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added missing subscript
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Richard Stanley
Richard Stanley
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We have $$ \frac{\zeta_1+\frac{1}{\zeta_1}}{\zeta_2+\frac{1}{\zeta_2}}=\sqrt{2}. $$ By applying Galois group automorphisms we can assume that $\zeta_2=e^{2\pi i/n}$ for some $n$, where we can take either value of $\sqrt{2}$. Solving for $\zeta_1$ gives $$ \sqrt{2}\zeta_1=\zeta_2+\frac{1}{\zeta_2} \pm\sqrt{\zeta_2^2+\frac{1}{\zeta^2}}. $$$$ \sqrt{2}\zeta_1=\zeta_2+\frac{1}{\zeta_2} \pm\sqrt{\zeta_2^2+\frac{1}{\zeta_2^2}}. $$ If $n>8$ then $\zeta_2^2+\frac{1}{\zeta_2^2}>0$. This implies that $\zeta_1$ is real, a contradiction. Thus there are no further solutions.

We have $$ \frac{\zeta_1+\frac{1}{\zeta_1}}{\zeta_2+\frac{1}{\zeta_2}}=\sqrt{2}. $$ By applying Galois group automorphisms we can assume that $\zeta_2=e^{2\pi i/n}$ for some $n$, where we can take either value of $\sqrt{2}$. Solving for $\zeta_1$ gives $$ \sqrt{2}\zeta_1=\zeta_2+\frac{1}{\zeta_2} \pm\sqrt{\zeta_2^2+\frac{1}{\zeta^2}}. $$ If $n>8$ then $\zeta_2^2+\frac{1}{\zeta_2^2}>0$. This implies that $\zeta_1$ is real, a contradiction. Thus there are no further solutions.

We have $$ \frac{\zeta_1+\frac{1}{\zeta_1}}{\zeta_2+\frac{1}{\zeta_2}}=\sqrt{2}. $$ By applying Galois group automorphisms we can assume that $\zeta_2=e^{2\pi i/n}$ for some $n$, where we can take either value of $\sqrt{2}$. Solving for $\zeta_1$ gives $$ \sqrt{2}\zeta_1=\zeta_2+\frac{1}{\zeta_2} \pm\sqrt{\zeta_2^2+\frac{1}{\zeta_2^2}}. $$ If $n>8$ then $\zeta_2^2+\frac{1}{\zeta_2^2}>0$. This implies that $\zeta_1$ is real, a contradiction. Thus there are no further solutions.

Source Link
Richard Stanley
Richard Stanley
  • 50.8k
  • 14
  • 155
  • 279

We have $$ \frac{\zeta_1+\frac{1}{\zeta_1}}{\zeta_2+\frac{1}{\zeta_2}}=\sqrt{2}. $$ By applying Galois group automorphisms we can assume that $\zeta_2=e^{2\pi i/n}$ for some $n$, where we can take either value of $\sqrt{2}$. Solving for $\zeta_1$ gives $$ \sqrt{2}\zeta_1=\zeta_2+\frac{1}{\zeta_2} \pm\sqrt{\zeta_2^2+\frac{1}{\zeta^2}}. $$ If $n>8$ then $\zeta_2^2+\frac{1}{\zeta_2^2}>0$. This implies that $\zeta_1$ is real, a contradiction. Thus there are no further solutions.