Re: [math-fun] Square hex numbers?
Thanks, Adam! I thought there should be some general method for solving P(n) = Q(K) where P, Q are quadratic polynomials over the integers, but didn't know where to go from there. * * * When there is an integer like 169 that belongs to two figurate numbers (H_7 = 169 = 13^2), I wonder if there is a set of 169 points in R^4 = R^2 x R^2 whose projection by p_12 onto one R^2 factor is the first figure and onto the other by p_34 is the second figure. I seem to recall asking this here some years ago for T_8 = 6^2, and that Michael Kleber showed that there's no set X of 36 points in R^4 with p_12(X) = triangle of side 8 and p_34(X) = square of side 6. Maybe there's a general statement about when such a thing can exist? —Dan —Dan Adam Goucher wrote: ----- Your equation is of the form: (quadratic in n) = (quadratic in K) ... ... -----
Heh, I was just going to say that I remember a similar question about 36! But it was two projections from R^3, right? (It must be easy for R^4, since you can just specify the coordinates independently.) I don't immediately see the answer for 169; it would certainly be neat if possible! --Michael On Wed, Apr 24, 2019 at 12:12 PM Dan Asimov <dasimov@earthlink.net> wrote:
Thanks, Adam! I thought there should be some general method for solving P(n) = Q(K) where P, Q are quadratic polynomials over the integers, but didn't know where to go from there.
* * *
When there is an integer like 169 that belongs to two figurate numbers (H_7 = 169 = 13^2), I wonder if there is a set of 169 points in R^4 = R^2 x R^2 whose projection by p_12 onto one R^2 factor is the first figure and onto the other by p_34 is the second figure.
I seem to recall asking this here some years ago for T_8 = 6^2, and that Michael Kleber showed that there's no set X of 36 points in R^4 with p_12(X) = triangle of side 8 and p_34(X) = square of side 6.
Maybe there's a general statement about when such a thing can exist?
—Dan
—Dan
Adam Goucher wrote: ----- Your equation is of the form:
(quadratic in n) = (quadratic in K)
... ... -----
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When there is an integer like 169 that belongs to two figurate numbers (H_7 = 169 = 13^2), I wonder if there is a set of 169 points in R^4 = R^2 x R^2 whose projection by p_12 onto one R^2 factor is the first figure and onto the other by p_34 is the second figure.
Rather than just a projection onto R^3, I think what would be more interesting, even if it's in more than 3 dimensions, is a point set in a higher dimension that displays a the symmetries of the two figures. If you have a set of 169 points in R^n, and projections down to the hexagonal and square figures, consider the symmetry group of the 169 points in R^n. The projections give maps of this symmetry group to the symmetry groups of the square and the hexagon. If both these mappings are surjective, then in some sense the two symmetries are "explained" by being two components of the higher symmetry of the set of points in R^n. The smaller n is, the more satisfying this "explanation" is; with N=169, we can take the vertices of a simplex, with symmetry group S_169, and find projections that keep all the symmetries. Is there any obvious lower-dimensional set with the desired properties? A set of k*169 points in a higher-dimensional space would satisfy me as an explanation too, as long as the inverse image of each point under each projection has exactly k points. Andy
I seem to recall asking this here some years ago for T_8 = 6^2, and that Michael Kleber showed that there's no set X of 36 points in R^4 with p_12(X) = triangle of side 8 and p_34(X) = square of side 6.
Maybe there's a general statement about when such a thing can exist?
—Dan
—Dan
Adam Goucher wrote: ----- Your equation is of the form:
(quadratic in n) = (quadratic in K)
... ... -----
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participants (3)
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Andy Latto -
Dan Asimov -
Michael Kleber