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Over $\mathbb{C}$, with $\alpha$ being a root of $1+x+x^2$, does this system of equations have any exact solutions when $\lambda_4\neq 0$? \begin{align} -\lambda_1^2+\lambda_2\lambda_6&= \alpha^2\lambda_4, &\label{lambda4}\\ -\lambda_2\lambda_5+\lambda_1\lambda_6&=\lambda_3, &\label{lambda3}\\ \lambda_5&=-\alpha\lambda_4, & \label{lambda5}\\ \lambda_4\lambda_5-\lambda_3\lambda_6&= \lambda_2, &\\ \lambda_4\lambda_6-\lambda_1\lambda_3&=1, &\\ -\lambda_6^2+\lambda_1\lambda_5&=\lambda_1\lambda_4-\lambda_2\lambda_3. & \end{align} There are precisely $7$ solutions when $\lambda_4 = 0$.

aliquot
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1 Answers1

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Yes, it has many solutions for other $\lambda_4\neq 0$, for example \begin{align} \lambda_1 & = -1,\cr \lambda_2 & = -\frac{1+\sqrt{-3}}{2},\cr \lambda_3 & = -\alpha,\cr \lambda_4 & = 1,\cr \lambda_5 & = -\alpha,\cr \lambda_6 & = -\lambda_2. \end{align}

Dietrich Burde
  • 130,978
  • Any idea about generic $\lambda_4?$ – aliquot Mar 23 '23 at 18:22
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    Yes, of course. You obtain a polynomial condition (of high degree in general), which makes it hard to write down a nice solution. But there is one. For example, writing $\lambda_i=x_i$, already for $x_4=3$ we obtain a more complicated solution, namely: – Dietrich Burde Mar 23 '23 at 19:20
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    $$x_2=(74917528803753634935x_1^{13} + 95480103523754706795x_1^{12} - 1047669995719087540344 x_1^{11} - 1676306179865227806666x_1^{10} + 6695884698243335938083x_1^9 + 62065024358890686756777 x_1^8 - 6930878885404393209030x_1^7 - 790583763174146169322617x_1^6 + 5634490277083093124774058 x_1^5 + 2980022636662406003978956x_1^4 - 51648212948235010351392474x_1^3 - 13397787275221502207378541 x_1^2 + 94408690214474921415013389*x_1 - 49150493250631221793624875)/468930515866794097289617094$$ etc., where $x_1$ is a root of a polynomial of degree $14$. – Dietrich Burde Mar 23 '23 at 19:22