五次方程

五次方程是一種最高次數為五次的多項式方程。本條目專指只含一個未知數的五次方程（一元五次方程），即方程形如 a x 5 + b x 4 + c x 3 + d x 2 + e x + f = 0 {\displaystyle ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0} 其中，a、b、c、d、e和f为复数域内的数，且a不为零。例如： x 5 − 4 x 4 + 2 x 3 − 3 x + 7 = 0 {\displaystyle x^{5}-4x^{4}+2x^{3}-3x+7=0} 二次方程很早就找到了公式解。經過數學家們的不斷努力，三次方程及四次方程在16世紀中有了解答，但是之后很长的一段时间里沒有人知道五次方程是否存在公式解。直到1824年，保羅·魯菲尼和尼爾斯·阿貝爾證明了一般的五次方程，不存在統一的根式解（即由方程的係數通過有限次的四則運算及根號組合而成的公式解）。認為一般的五次方程沒有公式解存在的看法其实是不正確的。事實上，利用一些超越函數，如Θ函数或戴德金η函數即可構造出五次方程的公式解。另外，若只需求得數值解，可以利用數值方法（如牛頓法）得到相當理想的解答。 證明一般五次及其以上的一元多项式方程無根式解的人是埃瓦里斯特·伽羅瓦，他巧妙地利用群論處理了上述的問題。 對於一般的五次方程式 x 5 + a 1 x 4 + a 2 x 3 + a 3 x 2 + a 4 x + a 5 = 0 {\displaystyle x^{5}+a_{1}x^{4}+a_{2}x^{3}+a_{3}x^{2}+a_{4}x+a_{5}=0\,} 可以藉由以下的多项式变换 y = x 4 + b 1 x 3 + b 2 x 2 + b 3 x + b 4 {\displaystyle y=x^{4}+b_{1}x^{3}+b_{2}x^{2}+b_{3}x+b_{4}\,} 得到一個 y {\displaystyle y\,} 的五次多項式，上述的轉換稱為契爾恩豪森轉換（英语：Tschirnhaus transformation）（Tschirnhaus transformation），藉由特別選擇的係數 b i {\displaystyle b_{i}\,} ，可以使 y 4 {\displaystyle y^{4}\,} , y 3 {\displaystyle y^{3}\,} , y 2 {\displaystyle y^{2}\,} 的係數為 0 {\displaystyle 0\,} ，從而得到如下的方程式： y 5 + m y + n = 0 {\displaystyle y^{5}+my+n=0\,} 以上的化簡方法是由厄蘭·塞缪爾·布靈（英语：Erland Samuel Bring）所發現，後來喬治·傑拉德（英语：George Jerrard）也獨立發現了此法，因此上式稱為布靈·傑拉德正規式（Bring-Jerrard normal form）。 其步驟如下： 首先令 x = y − a 1 5 {\displaystyle x=y-{\frac {a_{1}}{5}}\,} 可消去四次方項，得到 y 5 + a y 3 + b y 2 + c y + d = 0 {\displaystyle y^{5}+ay^{3}+by^{2}+cy+d=0\,} ； 其中， a = 5 a 2 − 2 a 1 2 5 {\displaystyle a={\frac {5a_{2}-2a_{1}^{2}}{5}}\,} b = 25 a 3 − 15 a 1 a 2 + 4 a 1 3 25 {\displaystyle b={\frac {25a_{3}-15a_{1}a_{2}+4a_{1}^{3}}{25}}\,} c = 125 a 4 − 50 a 1 a 3 + 15 a 1 2 a 2 − 3 a 1 4 125 {\displaystyle c={\frac {125a_{4}-50a_{1}a_{3}+15a_{1}^{2}a_{2}-3a_{1}^{4}}{125}}\,} d = 3125 a 5 − 625 a 1 a 4 + 125 a 1 2 a 3 − 25 a 1 3 a 2 + 4 a 1 5 3125 {\displaystyle d={\frac {3125a_{5}-625a_{1}a_{4}+125a_{1}^{2}a_{3}-25a_{1}^{3}a_{2}+4a_{1}^{5}}{3125}}\,} 接下來，令 z = y 2 + p y + q {\displaystyle z=y^{2}+py+q\,} ， 得到 z 5 + P z 4 + Q z 3 + A z 2 + B z + C = 0 {\displaystyle z^{5}+Pz^{4}+Qz^{3}+Az^{2}+Bz+C=0\,} ， 再令 P = Q = 0 {\displaystyle P=Q=0\,} ， 求得 p = − 15 b + 60 a 3 + 225 b 2 − 200 a c 10 a {\displaystyle p={\frac {-15b+{\sqrt {60a^{3}+225b^{2}-200ac}}}{10a}}\,} ； q = 2 a 5 . {\displaystyle q={2a \over 5}.\,} 第三步，利用契爾恩豪森想到的方法，令： X = z 4 + b 1 z 3 + b 2 z 2 + b 3 z + b 4 {\displaystyle X=z^{4}+b_{1}z^{3}+b_{2}z^{2}+b_{3}z+b_{4}\,} ， 代入 z 5 + A z 2 + B z + C = 0 {\displaystyle z^{5}+Az^{2}+Bz+C=0\,} ， 得到 X 5 + R X 4 + S X 3 + T X 2 + U X + V = 0 {\displaystyle X^{5}+RX^{4}+SX^{3}+TX^{2}+UX+V=0\,} ， 再令 R = S = T = 0 {\displaystyle R=S=T=0\,} ， 則得 b 4 = 3 A b 1 + 4 B 5 {\displaystyle b_{4}={\frac {3Ab_{1}+4B}{5}}\,} ， 若令 b 2 = − 4 B b 1 + 5 C 3 A {\displaystyle b_{2}=-{\frac {4Bb_{1}+5C}{3A}}\,} ， 則 b 1 {\displaystyle b_{1}\,} ， b 3 {\displaystyle b_{3}\,} 可由以下兩個方程解得： ( 27 A 4 − 160 B 3 + 300 A B C ) b 1 2 + ( 27 A 3 B − 400 B 2 C + 375 C 2 A ) b 1 + ( 18 A 2 B 2 − 45 A 3 C − 250 B C 2 ) = 0 ; {\displaystyle (27A^{4}-160B^{3}+300ABC)b_{1}^{2}+(27A^{3}B-400B^{2}C+375C^{2}A)b_{1}\,+(18A^{2}B^{2}-45A^{3}C-250BC^{2})=0;\,} 675 A 3 b 3 3 + ( 3375 A 2 C b 1 − 3600 A B 2 b 1 − 2025 A 4 − 4500 A B C ) b 3 2 {\displaystyle 675A^{3}b_{3}^{3}+(3375A^{2}Cb_{1}-3600AB^{2}b_{1}-2025A^{4}\,-4500ABC)b_{3}^{2}} + ( 675 A 3 B b 1 2 + 6000 B 2 C b 1 2 + 7200 A 2 B 2 b 1 − 4050 A 3 C b 1 + 15000 C 2 B b 1 + 9375 C 3 + 9675 A 2 B C + 2025 A 5 ) b 3 + {\displaystyle +(675A^{3}Bb_{1}^{2}+6000B^{2}Cb_{1}^{2}\,+7200A^{2}B^{2}b_{1}-4050A^{3}Cb_{1}+15000C^{2}Bb_{1}\,\!