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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