3 ms·
Excellent. Now I'm wondering if Fermat's Last Theorem has a bitwise analogue.
by Y_Y 4y ago
Excellent. Now I'm wondering if Fermat's Last Theorem has a bitwise analogue.
- kps 4y agoBitwise multiplication is bitwise ‘and’, so exponentiation (repeated multiplication) is the identity, so aⁿ+bⁿ=cⁿ iff a+b=c. I have a truly remarkable proof of this, but my filesystem is full.
- Someone 4y ago> 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.
- Y_Y 4y agoNeat proof!