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The proposition makes it easier to compute ϕ(n). For example, ϕ(12) = ϕ(22 ) · ϕ(3) = 2 · 2 = 4. Also, for n ≥ 1, we have ϕ(pn ) = pn − pn = pn − pn−1 = pn−1 (p − 1), p since ϕ(pn ) is the number of numbers less than pn minus the number of those that are divisible by p. , ϕ(389 · 112 ) = 388 · (112 − 11) = 388 · 110 = 42680. 1. 30 3. 5 Quickly Computing Inverses and Huge Powers This section is about how to solve ax ≡ 1 (mod n) when we know it has a solution, and how to efficiently compute am (mod n).

It’s easy to give an inefficient algorithm that solves the discrete log problem. , until we find an exponent n such that bn = a. For example, suppose a = 18, b = 5, and p = 23. We have b1 = 5, b2 = 2, b3 = 10, . . , b12 = 18, so n = 12. When p is large, computing the discrete log this way soon becomes impractical, because doubling the number of digits of the modulus makes the computation take much longer. 2 Realistic Diffie-Hellman Example In this section we present an example that uses bigger numbers.

Because R is a complete set of residues, this implies that x = x . Thus the elements of aR have distinct reductions modulo n. It follows, since #aR = n, that aR is a complete set of residues modulo n. 9. If gcd(a, n) = 1, then the equation ax ≡ b (mod n) has a solution, and the solution is unique modulo n. Proof. Let R be a complete set of residues modulo n, so there is a unique element of R that is congruent to b modulo n. 8, aR is also a complete set of residues modulo n, so there is a unique element ax ∈ aR that is congruent to b modulo n, and we have ax ≡ b (mod n).

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