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.  --  Gene 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 _______________________________________________