+ /* --- Unpicking the magic --- *
+ *
+ * Choose %$\rho \inr \gf{2^m}$% with %$\Tr(\rho) = 1$%. Let
+ * %$z = \sum_{0\le i<m} \rho^{2^i} \sum_{0\le j<i} x^{2^j} = {}$%
+ * %$\rho^2 x + \rho^4 (x + x^2) + \rho^8 (x + x^2 + x^4) + \cdots + {}$%
+ * %$\rho^{2^{m-1}} (x + x^2 + x^{2^{m-2}})$%. Then %$z^2 = {}$%
+ * %$\sum_{0\le i<m} \rho^{2^{i+1}} \sum_{0\le j<i} x^{2^{j+1}} = {}$%
+ * %$\sum_{1\le i\le m} \rho^{2^i} \sum_{1\le j\le i} x^{2^j}$% and,
+ * somewhat miraculously, %$z^2 + z = \sum_{0\le i<m} \rho^{2^i} x + {}$%
+ * %$\rho \sum_{1\le i<m} x^{2^i} = x \Tr(\rho) + \rho \Tr(x)$%. Again,
+ * this gives us the root we want whenever %$\Tr(x) = 0$%.
+ *
+ * The loop below calculates %$w = \Tr(\rho)$% and %$z$% simultaneously,
+ * since the same powers of %$\rho$% are wanted in both calculations.
+ */
+