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Example text

There is an integer k such that c = kd. Since d is a sum of multiples of a and b, we may write am + bn = d. Multiplying this equation by k, we get a(mk) + b(nk) = dk = c so that x = mk and y = nk is a solution. For the “only if” part, suppose x0 and y0 is a solution of the equation. Then ax0 + by0 = c. Since d | a and d | b, then d | c. 2 (Division Theorem). 6) where a is called the dividend, q the quotient, and r the remainder. If b a, then r satisfies the stronger inequalities 0 < r < a. 18 1.

Hence, r = r1 , and also q = q1 . 1. Let a and b be integers with a = 0. We say a divides b, denoted by a | b, if there exists an integer c such that b = ac. When a divides b, we say that a is a divisor (or factor) of b, and b is a multiple of a. If a does not divide b, we write a b. If a | b and 0 < a < b, then a is called a proper divisor of b. The largest divisor d such that d | a and d | b is called the greatest common divisor (gcd) of a and b. The greatest common divisor of a and b is denoted by gcd(a, b).

The algebraic formula for computing P3 (x3 , y3 ) = P1 (x1 , y1 ) + P2 (x2 , y2 ) on E is as follows: (x3 , y3 ) = (λ2 − x1 − x2 , λ(x1 − x3 ) − y1 ), where  3x21 + a    if P1 = P2 2y1 λ= y − y1    2 otherwise. x2 − x1 The idea for fast computing Q = kP over an elliptic curve E is similar to that of computing y = xk over N. For example, to compute Q = 105P , we first let k = 105 = (1101001)2 and then perform the operations as follows starting from e6 to e0 : 1: 1: 0: 1: 0: 0: 1: Q ← P + 2Q Q ← P + 2Q Q ← 2Q Q ← P + 2Q Q ← 2Q Q ← 2Q Q ← P + 2Q ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ Q←P ⇒ c ← P + 2P ⇒ Q ← 2(P + 2P ) ⇒ Q ← P + 2(2(P + 2P )) ⇒ Q ← 2(P + 2(2(P + 2P ))) ⇒ Q ← 2(2(P + 2(2(P + 2P )))) ⇒ Q ← P + 2(2(2(P + 2(2(P + 2P ))))) ⇒ Q=P Q = 3P Q = 6P Q = 13P Q = 26P Q = 52P Q = 105P .

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New Scientist (August 20, 2005)


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