# Equation $x_1^2+\cdots +x_k^2=-1$ Doesn’t Have a Solution in Number Field $\Q(\sqrt{2}e^{2\pi i/3})$ ## Problem 358

Let $\alpha= \sqrt{2}e^{2\pi i/3}$. Prove that $x_1^2+\cdots +x_k^2=-1$ has no solutions with all $x_i\in \Q(\alpha)$ and $k\geq 1$. Add to solve later

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## Proof.

Note that $\alpha= \sqrt{2}e^{2\pi i/3}$ is a root of the polynomial $x^3-2$.
The polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein’s criterion with prime $p=2$.

The roots of this polynomial are
$\sqrt{2}, \sqrt{2}e^{2\pi i/3}, \sqrt{2}e^{4\pi i/3}.$ Then it follows that we have an isomorphism
\begin{align*}
\Q(\alpha)=\Q(\sqrt{2}e^{2\pi i/3}) \cong \Q[x]/(x^3-2)\cong \Q(\sqrt{2}).
\end{align*}
Let us denote this isomorphism by $\phi:\Q(\alpha) \to \Q(\sqrt{2})$, which sends $\alpha$ to $\sqrt{2}$ and fixed $\Q$ elementwise.

Seeking a contradiction, we assume that there exist $x_1, \dots, x_k \in \Q(\alpha)$ such that
$x_1^2+\cdots +x_k^2=-1.$

Then we apply the isomorphism $\phi$ and obtain
\begin{align*}
-1&=\phi(-1)=\phi(x_1^2+\cdots +x_k^2)\\
&=\phi(x_1)^2+\cdots +\phi(x_k)^2.
\end{align*}

However, this equality does not hold since $\phi(x_i) \in \Q(\sqrt{2})$ and the field $\Q(\sqrt{2})$ consists of real numbers.

Thus, we have reached a contradiction, hence there is no solution of $x_1^2+\cdots +x_k^2=-1$ in the field $\Q(\alpha)$. Add to solve later

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