Techniques of Solving Equations à la Galois
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Name: scruta \(\quad\)
Daily life: mowing
Revision history
1st upload: 2023/06/17
revision2 : 2023/07/27
revision3 : 2024/12/22
revision4 : 2025/09/14
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[Ch.1] | Mastering Galois-style Equation Solving at Lightning Speed | \(C_2\) | \(x^2+x+1\) | 2pp. |
[Ch.2] | The Essence of Galois Theory Packed Into One | \( S_3 \) | \(x^3+3x+1 \) | 9pp. |
[Ch.3] | Exhausting! Packed with Cyclic Extensions! | \( S_4 \) | \( x^4+4x+2 \) | 12pp. |
[Ch.4] | Super Cool! The Magic Technique of Resultants | \(A_3\) | \(x^3-3x+1\) | 4pp. |
[Ch.5] | The Terror of Factorization: Trager Algorithm | \( A_3\) | \(x^3-3x+1 \quad (gcd)\) | 5pp. |
[Ch.6] | Solving Cyclic Equations \(\Phi_{5}(x)\) | \( C_4 \) | \(x^4+x^3+x^2+1 \) | 3pp. |
[Ch.7] | Galois-Style Solution for the Cyclotomic Equation \(\Phi_{17}(x)\) | \( C_{16} \) | \(x^{16}+x^{15}+...+x^2+x+1 \) | 6pp. |
[Ch.8] | A Treasure Trove of Group Theory Problems: Frobenius Groups | \( F_{20} \) | \( x^5+x^4+2x^3+4x^2+x+1\) | 10pp. |
[Ch.9] | What!? Is the Galois-Style Method Falling Apart? | ?,? | \(x^3-2, x^5-5x^3+5x+6 \) | 5pp. |
\begin{align} &f(x)=x^2+x+1 \quad v \equiv \alpha+2\beta \notag \\ \notag \\ &V(x) \equiv (x-v_{1})(x-v_{2})=g_0(x) \notag \\ &P_\alpha(x)=V(x) \bigl(\frac{\alpha }{x-{v_1}}+\frac{\beta }{x-{v_2}} \bigr) \notag \\ &\alpha=\frac{ P_\alpha(v_1)}{(v_1-v_2)} \notag \\ \notag \\ &\left\{ \begin{array}{l} h_0 = (x-v_1) \\ h_1 = (x-v_2) \\ \end{array} \right. \notag \\ &\begin{bmatrix} t_0 \\ t_1 \end{bmatrix} \equiv \frac{1}{2} \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} \cdot \begin{bmatrix} h_0 \\ h_1 \end{bmatrix} \notag \\ \notag \\ &\left\{ \begin{array}{l} B_0(x)=x^2-A_0=0 \\ t_1^2=A_0=-\frac{3}{4} \\ a_0=\sqrt{A_0} \ \in Q_1 \\ \end{array} \right. \notag \\ \notag \\ &\begin{bmatrix} \tilde{h_0} \\ \tilde{h_1 } \end{bmatrix} = \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} \cdot \begin{bmatrix} t_0 \\ \tilde{t_1} \end{bmatrix} = \begin{bmatrix} g_1(x) \\ t_0- \tilde{t_1} \end{bmatrix} \notag \\ \notag \\ &\left\{ \begin{array}{l} g_1(x)=0 \ \rightarrow \ v=-\frac{3}{2}-a_0 \\ \therefore \alpha=-v-2=-\frac{1}{2}+a_0 =\omega \end{array} \right. \notag \\ \end{align}
\begin{align} &f(x)=x^3-3x+1 \quad g(v)=v^3-9v-9 \notag \\ \notag \\ &Res(f(x+v) ,g(v),v) \notag \\ &=-(x^3-21x+37)(x^3-12x-8)(x^3-3x+1) \notag \\ \notag \\ \end{align}
\begin{align} Res&(f(x+v), g(v),v) \notag \\ =(-1)& \left\{\bbox[#FFFF00]{ (x+v_1-\alpha) }\bbox[#7FFF00]{(x+v_1-\beta)}\bbox[#00FFFF]{(x+v_1-\gamma)} \right \} \notag \\ \times &\left\{\bbox[#00FFFF]{ (x+v_4-\alpha)}\bbox[#FFFF00]{ (x+v_4-\beta)}\bbox[#7FFF00]{(x+v_4-\gamma)} \right \} \notag \\ \times &\left\{\bbox[#7FFF00]{(x+v_5-\alpha)}\bbox[#00FFFF]{(x+v_5-\beta)}\bbox[#FFFF00]{ (x+v_5-\gamma)} \right \} \notag \\ \notag \\ =(-1)& \left\{ \bbox[#FFFF00]{(x+v_1-\alpha)(x+v_4-\beta)(x+v_5-\gamma)} \right \} \notag \\ \times &\left\{\bbox[#7FFF00]{(x+v_1-\beta)(x+v_4-\gamma)(x+v_5-\alpha)} \right \} \notag \\ \times &\left\{\bbox[#00FFFF]{(x+v_1-\gamma)(x+v_4-\alpha)(x+v_5-\beta)} \right \} \notag \\ \notag \\ =(-1)& \cdot\bbox[#FFFF00]{H_1} \cdot \bbox[#7FFF00]{H_2} \cdot \bbox[#00FFFF]{H_3} \notag \\ \notag \\ \end{align}
\begin{align} &f(x+v) \notag \\ &=\bbox[#FFFF00]{(x+v-\alpha)} \cdot \bbox[#7FFF00]{(x+v-\beta)} \cdot \bbox[#00FFFF]{(x+v-\gamma)} \notag \\ \notag \\ &\left\{ \begin{array}{l} GCD\bigl(\left. \bbox[#FFFF00]{H_1} \right|_{v_1=v} \ ,f(x+v)\bigr)= \bbox[#FFFF00]{(x+v-\alpha)}=Y_1 \\ GCD\bigl(\left. \bbox[#7FFF00]{H_2} \right|_{v_1=v} \ ,f(x+v)\bigr)= \bbox[#7FFF00]{(x+v-\beta)}=Y_2 \\ GCD\bigl(\left. \bbox[#00FFFF]{H_3} \right|_{v_1=v} \ ,f(x+v)\bigr)= \bbox[#00FFFF]{(x+v+\gamma)}=Y_3 \\ \end{array} \right. \notag \\ \notag \\ \end{align}
\begin{align} f(x)&=Y_1(x-v) \cdot Y_2(x-v) \cdot Y_3(x-v) \notag \\ &=(x-x_1)(x-x_2)(x-x_3) \notag \\ \end{align}