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12.1. FERMAT’S THEOREM 233
Example: Since 13 is prime and doesn’t divide 5, we have (5 12 )%13 = 1.
Similarly, 37 and 101 are primes, and they don’t divide 52 and 133, respectively,
so (52 36 )%37 = 1, and (133 100 )%101 = 1.
⋄
This is a beautiful, deep and powerful theorem. The beauty lies in its simplicity:
Choose any prime number and any other number that is not a multiple
of that prime. Then, the second number, when raised to the prime minus first
power, will be one less than a multiple of the prime. In other words, the remainder
upon division is always 1. No special conditions, no separate cases, just a
1.
OK, there is one condition: that p doesn’t divide the number. But if p
does divide a, then a ≡ 0 (mod p) so a p−1 ≡ 0 (mod p). Thus, we perfectly
understand a p−1 %p for all a’s: If a is a multiple of p then the answer is 0,
otherwise it is 1.
Since 1 ∗ a ≡ a (mod p) we may multiply a p−1 ≡ 1 by a to see that a p ≡ a
(mod p) when a is not a multiple of p. And, since both sides of this equivalence
are 0 when a is a multiple of p we have the following corollary:
Corollary 1 If p is a prime number and a is any integer, then a p ≡ a (mod p).
The depth of this theorem is what it tells us about arithmetic. One might
think that raising integers to powers leads to somewhat arbitrary results. But
Fermat’s Theorem says otherwise: raising to powers modulo a prime number
produces a definite structure. 3
The power of the theorem lies our application of it. Suppose we wish to
compute 33 125 %41. Since 41 does not divide 33, by Fermat’s Theorem we
know that 33 40 ≡ 1 (mod 41). So ( 33 40) 2
≡ 1 2 ≡ 1 (mod 41) and, likewise,
( ) 33
40 3
≡ 1 3 ≡ 1 (mod 41). Because 125 = 3 ∗ 40 + 5, we then have
33 125 = 33 3∗40+5 = 33 3∗40 · 33 5 = ( 33 40) 3
· 33 5 ≡ 1 3 · 33 5 ≡ 33 5 (mod 41).
So 33 125 is the same as 33 5 modulo 41. Because 33 5 %41 = 32 is small enough to
compute on a calculator, we conclude that 33 125 %41 = 32. Fermat’s Theorem
has turned the seemingly impossible 33 125 (mod 41) into a calculation that is
easy.
Example: Compute 20 236 %59.
Since 59 − 1 = 58 and 236 = 4 ∗ 58 + 4, we have 20 236 = 20 4∗58+4 =
20 4∗58 · 20 4 = ( 20 58) 4
· 20 4 ≡ 1 4 · 20 4 ≡ 20 4 ≡ 51 (mod 59). ⋄
3 In fact, it can be said that Fermat’s Little Theorem is the first important result in the
subject of Abstract Algebra, a subject that, along with Analysis, constitutes most of modern
mathematics.