# Recitation 24

Today we discussed that the Chinese remainder theorem is not applicable unless all n_is are pairwise relatively prime, in which case we need to break each single linear congruence equation that causes the problem into equivalent equations.

For instance, suppose we want to solve x\equiv 3 \mod 6, x\equiv 4 \mod 7, x\equiv 5 \mod 8 for x. Then we could replace x\equiv 3 \mod 6 by x\equiv 3 \mod 2 and x\equiv 3 \mod 3. Also notice that x\equiv 5 \mod 8 implies x\equiv 3 \mod 2. So it is enough to solve x\equiv 3 \mod 3, x\equiv 4 \mod 7, x\equiv 5 \mod 8 , in which case we could apply the Chinese remainder theorem.

Also we covered an elegant proof of Fermat’s little theorem. In that proof, we considered two reduced residue systems, 1, 2, \ldots, p-1 and a, 2a, \ldots, (p-1)a. Since they are both reduced residue systems, the products should be the same in modulo p arithmetic. Hence (p-1)!\equiv (p-1)!a^{p-1} \mod p.

One can also use this idea to prove Euler’s theorem, which says the following.

If n and a are coprime positive integers, then a^{\phi(n)}\equiv 1 \mod n, where \phi(n) is Euler’s totient function.