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nirax call $\displaystyle A$ as the multiplicative group of squares. so we decompose $\displaystyle F = A \cup \{0\} \cup B$. $\displaystyle B$ is the set of non squares. suppose an element $\displaystyle a$ is not expressible as sum of two squares. consider $\displaystyle a - A$. this set must be the same as $\displaystyle B$. consider $\displaystyle x = a - a^2$. clearly $\displaystyle x \in B$
but $\displaystyle B$ was the coset of the multiplicative group $\displaystyle A$ in $\displaystyle F^*$. so $\displaystyle B = a.A$. it follows that there exists a $\displaystyle y$ s.t
$\displaystyle x = a.y^2 = a - a^2$
$\displaystyle \Rightarrow a (y^2 - 1 + a) = 0$
$\displaystyle \Rightarrow y^2 = a - 1 $
RHS belongs to $\displaystyle B$ whereas LHS belongs to $\displaystyle A$ which are disjoint.