Ok, so I understand the importance of using the product of two large primes, but why not use the product of three large primes instead?
Would this make the encryption weaker in some way?
If the answer is not a simple one, then I would appreciate a technical answer if possible.
Say you have a 1024-bit key. With two primes, each is about 512 bits; with three primes, each is about 341 bits. Current factorization methods require exponential time, so each additional bit requires about double the time to find the factor. Thus, two primes are substantially stronger than three primes.
To be specific, a 768-bit key has been factored, and reported in the mathematical literature, which means that a 341-bit factor can be found (the 768-bit factorization took about 2000 PC-years, so it's not easy, but possible). No one has yet reported factorization of a 1024-bit key (although there are doubtless people working on it).
The answer is simple, really. For numbers of a given size (e.g. 1024 bits) the toughest factoring problem is when the number factors into two primes (assuming that they are not too close to the square root of the overall number). Intuitively, it is easier to fish in oceans that contain more fish. It is easier to find one of three primes than one of two.
The public and the private key-generation algorithm is the most complex part of RSA cryptography. Two large prime numbers, p and q, are generated using the Rabin-Miller primality test algorithm. A modulus n is calculated by multiplying p and q. This number is used by both the public and private keys and provides the link between them.
Between sender and receiver you need 2 keys public and private. But for that you can use any number of primes but generally 2 is used.
