*
* Choose %$\rho \inr \gf{2^m}$% with %$\Tr(\rho) = 1$%. Let
* %$z = \sum_{0\le i<m} \rho^{2^i} \sum_{0\le j<i} x^{2^j} = {}$%
+ * %$\sum_{1\le i<m} \rho^{2^i} (x + \sum_{1\le j<i} x^{2^j} = {}$%
* %$\rho^2 x + \rho^4 (x + x^2) + \rho^8 (x + x^2 + x^4) + \cdots + {}$%
* %$\rho^{2^{m-1}} (x + x^2 + x^{2^{m-2}})$%. 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,
+ * %$\sum_{1\le i\le m} \rho^{2^i} \sum_{1\le j<i} x^{2^j} = {}$%
+ * %$\sum_{1\le i<m} \rho^{2^i} \sum_{1\le j<i} x^{2^j} + {}$%
+ * %$\rho^{2^m} \sum_{1\le j<m} x^{2^j}$%; and, somewhat miraculously,
+ * %$z^2 + z = \sum_{1\le i<m} \rho^{2^i} x + {}$%
+ * %$\rho \sum_{1\le i<m} x^{2^i} = x (\Tr(\rho) + \rho) + {}$%
+ * %$\rho (\Tr(x) + x) = x \Tr(\rho) + \rho \Tr(x)$%. Again,
* this gives us the root we want whenever %$\Tr(x) = 0$%.
*
* The loop below calculates %$w = \Tr(\rho)$% and %$z$% simultaneously,