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# Class Note for CSCI 162 with Professor Vora at GW (2)

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This 2 page Class Notes was uploaded by an elite notetaker on Saturday February 7, 2015. The Class Notes belongs to a course at George Washington University taught by a professor in Fall. Since its upload, it has received 21 views.

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Date Created: 02/07/15
CSCI 162 and CSCI 284NoIaGWU l CSCI 162284 Cryptography GCD and Multiplicative Inverses mod m De nition The greatest common divisor of two positive integers m and n is the largest positive integer that divides both m and n It is denoted m7 n or gcdm7 Examples 69 3 1236 12 59 1 De nition m and n are said to be relatively prime if m7 n 1 The following theorem characterizes all the elements with multiplicative inverses in Zm Essentially exactly those elements a such that gcdm7 a 1 are invertible There are two parts to showing this The rst is to show that if the element a has an inverse mod m then gcda7 m 1 The second is the other direction to show that if gcda7 m 1 then a 1 exists The rst is easier than the second The proof sketch of the second part is as follows The formal proof is provided later Suppose gcdm7 a 1 Consider the collection of all integers that are combinations of m and a that is are of the form Sm Ta Consider the smallest positive integer in this collection call it 9 Examine the remainders when the numbers in the collection are divided by y We can see that the remainders are also in the collection However because 9 is the smallest positive integer in the collection and the remainders are smaller than 9 and nonnegative they are all zeroHence g divides all numbers in the collection In particular it divides m and a which also belong to the collection m corresponds to S 1 and T 0 and a to S 0 and T 1 Hence 9 is a common factor ofa and m However as gcda7 m 1 g 1 is the only possible positive common factor Hence 9 1 is in the collection and can be expressed in the form 5m at that is 5m at 1 Looking at this equation mod m we get that at E 1 mod m and that a 1 mod m exists Theorem a has a multiplicative inverse mod m denoted a 1 mod m ltgt gcdm7 a 1 Proof gt Suppose a 1 exists Then there exists integer t such that at E 1 mod m That is there exist integers 57 such that at 5m 1 A common factor of a and m would divide both terms on the left hand side of the equation and hence would also divide the right hand side But as the right hand side is 1 the only positive integer divisor of it can be 1 hence the only positive common factor ofa andm is 1 Hence gcdam 1 4 Suppose gcdm7 n 1 Consider all integers of the form Sm Ta for integers S and T Let g Som Tea be the smallest such positive integer We show that g 1 and hence that St T073 SO such that 5m ta g 1 Looking at this equation mod m we see that it implies that at E 1 mod m that is that CSCI 162 and CSCI 284NoIaGWU 2 a 1 mod Tn exists In fact it a 1 mod Tn We proceed as follows Consider any arbitrary integer z Sm Ta Let T 1 Tem g be the remainder when I is divided by g It is the unique nonnegative integer smaller than 9 such that z 19 TThen 7 qu9 qEZ SmTaiqg SiqSOm TiqT0a and T is also a combination of Tn and a However 9 is the smallest positive integer of that form and T is smaller than 9 Hence T0 ngmTa7 VST glmgln 51T0S0Tl But gcdam 1 hence g 1 Hence there exists S SO and T To such that Som Tea 1 That is Toa E 1 mod Tn and a 1 mod Tn exists in fact its value is To B

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