MACM 101 Lecture 28: Lecture 28 Part 2_ Chinese Remainder Theorem

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browncricket987

21 Dec 2018

School

Simon Fraser University

Department

Math & Comp Sci

Course

MACM 101

Professor

Andrei Bulatov

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Therefore has the inverse modulo mi, that is yi such that. Let us set x = a1m1y1 + a2m2y2++akmkyk. Note that whenever i j, all terms except for the ith term in this sum are congruent to 0 modulo mi. As miyi 1(mod mi), we have x aimiyi ai(mod mi) Sun tzu"s puzzle x 2 (mod 3) x 3 (mod 5) x 2 (mod 7) a1 = 2 a2 = 3 a3 = 2 m1 = 3 m2 = 5 m3 = 7. 1(mod 7) y1 2-1 2(mod 3) y2 1-1 1(mod 5) y3 1-1 1(mod 7) x = a1m1y1 + a2m2y2 + a3m3y3. For any n > 2, the equation xn + yn = zn does not have integer solutions x,y,z > 0. It had remained unproven for 358 years (posed in 1637, proved in 1995). If p is prime and a is an integer not divisible by p, then ap-1 1(mod p)

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MACM 101 Lecture 28: Lecture 28 Part 2_ Chinese Remainder Theorem

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