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Squaring a number is like multiplying two $n$ bit numbers and using Gauss' observation, it takes $O(n^{1.59})$ time. Taking mod on the order of $2^o$ is simple because you do a shift that takes constant time and the answer is the bits you were shifting. So the cost of recombining is $O(n^{1.59}$. SO using Master Theorem:  \[T(n) = 1*T(\frac{n}{2}) + O(n^{1.59})\]  Using the theorem, lg(1) < 1.59m so cost of recursion is $n^{1.59}$, and we need to do recursion $n$ times. Since $y=2^n$, we need y = 1 until we can stop. So total time is:     \[O(n^{2.59})\]  \end{itemize}