+9375C^{3}+9675A^{2}BC+2025A^{5})b_{3}+\,} ( − 4770 A 3 B C − 1125 A 2 C 2 b 1 2 − 1500 B 2 C 2 − 320 A B 3 b 1 3 − 960 B 4 b 1 2 − 3843 A 3 B 2 b 1 + 1485 A 4 C b 1 − 54 A 5 b 1 3 {\displaystyle (-4770A^{3}BC-1125A^{2}C^{2}b_{1}^{2}-1500B^{2}C^{2}\,-320AB^{3}b_{1}^{3}-960B^{4}b_{1}^{2}-3843A^{3}B^{2}b_{1}+1485A^{4}Cb_{1}-54A^{5}b_{1}^{3}} − 6250 A C 3 − 2400 B 3 C b 1 − 108 A 2 B 3 − 675 A 6 − 756 A 4 B b 1 2 − 9375 A C 2 B b 1 − 3900 A B 2 C b 1 2 − 225 A 2 B C b 1 3 ) = 0. {\displaystyle -6250AC^{3}-2400B^{3}Cb_{1}-108A^{2}B^{3}-675A^{6}\,-756A^{4}Bb_{1}^{2}-9375AC^{2}Bb_{1}-3900AB^{2}Cb_{1}^{2}-225A^{2}BCb_{1}^{3})=0.\,} 若以函數的觀點來看，方程 X 5 + U X + V = 0 {\displaystyle X^{5}+UX+V=0\,} 的解有兩個自變數 U {\displaystyle U\,} , 和 V {\displaystyle V\,} 。 若再令 X = − U 4 ξ {\displaystyle X={\sqrt[{4}]{-U}}\xi \,} 則方程式可以進一步化簡為如下形式： ξ 5 − ξ + t = 0 {\displaystyle \xi ^{5}-\xi +t=0\,} 它的解 ξ {\displaystyle \xi \,} 是單一變數 t {\displaystyle t\,} 的函數。 雖然一般的五次方程不存在根式解，但是對於某些特殊的五次方程，滿足某些條件後還是有根式解的。 a x 5 + b x 4 + c x 3 + 15 a b c − 4 b 3 25 a 2 x 2 + 25 a 2 c 2 − 5 a b 2 c − b 4 125 a 3 x + f = 0 {\displaystyle {ax^{5}+bx^{4}+cx^{3}+{\frac {15abc-4b^{3}}{25a^{2}}}x^{2}+{\frac {25a^{2}c^{2}-5ab^{2}c-b^{4}}{125a^{3}}}x+f=0}} ，当 a ≠ 0 {\displaystyle a\neq 0} 时， x 1 = − 2 b + 176 b 5 − 1200 a b 3 c + 2000 a 2 b c 2 − 50000 a 4 f + 16 ( 11 b 5 − 75 a b 3 c + 125 a 2 b c 2 − 3125 a 4 f ) 2 − 4 ( 2 b 2 − 5 a c ) 5 5 + 176 b 5 − 1200 a b 3 c + 2000 a 2 b c 2 − 50000 a 4 f − 16 ( 11 b 5 − 75 a b 3 c + 125 a 2 b c 2 − 3125 a 4 f ) 2 − 4 ( 2 b 2 − 5 a c ) 5 5 10 a {\displaystyle {x_{1}={\frac {-2b+{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f+16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}+{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f-16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}}{10a}}}\,} x 2 = − b 5 a + − 1 + 5 + 10 + 2 5 i 40 a 176 b 5 − 1200 a b 3 c + 2000 a 2 b c 2 − 50000 a 4 f + 16 ( 11 b 5 − 75 a b 3 c + 125 a 2 b c 2 − 3125 a 4 f ) 2 − 4 ( 2 b 2 − 5 a c ) 5 5 + − 1 + 5 − 10 + 2 5 i 40 a 176 b 5 − 1200 a b 3 c + 2000 a 2 b c 2 − 50000 a 4 f − 16 ( 11 b 5 − 75 a b 3 c + 125 a 2 b c 2 − 3125 a 4 f ) 2 − 4 ( 2 b 2 − 5 a c ) 5 5 {\displaystyle {x_{2}=-{\frac {b}{5a}}+{\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f+16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}+{\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f-16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}}\,} x 3 = − b 5 a + − 1 − 5 + 10 − 2 5 i 40 a 176 b 5 − 1200 a b 3 c + 2000 a 2 b c 2 − 50000 a 4 f + 16 ( 11 b 5 − 75 a b 3 c + 125 a 2 b c 2 − 3125 a 4 f ) 2 − 4 ( 2 b 2 − 5 a c ) 5 5 + − 1 − 5 − 10 − 2 5 i 40 a 176 b 5 − 1200 a b 3 c + 2000 a 2 b c 2 − 50000 a 4 f − 16 ( 11 b 5 − 75 a b 3 c + 125 a 2 b c 2 − 3125 a 4 f ) 2 − 4 ( 2 b 2 − 5 a c ) 5 5 {\displaystyle {x_{3}=-{\frac {b}{5a}}+{\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f+16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}+{\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f-16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}}\,} x 