$-1$ is not a sum of two squares in $\mathbb{Q}(\sqrt{-7})$

An elementary solution that shows that $-1$ is not a sum of squares from $\mathbb{Z}[\sqrt{-7}]$. Assume $\sum_k (a_k + b_k \sqrt{-7})^2 = -1$ that is $\sum_k (a_k^2 - 7 b_k^2) =-1$ and $\sum_k 2 a_k b_k = 0$, adding the last two equalities we get $$\sum_k (a_k + b_k)^2 =-1 +8 \sum_k ( b_k)^2$$

so a sum of three squares $\equiv -1 \pmod 8$, not possible.

Tags:

Number Theory

Algebraic Number Theory

Sums Of Squares

Field Theory

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