本原勾股数的存在性

xbtianlang

将直角三角形整数边长的斜边分成两类：
一、12n+5或12n+1型素数，及其这些素数的幂或相互的乘积，比如5，25，125，13，17，29，65等等；
二、12n-5或12n-1型素数，及其这些素数的幂或含有这些素数的乘积，7，49，35，11，121，55等等。
求证：一类斜边有本原勾股数，二类斜边不存在本原勾股数。（勾股弦两两互质）

王守恩

现学现用。谢谢northwolves！

1. Table[Select[Table[{k[[2]]^2 - k[[1]]^2, 2 Times @@ k}, {k, PowersRepresentations[12 n + 5, 2, 2]}], CoprimeQ @@ # &], {n, 0, 60}]

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{{{3, 4}}, {{15, 8}}, {{21, 20}}, {{9, 40}}, {{45, 28}}, {{63, 16}, {33, 56}}, {}, {{39, 80}}, {{99, 20}}, {{15, 112}}, {{117, 44}}, {{105, 88}}, {{51, 140}}, {}, {{165, 52}}, {{153, 104}, {57, 176}}, {{195, 28}},
{}, {{171, 140}, {21, 220}}, {{105, 208}}, {}, {{255, 32}}, {{69, 260}}, {{231, 160}}, {{285, 68}}, {{273, 136}, {207, 224}}, {{75, 308}}, {}, {}, {{225, 272}}, {{357, 76}, {27, 364}}, {{345, 152}, {135, 352}},
{{189, 340}}, {{399, 40}}, {}, {{297, 304}, {87, 416}}, {}, {{351, 280}}, {{261, 380}}, {}, {{483, 44}, {93, 476}}, {}, {{459, 220}}, {{279, 440}}, {{525, 92}, {435, 308}}, {{513, 184}, {33, 544}}, {{165, 532}},
{{231, 520}}, {},{{465, 368}}, {}, {{105, 608}}, {{621, 100}, {429, 460}}, {{609, 200}}, {{315, 572}}, {}, {{675, 52}}, {{561, 400}, {111, 680}}, {{651, 260}}, {}, {{627, 364}, {333, 644}}}

1. Table[Select[Table[{k[[2]]^2 - k[[1]]^2, 2 Times @@ k}, {k, PowersRepresentations[12 n + 1, 2, 2]}], CoprimeQ @@ # &], {n, 1, 60}]

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{{{5, 12}}, {{7, 24}}, {{35, 12}}, {}, {{11, 60}}, {{55, 48}}, {{77, 36}, {13, 84}}, {{65, 72}}, {{91, 60}}, {}, {}, {{143, 24}, {17, 144}}, {{85, 132}}, {{119, 120}}, {{19, 180}}, {{95, 168}}, {{187, 84}},
{133, 156}}, {}, {{221, 60}}, {{209, 120}}, {}, {{247, 96}, {23, 264}}, {{115, 252}}, {{161, 240}}, {}, {{25, 312}}, {{323, 36}, {253, 204}}, {{175, 288}}, {{299, 180}}, {}, {{275, 252}}, {}, {{325, 228}},
{{391, 120}}, {{29, 420}}, {{145, 408}}, {{437, 84}, {203, 396}}, {{425, 168}}, {}, {{319, 360}, {31, 480}}, {{475, 132}, {155, 468}}, {{377, 336}, {217, 456}}, {}, {}, {{341, 420}}, {}, {{493, 276},
{403, 396}}, {{575, 48}}, {}, {{551, 240}}, {{35, 612}}, {{527, 336}}, {}, {}, {{589, 300}}, {{385, 552}}, {{667, 156}, {37, 684}}, {{455, 528}, {185, 672}}, {{259, 660}}, {}}

1. Table[Select[Table[{k[[2]]^2 - k[[1]]^2, 2 Times @@ k}, {k, PowersRepresentations[12 n - 5, 2, 2]}], CoprimeQ @@ # &], {n, 0, 120}]

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{{}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {},{}, {}, {},
{}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}}

1. Table[Select[Table[{k[[2]]^2 - k[[1]]^2, 2 Times @@ k}, {k, PowersRepresentations[12 n - 1, 2, 2]}], CoprimeQ @@ # &], {n, 0, 120}]

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{{}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {},
{}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}}

不仅是 12n-5,12n-1 无解,   好像 12n-4,12n-3,12n-2,12n-0,12n+2,12n+3,12n+4 也是无解？

xbtianlang

第一步，抛砖引玉，先排除6n+3或者写成12n±3的斜边。
如果$a^2+b^2=c^2$是一组本原勾股数，则必有一奇一偶的两个数u、v
使得$a=|u^2-v^2|,b=2uv,c=u^2+v^2$,不妨设u奇v偶。
此时，假设c是12n±3型的。则b=2uv中u,v都不是3的倍数。
必有，$u=6i±1，v=6j±2，c=u^2+v^2=36(i^2+j^2)±12(i±2j)+5$
与假设c是12n±3型的矛盾。

northwolves

1640年，Fermat证明形式为4n+1的素数作为斜边有且仅有一个具有整数勾股三角形。

xbtianlang

现在已知，斜边不可能是2的倍数或3的倍数。
第二步，要证明第二类斜边没有本原勾股数；
第三步，要证明第一类斜边有本原勾股数。
看起来有点难，不知道如何下手……

xbtianlang

第二步，小试牛刀，先证明斜边不可能是7的倍数。
如果 $a^2+b^2=c^2$ 是一组本原勾股数，则必有一奇一偶的两个数u、v
使得, $a=|u^2-v^2|, b=2uv, c=u^2+v^2$ ,不妨设u奇v偶。
此时，若b是7的倍数，则c肯定不是7的倍数；
假设b不是7的倍数。则b=2uv中u,v都不是7的倍数。
必有，$u=14i±1,±3,±5, v=14j±2,±4±,±6, c=u^2+v^2=196(i^2+j^2)±28(i,3i,5i±2j,4j,6j)+k$
k=5,17,37,13,25,45,29,41,61，所以c不可能是7的倍数。
同理可证，c也不可能是11，19，23，…的倍数。