> Now I'm wondering if Fermat's Last Theorem has a bitwise analogue. It hasn’t in this sense. The ^n operator flips the ‘one’ bits in n in its argument, so for

> Now I'm wondering if Fermat's Last Theorem has a bitwise analogue. It hasn’t in this sense. The ^n operator flips the ‘one’ bits in n in its argument, so for any n, we have a = a^n^n So, for all a and b, a^n + b^n = ((a^n + b^n)^n)^n that makes c = (a^n + b^n)^n a solution to a^n + b^n = c^n and that means we can find a c for every n, a and b.