On Aug 9, 2017 04:32, "Dan Asimov" <dasimov@earthlink.net> wrote: And paradoxically, the conformal bijection z |—> -1/(1+z) of the Riemann sphere S^2 = C u {oo} to itself that preserves the integer points Z^2 ... In what sense does this preserve the integer points? F(17) = -1/18 Andy (i.e., a fractional linear transformation of form z |—> (az + b)/(cz + d), a,b,c,d in Z, ad - bc = 1, i.e., a member of the group PSL(2,Z) = SL(2,Z) / {+-I}) ... is of order three: f(f(f(z))) == z (for example). —Dan

From: Eugene Salamin via math-fun <math-fun@mailman.xmission.com> Sent: Aug 8, 2017 5:17 PM Suppose, to the contrary, that such a triangle exists. We may translate one vertex to the origin, the remaining vertices being (x1, y1) and (x2, y2). If all 4 of these numbers are even, there exists a half-size triangle. So we may assume that one of these numbers is odd, and by rotation and reflection, we can take x1 to be odd. We have

x1^2 + y1^2 = x2^2 + y2^2 = (x1 - x2)^2 + (y1 - y2)^2 = s^2, and thus that 2 (x1 x2 + y1 y2) = s^2. Since s^2 is even, x1 and y1 have the same parity, and likewise for x2 and y2. So y1 is odd. But then s^2 = x1^2 + y1^2 = 2 mod 4. Hence x2 and y2 must also be odd. With all 4 numbers odd, x1 x2 + y1 y2 = 2 mod 4. But s^2 is twice this, so s^2 = 0 mod 4, and we have a contradiction. On Monday, August 7, 2017, 4:04:20 AM PDT, Bill Gosper <billgosper@gmail.com> wrote: More generally, there is no threefold rotational symmetry. Julian's function can produce only gridpoints: ListPlot[{3, 1} # & /@ NestWhileList[{#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &@ {#[[1]], #[[2]] + Floor[381 #[[1]]/128]} &@ {#[[1]] - Floor[64 #[[2]]/127], #[[2]]} &, {2, 1}, # != -{1, 4} &], AspectRatio -> 1, Axes -> False, Frame -> True] gosper.org/trinsky3spiral.png And all the arguments to the Floors are merely rational. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun