素数の規則を見つけたい。。。

素数の規則を見つけたい。。。 (701ﾚｽ)
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ﾘﾛｰﾄﾞ規制です｡10分ほどで解除するので､他のﾌﾞﾗｳｻﾞへ避難してください｡

200: １３２人目の素数さん [sage] 2023/11/26(日) 00:24:18.90 ID:5ylX1SN5

x^4 - 2 x^2 y^2 + 2 x^2 z^2 + y^4 + 2 y^2 z^2 + z^4=√((x+y)^2+z^2)^2*√((x-y)^2+z^2)^2*e^(i*arcsin(z/(x+y)))*e^(i*arcsin(-z/(x+y)))*e^(i*arcsin(+z/(x-y)))*e^(i*arcsin(-z/(x-y)))
x^4 - 2 x^2 y^2 + 2 x^2 z^2 + y^4 + 2 y^2 z^2 + z^4=((x+y+i^(2n+1)*z)*(x+y-i^(2n+1)*z)*(x-y+i^(2n+1)*z)*(x-y-i^(2n+1)*z))

x^4 - 2 x^2 y^2 - 2 x^2 z^2 + y^4 - 2 y^2 z^2 + z^4=((x+y+z)*(x+y-z)*(x-y+z)*(x-y-z))*e^(i*arcsin(iz/(x+y)))*e^(i*arcsin(-iz/(x+y)))*e^(i*arcsin(+iz/(x-y)))*e^(i*arcsin(-iz/(x-y)))
x^4 - 2 x^2 y^2 - 2 x^2 z^2 + y^4 - 2 y^2 z^2 + z^4=((x+y+i^2n*z)*(x+y-i^2n*z)*(x-y+i^2n*z)*(x-y-i^2n*z))

x^12 - 2 x^6 y^6 - 2 x^6 z^6 + y^12 - 2 y^6 z^6 + z^12=((x^3+y^3+i^2*z^3)*(x^3+y^3-i^2*z^3)*(x^3-y^3+i^2*z^3)*(x^3-y^3-i^2*z^3))=0
x^12 - 2 x^6 y^6 - 2 x^6 z^6 + y^12 - 2 y^6 z^6 + z^12≠0

cos(2pi*((2*a+1)/2^3-(3*b+1)/3^3-c/5^3-d/7^3)) > cos(2pi*(11^2/210^3))
a = 4 n_1, b = 9 n_2, c = 125 n_3 + 97, d = 343 n_4 + 107,           cos(2pi*((2*4+1)/2^3-(3*9+1)/3^3-97/5^3-107/7^3)) =cos((89 π)/4630500)
a = 4 n_1, b = 3 (3 n_2 + 1), c = 5 (25 n_3 + 22), d = 343 n_4 + 300,  cos(2pi*((2*4+1)/2^3-(3*3+1)/3^3-110/5^3-300/7^3)) =cos((55 π)/4630500) ←110が5を持つため非素数
a = 4 n_1, b = 3 (3 n_2 + 2), c = 125 n_3 + 41, d = 343 n_4 + 32, cos(2pi*((2*3+1)/2^3-(3*6+1)/3^3-41/5^3-32/7^3)) =sin((17 π)/4630500)
a = 4 n_1, b = 9 n_2 + 1, c = 125 n_3 + 31, d = 343 n_4 + 250,    cos(2pi*((2*3+1)/2^3-(3*1+1)/3^3-31/5^3-250/7^3)) =-sin((103 π)/4630500)
a = 4 n_1, b = 9 n_2 + 1, c = 125 n_3 + 74, d = 343 n_4 + 132, 　cos(2pi*((2*3+1)/2^3-(3*1+1)/3^3-74/5^3-132/7^3)) =sin((113 π)/4630500)

http://rio2016.5ch.net/test/read.cgi/math/1640355175/200

201: １３２人目の素数さん [sage] 2023/11/26(日) 00:35:25.83 ID:5ylX1SN5

↓３次元では書けないベクトル和（((x+y+i^m*z)*(x+y-i^m*z)*(x-y+i^m*z)*(x-y-i^m*z)) mが３以上のベクトル和をかけない）　
√(x^4 - 2 x^2 y^2 + 2 x^2 z^2 + y^4 + 2 y^2 z^2 + z^4)=√(((x+y+i^(2n+1)*z)*(x+y-i^(2n+1)*z)*(x-y+i^(2n+1)*z)*(x-y-i^(2n+1)*z)))　　　
√(x^4 - 2 x^2 y^2 - 2 x^2 z^2 + y^4 - 2 y^2 z^2 + z^4)=√(((x+y+i^2n*z)*(x+y-i^2n*z)*(x-y+i^2n*z)*(x-y-i^2n*z)))

cos(2pi*((2*a+1)/2^3-(3*b+1)/3^3-c/5^3-d/7^3+e/11^3)) > cos(2pi*(13^2/2310^3))

a = 4 n_1, b = 9 n_2, c = 125 n_3, d = 343 n_4 + 83, e = 1331 n_5 + 205,　
a = 4 n_1, b = 9 n_2, c = 125 n_3 + 53, d = 7 (49 n_4 + 29), e = 1331 n_5 + 1235, cos(2pi*((2*4+1)/2^3-(3*9+1)/3^3-53/5^3-7*29/7^3+1235/11^3))=cos((91 π)/6163195500) ←7*29が7をもつため非素数
a = 4 n_1, b = 3 (3 n_2 + 1), c = 125 n_3 + 77, d = 343 n_4 + 163, e = 1331 n_5 + 448, 　cos(2pi*((2*4+1)/2^3-(3*3+1)/3^3-77/5^3-163/7^3+448/11^3))＝cos((19 π)/6163195500)
a = 4 n_1, b = 3 (3 n_2 + 2), c = 125 n_3 + 29, d = 343 n_4 + 243, e = 1331 n_5 + 691,　cos(2pi*((2*4+1)/2^3-(3*6+1)/3^3-29/5^3-243/7^3+691/11^3))=cos((163 π)/6163195500)
a = 4 n_1, b = 3 (3 n_2 + 2), c = 125 n_3 + 101, d = 343 n_4 + 123, e = 1331 n_5 + 992,  cos(2pi*((2*4+1)/2^3-(3*6+1)/3^3-101/5^3-123/7^3+992/11^3))=cos((53 π)/6163195500)

http://rio2016.5ch.net/test/read.cgi/math/1640355175/201

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