Choosing p , q to rectify mod problems :

We encrypt by calculating message X (mod pq). In practice, we are forced to break X into a series of smaller packets, and encrypt these packets individually. Otherwise X could be an arbitrarily large integer, which becomes difficult to deal with. Our current version uses base-10 numbers, and we are forced to fix the size of the packet to some number of digits. We chose 8 for our packet size. This means that our packets can only represent numbers from 0 to . Therefore, we would like to choose pq to be as close to this number as possible. If 200,000,000, we would encrypt just fine. However, on decryption, we could obtain a packet with value more than 99,999,999. In fact, these “overflowed packets” would become very common, if we assume each packet value to be equally likely. A full one half of all of the packets we receive would fail to be correctly decrypted. This would lead to a 50% packet loss and an incomprehensible message. By carefully choosing p and q, we can eliminate almost all of the packet loss and maintain a better exchange of packets. For example, take pq = (10061)(9941) = 100,016,401. This allows only 16,402 packets in 100,000,000 to be mistranslated, which is 0.016402% packet loss, a great improvement. This is why we have chosen these primes as our default values in the program.

If we were to manipulate packets in binary, our pq must be near a power of two, which is much simpler to find. Primes of the form and are well known. Naturally, a product of primes of these forms together would be reasonably close to a power of two. The problem with choosing primes in this manner is that, given a public key, finding such formulaic primes to break the encryption would be much simpler than finding other primes.