I also fine this confusing. From the video, it sounds like the claim is that sqrt(x) = x^(2^(k-1)), which you verify by checking that x=x^(2^k) [0]. However, s

I also fine this confusing. From the video, it sounds like the claim is that sqrt(x) = x^(2^(k-1)), which you verify by checking that x=x^(2^k) [0]. However, since the order of the group is 2^k, it is an elementary result of group theory that x^(2^k)=1 [1] [0] That is, you square the claimed root which should be x^(2^(k-1)). EDIT: watching the video some more for clues, he does mention that he uses a particular kind of group, so the claimed property does not need to hold for any group of order 2^k. However, as I argue above, I still don't see how it is possible for any group of order 2^k. [1] This is a direct result of Lagranges theorem. https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)#Applications https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_th...