4 = − b 5 a + − 1 − 5 − 10 − 2 5 i 40 a 176 b 5 − 1200 a b 3 c + 2000 a 2 b c 2 − 50000 a 4 f + 16 ( 11 b 5 − 75 a b 3 c + 125 a 2 b c 2 − 3125 a 4 f ) 2 − 4 ( 2 b 2 − 5 a c ) 5 5 + − 1 − 5 + 10 − 2 5 i 40 a 176 b 5 − 1200 a b 3 c + 2000 a 2 b c 2 − 50000 a 4 f − 16 ( 11 b 5 − 75 a b 3 c + 125 a 2 b c 2 − 3125 a 4 f ) 2 − 4 ( 2 b 2 − 5 a c ) 5 5 {\displaystyle {x_{4}=-{\frac {b}{5a}}+{\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f+16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}+{\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f-16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}}\,} x 5 = − b 5 a + − 1 + 5 − 10 + 2 5 i 40 a 176 b 5 − 1200 a b 3 c + 2000 a 2 b c 2 − 50000 a 4 f + 16 ( 11 b 5 − 75 a b 3 c + 125 a 2 b c 2 − 3125 a 4 f ) 2 − 4 ( 2 b 2 − 5 a c ) 5 5 + − 1 + 5 + 10 + 2 5 i 40 a 176 b 5 − 1200 a b 3 c + 2000 a 2 b c 2 − 50000 a 4 f − 16 ( 11 b 5 − 75 a b 3 c + 125 a 2 b c 2 − 3125 a 4 f ) 2 − 4 ( 2 b 2 − 5 a c ) 5 5 {\displaystyle {x_{5}=-{\frac {b}{5a}}+{\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f+16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}+{\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{40a}}{\sqrt[{5}]{176b^{5}-1200ab^{3}c+2000a^{2}bc^{2}-50000a^{4}f-16{\sqrt {\left(11b^{5}-75ab^{3}c+125a^{2}bc^{2}-3125a^{4}f\right)^{2}-4\left(2b^{2}-5ac\right)^{5}}}}}}\,} ( c 2 + 1 ) x 5 + 5 d 4 ( 3 ∓ c ) x − 4 d 5 ( ± 11 + 2 c ) = 0 {\displaystyle {(c^{2}+1)x^{5}+5d^{4}(3\mp c)x-4d^{5}(\pm 11+2c)=0}} ，当 c ≠ ± i {\displaystyle c\neq {\pm {i}}} 时， x 1 = d [ A + B + C + D ] {\displaystyle x_{1}=d\left[A+B+C+D\right]\,} x 2 = d [ ( − 1 + 5 ) + 10 + 2 5 i 4 A + ( − 1 − 5 ) + 10 − 2 5 i 4 B ] + d [ ( − 1 − 5 ) − 10 − 2 5 i 4 C + ( − 1 + 5 ) − 10 + 2 5 i 4 D ] {\displaystyle x_{2}=d\left[{\frac {(-1+{\sqrt {5}})+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}A+{\frac {(-1-{\sqrt {5}})+{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}B\right]+d\left[{\frac {(-1-{\sqrt {5}})-{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}C+{\frac {(-1+{\sqrt {5}})-{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}D\right]\,} x 3 = d [ ( − 1 − 5 ) + 10 − 2 5 i 4 A + ( − 1 − 5 ) − 10 − 2 5 i 4 B ] + d [ ( − 1 + 5 ) − 10 + 2 5 i 4 C + ( − 1 + 5 ) + 10 + 2 5 i 4 D ] {\displaystyle x_{3}=d\left[{\frac {(-1-{\sqrt {5}})+{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}A+{\frac {(-1-{\sqrt {5}})-{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}B\right]+d\left[{\frac {(-1+{\sqrt {5}})-{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}C+{\frac {(-1+{\sqrt {5}})+{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}D\right]\,} x 4 = d [ ( − 1 − 5 ) − 10 − 2 5 i 4 A + ( − 1 + 5 ) − 10 + 2 5 i 4 B ] + d [ ( − 1 + 5 ) + 10 + 2 5 i 4 C + ( − 1 − 5 ) + 10 − 2 5 i 4 D ] {\displaystyle x_{4}=d\left[{\frac {(-1-{\sqrt {5}})-{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}A+{\frac {(-1+{\sqrt {5}})-{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}B\right]+d\left[{\frac {(-1+{\sqrt {5}})+{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}C+{\frac {(-1-{\sqrt {5}})+{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}D\right]\,} x 5 = d [ ( − 1 + 5 ) − 10 + 2 5 i 4 A + ( − 1 + 5 ) + 10 + 2 5 i 4 B ] + d [ ( − 1 − 5 ) + 10 − 2 5 