____ / \__ ____/ \ / __/ \__ __/ \__ \ \__/ \ \__ __/ __/ \__/ / \ \____/ \ __/ \____/ / \ \ \__/ \ \ __/ / \__ \ \ ____ / / \____/ / __/ \____/ __/ \__/ / / \____ \__/ / \__/ / / \__/ / \ \ / __/ \__ / \__/ __/ __/ __/ __/ \ \ \__/ \__/ \__ \____ \ \__/ / \__/ \__/ \ \____/ / \ / \ \____/ \__ \____/ / / / \__/ \__/ / / / __/ \__ \____ \ \__ \__ / \____/ \ \ \ \__/ \__ \____/ \__ \__/ \__/ \__/ / \ \__/ / / \ \____/ \__ \__ \ \ \__/ / / \ \__/ / __/ \__ \____/ \__ \__ \____/ \__ \__/ / / \ \__/ \__ \____/ \ \__ \__/ \ \__/ \ \ / / / \ \____/ \ \____/ \__ \__ \____/ / / \ \__/ / __/ \ \____/ \ \ \__/ \__ \__/ / / \ \__/ \__ \____/ \ \__ \__ \__ \__/ \ \__/ / / \ \__/ \ \____/ \__/ \__ \____/ / / \ \__/ / / \ \____/ \ \ \ \____/ / / / \ \____/ \____/ \ \__ \__ \__ \ \__/ \__/ \____/ \ \____/ \__/ \__/ \__ / / \ / \ / \ \____/ \ \ \ \ \__/ \__/ \__/ \____/ \__ \__ \__ \__ \__ / \ / \ / \ / \ / \__/ \__/ \__/ \__/ \ \ \__/ \__/ \____/ \__/ \ \ \ \ \ / / \ / \ / \ / \ __/ __/ __/ __/ / \__/ \__/ \__/ \____/ \__/ \__/ \__/ \__/ \__/ / \ / \ / \ / \____/ / / / / \ \__/ \__/ \____/ / \____/ __/ __/ __/ / / \ / \____/ \____/ / \__/ \__/ \__/ \ \ \__/ / \ / \____/ / / / \__/ / \ \ \__/ / \____/ __/ __/ / \ \ \__/ __/ \____/ / \__/ \__/ \ \__/ / \__/ / \____/ / / / / \ \ \____/ / \____/ __/ \ \ \__/ __/ \____/ / \__/ \__/ / \__/ __/ \____/ __/ / \ \ \____/ __/ \__/ \ \__/ __/ \____/ __/ / / \__/ __/ \____/ \ \ \____/ __/ \__/ __/ \____/ / \__/ __/ \ \____/ \____/ By the way, did anyone figure out toral 4x4 tic-tac-toe? Dan Hoey(a)AIC.NRL.Navy.Mil

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[math-fun] math writing
by Richard Schroeppel 01 May '03

01 May '03

Having identified deficiencies in popular math writing, the obvious next step is to do better. If the people doing the writing have better examples to work from, maybe the result will be more accurate. Anyone care to tackle Dan's comment > (such as assuming that n-manifolds are > subsets of n+1 space) and come up with an accurate explanation of manifold, understandable at the level of a college graduate who majored in, say, chemistry? Rich rcs(a)cs.arizona.edu

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[math-fun] Re: nonlinear constraints on hypergeometric parameters--threat or menace?
by Robert S. Maier 01 May '03

01 May '03

A couple of years ago on this list, Bill Gosper asked about hypergeometric identities in which the parameters are constrained nonlinearly, instead of linearly. The example he gave was the summation formula > c + b + a + 4 >hyper_f ([c - a + 1, c + b, c - b + 1, c + a, -------------], > 5, 4 3 > > 3 c + b + a + 1 1 > [b + a, c, c + -, -------------], -) > 2 3 4 > > (b + a - 1)! (2 c + 1)! > = -------------------------------------------, > c (c + b + a + 1) (c + a - 1)! (c + b - 1)! > which holds if c = a*b. I've just written a paper on how Euler's transformation can be extended from the Gauss hypergeometric function to hypergeometric functions of higher order. It's at http://www.arxiv.org/math.CA/0302084 . The natural analogues of Euler's transformation turn out to involve, lo and behold, nonlinear parametric constraints. As a corollary, I've derived a curious nonlinearly parametrized summation formula for 2F1(-1), namely 2F1 (6t^2 - 11t + 6, 4; -6t^2 + 5t + 3; -1) = -t (6t+1) (6t^2 - 5t - 2). This can alternatively be derived from classical contiguous function relations, by uniformizing a genus-0 algebraic curve. But it certainly looks mysterious. In the course of writing the paper, I discovered an even more intriguing identity discovered by Niblett (a student of Chaundy's, I think) in the 1950s. The citation is J. D. Niblett, "Some Hypergeometric Identities", Pacific J. Math. 2 (1953), 219-224. Niblett's identity is a quadratic function transformation, which expresses 5F4(x) in terms of 3F2(-4x/(1-x)^2) whenever the hypergeometric parameters satisfy a certain nonlinear condition. There are three free parameters. Does anyone have any other examples of hypergeometric transformations or summation formulas with nonlinearly constrained parameters? How to verify any such identity algorithmically is an interesting problem. --Robert

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Re: [math-fun] cubic quaternions
by Mike Stay 01 May '03

01 May '03

>What you are doing >wrong is not to notice that if w is the exchange factor to move R >across Q in one direction it is (w inverse} which you need to move >it in the other direction, not w. It doesn't matter whenever >w = (w inverse), e.g. quaternions with w = -1. Ah! Thank you. >regards, > >Harold V. McIntosh -- Mike Stay staym(a)clear.net.nz http://www.xaim.com/staym

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Re: [math-fun] cubic quaternions
by Mike Stay 30 Apr '03

30 Apr '03

> >Mensaje citado por: Mike Stay <staym(a)clear.net.nz>: > >> But if Q^3 is a scalar q, you have >> >> QRQQ = w RQQQ = w q R >> QRQQ = ww QQQR = w^2 q R >> so w must equal w^2 -> w can only be 0,1 >> >wrong! > QRQQ = wwQQRQ > = wwwwQQQR Huh? RQ = w QR QRQQ = w QQRQ What am I doing wrong? -- Mike Stay staym(a)clear.net.nz http://www.xaim.com/staym

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Re: [math-fun] cubic quaternions
by Mike Stay 30 Apr '03

30 Apr '03

But if Q^3 is a scalar q, you have QRQQ = w RQQQ = w q R QRQQ = ww QQQR = w^2 q R so w must equal w^2 -> w can only be 0,1 I was looking at the same thing you described while trying to get an extension of geometric algebra (the even grades of 3-d geometric algebra are the quaternions) where instead of the product rule e_i^2 = (e_i . e_i) + (e_i /\ e_i) = 1+0 = 1 we have e_i^3 = 1 but I spotted the problem above and couldn't think of a solution for it (the commutative case being boring...) It reminded me of anyons, which only work in 2-D, so I think it might be possible to get an arbitrary phase on the exchange operator if you restrict yourself to powers of e_1 and e_2. Rich wrote: >I've been wondering about doing a cubic analog of the quaternions. > >The idea is to have two generators Q and R with Q = cbrt(q) and >R = cbrt(r), and the non-commutative multiplication rule R*Q = w Q*R >(where w = cbrt(1) = (-1 + i sqrt3)/2 = e^(2 pi i/3)). -- Mike Stay staym(a)clear.net.nz http://www.xaim.com/staym

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