I get that 41616 = (288 x 289)/2 = (204)^2 is a square triangular number. Is that special enough, or is it even specialer? * * * I imagine a square triangular numbers β say 36 β represented as a set of points in 4-space {(s_n, t_n) in C^2} where C is the complex plane, such that {s_n} forms a 6x6 square in C, and {t_n} forms the triangle of side 8, each centered at the origin. And then computer graphically, this set of 36 points in 4-space rotates around, perhaps randomly, at each step being projected onto the first 2 (real) coordinates and displayed on a computer screen. It should appear to be a square morphing into a triangle and back again. By varying the one-to-one correspondence between the 36 indices of {s_n} and {t_n} (i.e., considering {(s_n, t_π(n)} instead of {(s_n, t_n)} where π is a permutation of {1,2,...,36}, the exact set of points in 4-space will vary. I wonder if there are criteria for an "optimal" one-to-one correspondence. This suggests considering a homeomorphism from a square S = [-1,1]^2 in R^2 to a triangle T = C * convex hull of {(2,-1-1), (-1,2,-1), -1,-1,2)} in R^3, where C = (64/243)^(1/4) is a linear scaling factor to make their areas equal. Assuming that a homeomorphism h : S β> T is angle-preserving (conformal) on their interiors and takes the center of the square to the center of the triangle, there is only one real parameter remaining β one degree of freedom. That can be specified by the image of the positive x-direction in R^3, i.e., Dh(d/dx). What should it be for the optimal homeomorphism? βDan ----- If we allow the sum to be a different power ... The sum of the first N cubes is the square of a triangle. 1^3 + 2^3 + 3^3 + 4^3 = 10^2 Sometimes a triangle is also a square: T(8) = 36 = 6^2. The square of a square is a 4th power: 1^3 + 2^3 + ... + 8^3 = 1296 = 36^2 = 6^4 The next triangle-square is 1225. 1^3 + 2^3 + ... + 49^3 = 1225^2 = 35^4 There's a hidden fact in the Ramanujan-Hardy 1729 story, but it's never mentioned. 1729 is a (base 2) pseudoprime: 2^1728 (mod 1729) = 1. Here's how I flunked my Ramanujan test. My University of Arizona employee number was 41616. I didn't look at it very often-- mainly when I used my ID card to check out a library book. Years later, I noticed the number was special. In fact, very special. Puzzle: Why is it special? Puzzle2: Why is it very special? -----