i 4 C + ( − 1 − 5 ) − 10 − 2 5 i 4 D ] {\displaystyle x_{5}=d\left[{\frac {(-1+{\sqrt {5}})-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}A+{\frac {(-1+{\sqrt {5}})+{\sqrt {10+2{\sqrt {5}}}}{\mathrm {i} }}{4}}B\right]+d\left[{\frac {(-1-{\sqrt {5}})+{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}C+{\frac {(-1-{\sqrt {5}})-{\sqrt {10-2{\sqrt {5}}}}{\mathrm {i} }}{4}}D\right]\,} 其中 A = ( c 2 + 1 + c 2 + 1 ∓ c 2 + 1 ) 2 ( − c 2 + 1 + c 2 + 1 ± c 2 + 1 ) ( c 2 + 1 ) 2 5 {\displaystyle A={\sqrt[{5}]{\frac {\left({\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)^{2}\left(-{\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}\,} B = ( c 2 + 1 + c 2 + 1 ∓ c 2 + 1 ) 2 ( − c 2 + 1 + c 2 + 1 ± c 2 + 1 ) ( c 2 + 1 ) 2 5 {\displaystyle B={\sqrt[{5}]{\frac {\left({\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)^{2}\left(-{\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}\,} C = ( c 2 + 1 + c 2 + 1 ± c 2 + 1 ) 2 ( c 2 + 1 − c 2 + 1 ∓ c 2 + 1 ) ( c 2 + 1 ) 2 5 {\displaystyle C={\sqrt[{5}]{\frac {\left({\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)^{2}\left({\sqrt {c^{2}+1}}-{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}\,} D = − ( c 2 + 1 − c 2 + 1 ∓ c 2 + 1 ) 2 ( − c 2 + 1 + c 2 + 1 ± c 2 + 1 ) ( c 2 + 1 ) 2 5 {\displaystyle D=-{\sqrt[{5}]{\frac {\left({\sqrt {c^{2}+1}}-{\sqrt {c^{2}+1\mp {\sqrt {c^{2}+1}}}}\right)^{2}\left(-{\sqrt {c^{2}+1}}+{\sqrt {c^{2}+1\pm {\sqrt {c^{2}+1}}}}\right)}{(c^{2}+1)^{2}}}}\,} a 2 x 5 + 5 a b x 3 + 5 b 2 x + a c = 0 {\displaystyle {a^{2}x^{5}+5abx^{3}+5b^{2}x+ac=0}} ，当 a ≠ 0 {\displaystyle a\neq 0} 时， x 1 = − c 2 a + ( c 2 a ) 2 + ( b a ) 5 5 + − c 2 a − ( c 2 a ) 2 + ( b a ) 5 5 {\displaystyle {x_{1}={\sqrt[{5}]{-{\frac {c}{2a}}+{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}+{\sqrt[{5}]{-{\frac {c}{2a}}-{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}}\,} x 2 = − 1 + 5 + 10 + 2 5 i 4 − c 2 a + ( c 2 a ) 2 + ( b a ) 5 5 + − 1 + 5 − 10 + 2 5 i 4 − c 2 a − ( c 2 a ) 2 + ( b a ) 5 5 {\displaystyle {x_{2}={\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}+{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}+{\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}-{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}}\,} x 3 = − 1 − 5 + 10 − 2 5 i 4 − c 2 a + ( c 2 a ) 2 + ( b a ) 5 5 + − 1 − 5 − 10 − 2 5 i 4 − c 2 a − ( c 2 a ) 2 + ( b a ) 5 5 {\displaystyle {x_{3}={\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}+{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}+{\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}-{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}}\,} x 4 = − 1 − 5 − 10 − 2 5 i 4 − c 2 a + ( c 2 a ) 2 + ( b a ) 5 5 + − 1 − 5 + 10 − 2 5 i 4 − c 2 a − ( c 2 a ) 2 + ( b a ) 5 5 {\displaystyle {x_{4}={\frac {-1-{\sqrt {5}}-{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}+{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}+{\frac {-1-{\sqrt {5}}+{\sqrt {10-2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}-{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}}\,} x 5 = − 1 + 5 − 10 + 2 5 i 4 − c 2 a + ( c 2 a ) 2 + ( b a ) 5 5 + − 1 + 5 + 10 + 2 5 i 4 − c 2 a − ( c 2 a ) 2 + ( b a ) 5 5 {\displaystyle {x_{5}={\frac {-1+{\sqrt {5}}-{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}+{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}+{\frac {-1+{\sqrt {5}}+{\sqrt {10+2{\sqrt {5}}}}{\rm {i}}}{4}}{\sqrt[{5}]{-{\frac {c}{2a}}-{\sqrt {\left({\frac {c}{2a}}\right)^{2}+\left({\frac {b}{a}}\right)^{5}}}}}}\,} 在 Tschirnhaus 變換的幫助下，所有五次方程都可以在初等數學函數表達式的幫助下轉換為 Bring-Jerrard 形式。 