No, not really. The most complex math AES uses is operations in the finite field GF(256). Given that addition in a power-of-2 finite field is simply XOR, I expe

No, not really. The most complex math AES uses is operations in the finite field GF(256). Given that addition in a power-of-2 finite field is simply XOR, I expect any good programmer to be able to get AES.