[math-fun] Fwd: There are no equilateral triangles in the Integer Grid.
gosper.org/rectarith12.pdf (penultimate page). --rwg On 2017-08-08 17:17, Eugene Salamin via math-fun wrote:
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
Here's a beautiful essay by Joel Hamkins that handles 5-gons, 6-gons, 7-gons, etc. with a uniform argument and then settles the case of 3-gons by appealing to the fact about 6-gons. http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/ Jim Propp On Wed, Aug 9, 2017 at 2:16 PM, Bill Gosper <billgosper@gmail.com> wrote:
gosper.org/rectarith12.pdf (penultimate page). --rwg
On 2017-08-08 17:17, Eugene Salamin via math-fun wrote:
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
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Those arguments are very nice! Is there anything similar to exorcise regular pentagons in Z^3? (Z^3 has equilateral triangles, squares, and regular hexagons, but I think has no other regular polygons.) On Wed, Aug 9, 2017 at 3:03 PM, James Propp <jamespropp@gmail.com> wrote:
Here's a beautiful essay by Joel Hamkins that handles 5-gons, 6-gons, 7-gons, etc. with a uniform argument and then settles the case of 3-gons by appealing to the fact about 6-gons.
http://jdh.hamkins.org/no-regular-polygons-in-the-integer-lattice/
Jim Propp
On Wed, Aug 9, 2017 at 2:16 PM, Bill Gosper <billgosper@gmail.com> wrote:
gosper.org/rectarith12.pdf (penultimate page). --rwg
On 2017-08-08 17:17, Eugene Salamin via math-fun wrote:
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
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participants (3)
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Allan Wechsler -
Bill Gosper -
James Propp