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234 CHAPTER 12. RSA
In both of these examples the quotient played an insignificant role. What,
then, of the old exponent remains The remainder. Both 125 and 236 were
replaced by their remainder when divided by p − 1. But “replacing by the
remainder” is just another way of saying that we are doing modular arithmetic!
We can summarize this as the following:
Theorem 4 Fermat’s Theorem Restated: If p is a prime number and p does
not divide a, then a b ≡ a b′ (mod p), where b%(p − 1) = b ′ .
Less formally, when doing powers modulo p, we may work on the exponent
modulo p − 1.
Examples:
(1) Compute 6 191 %19.
Since 19 doesn’t divide 6 we can use Fermat’s theorem. Since 191%18 =
11, we know that 6 191 ≡ 6 11 (mod 19). This last is easily computed to be
17, which is our answer.
(2) Compute 12 360 (mod 17).
17 doesn’t divide 12, and 360%16 = 8, so 12 360 ≡ 12 8 ≡ 15 (mod 17). So
the answer is 15.
(3) Compute 23 465 (mod 43).
23 465 ≡ 23 3 ≡ 41 (mod 43). The answer is 41.
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄
12.2 Complication I: a small one
The examples we’ve done were carefully chosen so that we ended up with a fairly
small number raised to a fairly small number. What if the base was too large for
a calculator to handle this computation For example, what if we wanted 233 125
(mod 41) By Fermat’s Theorem this is the same as 233 5 (mod 41). But 233 5
is still too large for most calculators. What can we do Easy: we are doing
modular arithmetic, and 233%41 = 28. We then have the string of equivalences
233 125 ≡ 233 5 ≡ 28 5 ≡ 3 (mod 41).
We are now doing “double modular arithmetic”, modulo p on the base and
modulo p − 1 on the exponent.