Cryptology - Unofficial St. Mary's College of California Web Site

232 CHAPTER 12. RSA
At this point in discussing a new cipher there are always two questions we
ask: “How to decipher” and, “What is the security” Answering the first, if
you think about it for a bit, will be difficult. To undo cubing we must take the
cube root, But what we want to take the cube root of is not the ciphernumber
but the cube of the plainnumber. So we have the problem of trying to determine
the cubed plainnumber from the ciphernumber.
As for the security, just as always adding 3 or always multiplying by 3 is
insecure, always raising to the 3rd power is insecure. We should choose as our
key some integer to take powers with. And large integers would have to be
a possibility if we are to prevent the enemy from simply trying each possible
power. But this leads to a new problem: what letter does the power 33 encipher
S to
19 33 = 1580770532156861979997149793605296459437459
consists of 42 digits, and so is far too big for a calculator to handle. Similarly,
19 125 has 159 digits, and 125 is not a very large number. Or how can we
determine the remainder when 26 divides 19 392 , a number of over 500 digits!
There is a third, more devious, problem. 1 3 = 1 so a is enciphered to A,
but 3 3 = 27 ≡ 1 (mod 26) so c is also enciphered to A. Similarly, d and f both
become L. Why is this, and how can we prevent this
Despite these difficulties, the idea of developing a cryptosystem around the
raising of numbers to powers is a very good one. It will lead us to one of the
most popular cryptosystem of all time. But before we get there we need to
surmount the difficulties we’ve just discussed.
12.1 Fermat’s Theorem
Our first need is to learn how to compute values like 133 719 %101. As you know,
the “mod” just is shorthand for “find the remainder when divided by”, so in
some sense we are not learning anything new. All one has to do is to compute
133 719 , divide by 101, and find the remainder. However, 133 719 is galactically
far too large for you calculator to directly compute. 1 How then to do this
Some help is provided by a 360 year-old theorem.
Theorem 3 Fermat’s Little Theorem 2 (1640). If p is a prime number and
p does not divide a, then a p−1 ≡ 1 (mod p).
1 It has over 1500 digits in it. For comparison, the traditional estimate for the number of
elementary particles in the universe is 10 80 , i.e., a number of 80 digits.
2 Pierre (de) Fermat, 1601-1665, was one of the last truly great amateur mathematicians.
During the day he was a lawyer/councilor for his local parliament. At night he read famous
math books and added his thoughts and comments to the margins. It took over 350 years
before one of these notes, Fermat’s Last Theorem, was determined to actually be true. Another,
Fermat’s Little Theorem (to differentiate it from the Last Theorem) is what interests
us here.