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12.3. COMPLICATION II: A SUBSTANTIAL ONE 237
(2) For 57:
So 57 = 32 + 16 + 8 = +1.
remainder power of 2
57 ÷2 = 28.5 1 1
28 ÷2 = 14 0 2
14 ÷2 = 7 0 4
7 ÷2 = 3.5 1 8
3 ÷2 = 1.5 1 16
1 ÷2 = .5 1 32
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In fact, every positive integer can be written as a sum of the powers of 2: 1,
2, 4, 8, ..., using each power at most once. This is the binary expansion of
the number, and the process we have just described finds this expansion.
We can use the two processes, squaring using the powers of two, and the
binary expansion process, together.
Example: Find the binary expansion of 159. Use it to compute 23 159 %171.
159 ÷ 2 = 79.5 1 1
79 ÷ 2 = 39.54 1 2
39 ÷ 2 = 19.5 1 4
19 ÷ 2 = 9.5 1 8
9 ÷ 2 = 4.5 1 16
4 ÷ 2 = 2 0 32
2 ÷ 2 = 1 0 64
1 ÷ 2 = .5 1 128
Thus 159 = 128 + 16 + 8 + 4 + 2 + 1. Next, compute the required powers of 23:
23 ≡ 23 (mod 171),
23 2 = 23 2 ≡ 16 (mod 171),
23 4 = (23 2 ) 2 ≡ (16) 2 ≡ 85 (mod 171),
23 8 = (23 4 ) 2 ≡ (85) 2 ≡ 43 (mod 171),
23 16 = (23 8 ) 2 ≡ (43) 2 ≡ 139 (mod 171),
23 32 = (23 16 ) 2 ≡ (139) 2 ≡ 169 (mod 171),
23 64 = (23 32 ) 2 ≡ (169) 2 ≡ 4 (mod 171), and
23 128 = (23 64 ) 2 ≡ (4) 2 ≡ 16 (mod 171).
Using the powers that we need gives
23 159 = 23 128 · 23 16 · 23 8 · 23 4 · 23 2 · 23 ≡ 16 · 43 · 85 · 16 · 23 ≡ 163 (mod 171).
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