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12.5. COMPLICATION IV: THE LAST ONE 239
large to multiply together all at once, simply group them into a families of two
or three, multiply and reduce them, and then multiply and reduce the results.
Example: Compute 372 · 361 · 19 · 281 · 107 · 81 · 239 · 301 (mod 401)
372 · 361 · 19 · 281 · 107 · 81 · 239 · 301 (mod 401)
≡ (372 · 361 · 19) · (281 · 107 · 81) · (239 · 301) (mod 401)
≡ 386 · 154 · 160 (mod 401)
≡ 122 (mod 401).
⋄
12.5 Complication IV: the last one
The reason we are interested in computations like 133 172 %323 in the first place
is that they form the heart of the most popular public key cryptography system
in use today. The final complication is that the computations in this system
are not of the form a b %p, where p is prime, but like a b %pq, where p and q are
different primes. How does the change from a prime as the modulus to two
prime affect what we’ve done so far
We know that 12 6 ≡ 1 (mod 7) and 12 10 ≡ 1 (mod 11). How should we
complete 12 □ ≡ 1 (mod 77) It turns out this is equivalent 6 to finding a value
k so that 12 k ≡ 1 (mod 7) and 12 k ≡ 1 (mod 11). From Fermat’s theorem
12 6 ≡ 1 (mod 7) and 12 10 ≡ 1 (mod 11). In fact, any exponent that is a
multiple of 6 or 10, respectively, produces the same result: 12 6n ≡ ( 12 6) n
≡
1 n ≡ 1 (mod 7) and 12 10m ≡ ( 12 10) m
≡ 1 m ≡ 1 (mod 11). The simplest such
multiple 7 is the product, (7 − 1)(11 − 1) = 60 in this case. So 12 60 ≡ 1 (mod 7)
and 12 60 ≡ 1 (mod 11), hence 12 60 ≡ 1 (mod 77).
Now there is nothing special about 7 and 11. If p and q are distinct prime
numbers, neither of which divides a, then a k ≡ 1 (mod p) and a k ≡ 1 (mod q)
whenever k is a multiple of both p − 1 and q − 1. In particular, this is true when
k = (p − 1)(q − 1). This gives us
6 From Theorem 1 way back in Chapter 3, a ≡ b (mod pq) is the same as pq dividing a − b.
But when p and q are relatively prime, this needs both p and q to divide a − b. That is, a ≡ b
modulo both p and q.
7 The smallest such multiple is the least common multiple, which we recall from Section
8.4. For simplicity we will use the product.