Bring-Jerrard 形式包含五次項、線性項和絕對項。 但是四次、三次和二次項在這種形式的方程中根本不存在。 Bring-Jerrard 形式的廣義橢圓解將在以下段落中討論。根據數學家 Glashan、Young 和 Runge 發現的參數化公式，可以從方程和實解中導出以下一對公式: x 5 + x = 2 5 y − 5 / 4 ( 1 + y − y 2 ) 2 + 2 y 2 10 + 15 y − 10 y 2 4 {\displaystyle x^{5}+x={\frac {2}{5}}y^{-5/4}{\frac {(1+y-y^{2}){\sqrt {2+2y^{2}}}}{\sqrt[{4}]{10+15y-10y^{2}}}}} x = 2 5 y − 1 / 4 10 + 15 y − 10 y 2 4 cosh ⁡ { 1 5 arcosh [ 5 5 + 5 y 2 ( 1 + 2 y ) 4 + 6 y − 4 y 2 ] } − {\displaystyle x={\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {5{\sqrt {5+5y^{2}}}}{(1+2y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}-} − 2 5 y − 1 / 4 10 + 15 y − 10 y 2 4 sinh ⁡ { 1 5 arsinh [ 5 y 5 + 5 y 2 ( 2 − y ) 4 + 6 y − 4 y 2 ] } {\displaystyle -{\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {5y{\sqrt {5+5y^{2}}}}{(2-y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}} 這對公式對所有值 0 < y < 2 都有效。對於要用這種方法求解的 Bring-Jerrard 的一般形式，需要一個橢圓鍵。 這個橢圓密鑰可以根據 卡爾·雅可比 (Carl Gustav Jakob Jacobi) 使用 Θ函數 生成: x 5 + x = w {\displaystyle x^{5}+x=w} x = 2 5 y − 1 / 4 10 + 15 y − 10 y 2 4 cosh ⁡ { 1 5 arcosh [ 5 5 + 5 y 2 ( 1 + 2 y ) 4 + 6 y − 4 y 2 ] } − {\displaystyle x={\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\cosh {\biggl \{}{\frac {1}{5}}{\text{arcosh}}{\biggl [}{\frac {5{\sqrt {5+5y^{2}}}}{(1+2y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}-} − 2 5 y − 1 / 4 10 + 15 y − 10 y 2 4 sinh ⁡ { 1 5 arsinh [ 5 y 5 + 5 y 2 ( 2 − y ) 4 + 6 y − 4 y 2 ] } {\displaystyle -{\frac {2}{5}}y^{-1/4}{\sqrt[{4}]{10+15y-10y^{2}}}\sinh {\biggl \{}{\frac {1}{5}}{\text{arsinh}}{\biggl [}{\frac {5y{\sqrt {5+5y^{2}}}}{(2-y){\sqrt {4+6y-4y^{2}}}}}{\biggr ]}{\biggr \}}} y = 5 ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } 5 ⟩ 2 2 ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } ⟩ 2 − 1 2 {\displaystyle y={\frac {5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }^{2}}{2\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}-{\frac {1}{2}}} 現在在下面精確地解釋這個解決過程。 本段上式的等式刻度的右側取值 w: w = 2 5 y − 5 / 4 ( 1 + y − y 2 ) 2 + 2 y 2 10 + 15 y − 10 y 2 4 {\displaystyle w={\frac {2}{5}}y^{-5/4}{\frac {(1+y-y^{2}){\sqrt {2+2y^{2}}}}{\sqrt[{4}]{10+15y-10y^{2}}}}} 必須為值 y 求解該方程。 這需要一個橢圓模函數表達式，在這種情況下包括 Jacobi theta 函數: y = 5 ϑ 00 { q [ ( 50 5 w 2 + 32 + 2 3125 w 4 + 256 ) − 1 / 2 ( 3125 w 4 + 256 + 16 + 5 5 4 w ) ] 5 } 2 2 ϑ 00 { q [ ( 50 5 w 2 + 32 + 2 3125 w 4 + 256 ) − 1 / 2 ( 3125 w 4 + 256 + 16 + 5 5 4 w ) ] } 2 − 1 2 {\displaystyle y={\frac {5\,\vartheta _{00}{\bigl \{}q{\bigl [}{\bigl (}50{\sqrt {5}}\,w^{2}+32+2{\sqrt {3125w^{4}+256}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {3125w^{4}+256}}+16}}+5{\sqrt[{4}]{5}}\,w{\bigr )}{\bigr ]}^{5}{\bigr \}}^{2}}{2\,\vartheta _{00}{\bigl \{}q{\bigl [}{\bigl (}50{\sqrt {5}}\,w^{2}+32+2{\sqrt {3125w^{4}+256}}{\bigr )}^{-1/2}{\bigl (}{\sqrt {{\sqrt {3125w^{4}+256}}+16}}+5{\sqrt[{4}]{5}}\,w{\bigr )}{\bigr ]}{\bigr \}}^{2}}}-{\frac {1}{2}}} 此解表達式與以下表達式一致: y = 5 ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } 5 ⟩ 2 2 ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } ⟩ 2 − 1 2 {\displaystyle y={\frac {5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }^{2}}{2\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}-{\frac {1}{2}}} 現在必須定義此表達式中指定的函數。 