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12.9. RSA AND PUBLIC KEYS 245
There are at least two interesting things about these examples. First, because
of the choice of small numbers for P and Q in Example (1), the RSA
cipher became a complicated mono-graphic cipher. Frequency analysis of the
type we practiced in Chapter 6 would easily break this cipher. However, by
choosing P and Q so that N was larger than 2626 in Examples (2) and (3) we
used RSA as a digraphic cipher. And if we had chosen P = 487, Q = 541 then
N = 263467 would have let us use RSA as trigraphic cipher.
The second thing was how encryption and decryption, while very similar
processes, involve very different amounts of knowledge. To encipher we only
need to know e and N, for the only thing we must do is to compute M e %N. In
particular, we do not need either P or Q. To decipher, however, we do need to
know P and Q, because to find d we must use (P −1)(Q−1). The application of
this differential knowledge is what allows RSA to be public key cipher system.
12.9 RSA and Public Keys
The information needed to use any particular RSA cipher is very different for
the encryptor than it is for the decryptor. To encipher a message one only
needs the power e and the modulus N. The values of d and (P − 1)(Q − 1)
are unnecessary. When deciphering we need N, but also need d, and need e
and (P − 1)(Q − 1) to determine d. That is, the decipherer needs P and Q to
determine d.
This allows RSA to be used as a public key code. Alice chooses the two
primes P and Q and computes their product N. Then she chooses e and uses
P and Q to compute d. Alice then makes public the values N and e. Since
all that is needed to encipher a message is e and N, anyone can send Alice a
message using her system.
Alice Anderson
Phone: 1-800-CALL-ALC
Email: alice a○mymail.com
I use RSA. My public keys are
e = 17 and N = 549992441.
Conversely, Alice keeps P , Q, (P − 1)(Q − 1) and d all secret. Since she
knows d she can decipher any message sent to her. 14
12.10 How to break RSA
Suppose we capture an enciphered message E that is intended for our enemy.
How can we read the message
14 Since to decipher she only need to raise to the d-th power modulo N, she should throw P ,
Q and (P − 1)(Q − 1) away, erase them from any computers they are on and burn any papers
they are written on.