P(S(M)) =
(*M*^{d} *mod n*)^{e} mod n =

S(P(M)) =
(*M*^{e} *mod n*)^{d} mod n =

Since d = *e*^{-1} mod [(p-1)(q-1)]

Then ed = 1 + k(p-1)(q-1)

If M 0 (mod p), then

If M
0 (mod p), then

Fermat's Theorem

.

Similarly for q, thus

p and q are prime, n = pq.

Thus, by the Corollary to the Chinese Remainder Theorem, .

If the adversary can factor n into p and q, then the code is broken, but this is
hard.