所示的主要 theta 函數具有以下總和定義和以下等效乘積定義: ϑ 00 ( z ) = 1 + 2 ∑ k = 1 ∞ z k 2 {\displaystyle \vartheta _{00}(z)=1+2\sum _{k=1}^{\infty }z^{k^{2}}} ϑ 00 ( z ) = ∏ k = 1 ∞ ( 1 − z 2 k ) ( 1 + z 2 k − 1 ) 2 {\displaystyle \vartheta _{00}(z)=\prod _{k=1}^{\infty }(1-z^{2k})(1+z^{2k-1})^{2}} 字母 q 描述了數學橢圓 nome 函數: q ( ε ) = exp ⁡ [ − π K ( 1 − ε 2 ) K ( ε ) − 1 ] {\displaystyle q(\varepsilon )=\exp[-\pi K({\sqrt {1-\varepsilon ^{2}}})K(\varepsilon )^{-1}]} 內商中顯示的字母 K 表示完整的第一類 椭圆积分: K ( r ) = ∫ 0 π / 2 1 1 − r 2 sin ⁡ ( φ ) 2 d φ {\displaystyle K(r)=\int _{0}^{\pi /2}{\frac {1}{\sqrt {1-r^{2}\sin(\varphi )^{2}}}}\mathrm {d} \varphi } K ( r ) = 2 ∫ 0 1 1 ( u 2 + 1 ) 2 − 4 r 2 u 2 d u {\displaystyle K(r)=2\int _{0}^{1}{\frac {1}{\sqrt {(u^{2}+1)^{2}-4r^{2}u^{2}}}}\mathrm {d} u} 縮寫 ctlh 表示函數 双曲双扭线餘切函数 (Hyperbolic lemniscate cotangent)。 而縮寫 aclh 表示函數 双曲双扭线 面積餘弦函数 (Hyperbolic lemniscate Areacosine)。 這些函數與 卡爾·弗里德里希·高斯(Carl Friedrich Gauss) 建立的 双扭线函数 (Lemniscate elliptic functions) sl 和 cl 在代數上相關，並且可以使用這兩個函數來定義： s l ( φ ) = tan ⁡ ⟨ 2 arctan ⁡ { 4 G sin ⁡ ( φ G ) ∑ k = 1 ∞ cosh ⁡ [ ( 2 k − 1 ) π ] cosh ⁡ [ ( 2 k − 1 ) π ] 2 − cos ⁡ ( φ / G ) 2 } ⟩ {\displaystyle \mathrm {sl} (\varphi )=\tan {\biggl \langle }2\arctan {\biggl \{}{\frac {4}{G}}\sin {\bigl (}{\frac {\varphi }{G}}{\bigr )}\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\cos(\varphi /G)^{2}}}{\biggr \}}{\biggr \rangle }} c l ( φ ) = tan ⁡ ⟨ 2 arctan ⁡ { 4 G cos ⁡ ( φ G ) ∑ k = 1 ∞ cosh ⁡ [ ( 2 k − 1 ) π ] cosh ⁡ [ ( 2 k − 1 ) π ] 2 − sin ⁡ ( φ / G ) 2 } ⟩ {\displaystyle \mathrm {cl} (\varphi )=\tan {\biggl \langle }2\arctan {\biggl \{}{\frac {4}{G}}\cos {\bigl (}{\frac {\varphi }{G}}{\bigr )}\sum _{k=1}^{\infty }{\frac {\cosh[(2k-1)\pi ]}{\cosh[(2k-1)\pi ]^{2}-\sin(\varphi /G)^{2}}}{\biggr \}}{\biggr \rangle }} [ sl ( φ ) 2 + 1 ] [ cl ( φ ) 2 + 1 ] = 2 {\displaystyle [{\text{sl}}(\varphi )^{2}+1][{\text{cl}}(\varphi )^{2}+1]=2} ctlh ( ϱ ) = cl ⁡ ( 1 2 2 ϱ ) [ sl ⁡ ( 1 2 2 ϱ ) 2 + 1 sl ⁡ ( 1 2 2 ϱ ) 2 + cl ⁡ ( 1 2 2 ϱ ) 2 ] 1 / 2 {\displaystyle {\text{ctlh}}(\varrho )=\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho ){\biggl [}{\frac {\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}+1}{\operatorname {sl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}+\operatorname {cl} ({\tfrac {1}{2}}{\sqrt {2}}\varrho )^{2}}}{\biggr ]}^{1/2}} ctlh ( ϱ ) = cd ( ϱ ; 1 2 2 ) cd ( ϱ ; 1 2 2 ) 4 + sn ( ϱ ; 1 2 2 ) 4 4 {\displaystyle {\text{ctlh}}(\varrho )={\frac {{\text{cd}}(\varrho ;{\tfrac {1}{2}}{\sqrt {2}})}{\sqrt[{4}]{{\text{cd}}(\varrho ;{\tfrac {1}{2}}{\sqrt {2}})^{4}+{\text{sn}}(\varrho ;{\tfrac {1}{2}}{\sqrt {2}})^{4}}}}} a c l h ( s ) = 1 2 F [ 2 arccot ⁡ ( s ) ; 1 2 2 ] {\displaystyle \mathrm {aclh} (s)={\tfrac {1}{2}}F[2\operatorname {arccot}(s);{\tfrac {1}{2}}{\sqrt {2}}]} aclh ( s ) = 1 2 2 π G − ∫ 0 1 s s 4 t 4 + 1 d t {\displaystyle {\text{aclh}}(s)={\frac {1}{2}}{\sqrt {2}}\,\pi \,G-\int _{0}^{1}{\frac {s}{\sqrt {s^{4}t^{4}+1}}}\,\mathrm {d} t} G = 1 2 2 π Γ ( 3 4 ) − 2 {\displaystyle G={\tfrac {1}{2}}{\sqrt {2\pi }}\,\Gamma ({\tfrac {3}{4}})^{-2}} ctlh [ 1 2 a c l h ( s ) ] 2 = ( 2 s 2 + 2 + 2 s 4 + 1 ) − 1 / 2 ( s 4 + 1 + 1 + s ) {\displaystyle {\text{ctlh}}{\bigl [}{\tfrac {1}{2}}\mathrm {aclh} (s){\bigr ]}^{2}=(2s^{2}+2+2{\sqrt {s^{4}+1}})^{-1/2}({\sqrt {{\sqrt {s^{4}+1}}+1}}+s)} sl [ 1 2 2 a c l h ( s ) ] = s 4 + 1 − s 2 {\displaystyle {\text{sl}}{\bigl [}{\tfrac {1}{2}}{\sqrt {2}}\,\mathrm {aclh} (s){\bigr ]}={\sqrt {{\sqrt {s^{4}+1}}-s^{2}}}} 字母G代表高斯常數，可以用伽馬函數用剛才所示的方式表示。 连分数是拉马努金 (Rogers-Ramanujan continued fraction) 允許以 Bring-Jerrard 形式對廣義五次方程進行非常緊湊的解。 這個連分數函數和交替連分數可以定義如下： R ( z ) = z 1 / 5 ( z ; z 5 ) ∞ ( z 4 ; z 5 ) ∞ ( z 2 ; z 5 ) ∞ ( z 3 ; z 5 ) ∞ {\displaystyle R(z)=z^{1/5}{\frac {(z;z^{5})_{\infty }(z^{4};z^{5})_{\infty }}{(z^{2};z^{5})_{\infty }(z^{3};z^{5})_{\infty }}}} R ( z ) = tan ⁡ { 1 2 arctan ⁡ [ ϑ 00 ( z 1 / 2 ) 2 2 ϑ 00 ( z 5 / 2 ) 2 − 1 2 ] } 2 / 5 tan ⁡ { 1 2 arccot ⁡ [ ϑ 00 ( z 1 / 2 ) 2 2 ϑ 00 ( z 5 / 2 ) 2 − 1 2 ] } 1 / 5 {\displaystyle R(z)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z^{1/2})^{2}}{2\vartheta _{00}(z^{5/2})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z^{1/2})^{2}}{2\vartheta _{00}(z^{5/2})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}} R ( z 2 ) = tan ⁡ { 1 2 arctan ⁡ [ ϑ 00 ( z ) 2 2 ϑ 00 ( z 5 ) 2 − 1 2 ] } 2 / 5 tan ⁡ { 1 2 arccot ⁡ [ ϑ 00 ( z ) 2 2 ϑ 00 ( z 5 ) 2 − 1 2 ] } 1 / 5 {\displaystyle R(z^{2})=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}\tan {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}} S ( z ) = tan ⁡ { 1 2 arctan ⁡ [ ϑ 00 ( z ) 2 2 ϑ 00 ( z 5 ) 2 − 1 2 ] } 1 / 5 cot ⁡ { 1 2 arccot ⁡ [ ϑ 00 ( z ) 2 2 ϑ 00 ( z 5 ) 2 − 1 2 ] } 2 / 5 {\displaystyle S(z)=\tan {\biggl \{}{\frac {1}{2}}\arctan {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{1/5}\cot {\biggl \{}{\frac {1}{2}}\operatorname {arccot} {\biggl [}{\frac {\vartheta _{00}(z)^{2}}{2\vartheta _{00}(z^{5})^{2}}}-{\frac {1}{2}}{\biggr ]}{\biggr \}}^{2/5}} S ( z ) = R ( z 4 ) R ( z 2 ) R ( z ) {\displaystyle S(z)={\frac {R(z^{4})}{R(z^{2})R(z)}}} 括號，每個都有兩個條目，形成所謂的 Pochhammer-符號 (Pochhammer symbol) 並因此代表產品系列。 基於這些定義，可以為實際解建立以下壓縮精確解公式: x 5 + x = w {\displaystyle x^{5}+x=w} x = S ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } ⟩ 2 − R ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } 2 ⟩ S ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } ⟩ 2 × {\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{2}}}\times } × 1 − R ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } 2 ⟩ S ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } ⟩ R ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } 2 ⟩ 2 × {\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{2}{\bigr \rangle }^{2}}}\times } × ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } 5 ⟩ ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } 1 / 5 ⟩ 2 − 5 ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } 5 ⟩ 3 2 20 4 sl [ 1 2 2 aclh ( 5 4 5 4 w ) ] ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 5 4 5 4 w ) ] 2 } ⟩ 3 {\displaystyle \times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {5}{4}}{\sqrt[{4}]{5}}\,w)]^{2}\}{\bigr \rangle }^{3}}}} 分配给非初等数学实解的第一个自然数 w 是数字 w = 3: x 5 + x = 3 {\displaystyle x^{5}+x=3} x = S ⟨ q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } ⟩ 2 − R ⟨ q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } 2 ⟩ S ⟨ q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } ⟩ 2 × {\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}}}\times } × 1 − R ⟨ q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } 2 ⟩ S ⟨ q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } ⟩ R ⟨ q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } 2 ⟩ 2 × {\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }^{2}}}\times } × ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } 5 ⟩ ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } 1 / 5 ⟩ 2 − 5 ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } 5 ⟩ 3 2 20 4 sl [ 1 2 2 aclh ( 15 4 5 4 ) ] ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } ⟩ 3 {\displaystyle \times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{3}}}} q { ctlh [ 1 2 aclh ( 15 4 5 4 ) ] 2 } ≈ 0.452374059450344348576600264284387826377845763909 {\displaystyle q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {15}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.452374059450344348576600264284387826377845763909} x ≈ 1.132997565885065266721141634288532379816526027727 {\displaystyle x\approx 1.132997565885065266721141634288532379816526027727} 與此類似，數字 w = 7 僅分配給非基本解: x 5 + x = 7 {\displaystyle x^{5}+x=7} x = S ⟨ q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } ⟩ 2 − R ⟨ q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } 2 ⟩ S ⟨ q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } ⟩ 2 × {\displaystyle x={\frac {S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }}{S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{2}}}\times } × 1 − R ⟨ q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } 2 ⟩ S ⟨ q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } ⟩ R ⟨ q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } 2 ⟩ 2 × {\displaystyle \times {\frac {1-R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }\,S{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }}{R{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{2}{\bigr \rangle }^{2}}}\times } × ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } 5 ⟩ ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } 1 / 5 ⟩ 2 − 5 ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } 5 ⟩ 3 2 20 4 sl [ 1 2 2 aclh ( 35 4 5 4 ) ] ϑ 00 ⟨ q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } ⟩ 3 {\displaystyle \times {\frac {\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{1/5}{\bigr \rangle }^{2}-5\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}^{5}{\bigr \rangle }^{3}}{2{\sqrt[{4}]{20}}\,{\text{sl}}[{\tfrac {1}{2}}{\sqrt {2}}\,{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]\,\vartheta _{00}{\bigl \langle }q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}{\bigr \rangle }^{3}}}} q { ctlh [ 1 2 aclh ( 35 4 5 4 ) ] 2 } ≈ 0.53609630892200161460073096549143569900990236 {\displaystyle q\{{\text{ctlh}}[{\tfrac {1}{2}}{\text{aclh}}({\tfrac {35}{4}}{\sqrt[{4}]{5}})]^{2}\}\approx 0.53609630892200161460073096549143569900990236} x ≈ 1.4108138510595771319852918753499397839215989 {\displaystyle x\approx 1.4108138510595771319852918753499397839215989} 方程式論 布靈根式 阿米爾·艾克塞爾（Amir D. Aezel）. 費馬最後定理. 台北: 時報出版. 1998: p.87. ISBN 957-13-2648-8.  引文格式1维护：冗余文本 (link) 存档副本. [2022-02-18]. （原始内容存档于2022-02-18）.  存档副本. [2022-02-18]. （原始内容存档于2022-02-18）.  存档副本. [2022-02-19]. （原始内容存档于2022-02-19）.