The Euclidean algorithm for integers leads to the notion of congruence of two integers modulo a given integer.

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1 Integers Modulo m The Euclidean algorithm for integers leads to the notion of congruence of two integers modulo a given integer. Congruence Modulo m Two integers a and b are congruent modulo m if and only if a b is a multiple of m, in which case we write a b mod m. Thus, mod 9, because is a multiple of 9. Given integers a and m, themod function is given by amod m b if and only if a b mod m and 0 b m 1; hence, amod m is the smallest nonnegative residue of a modulo m. The underlying computer algebra system does not understand the congruence notation a b mod m, but it does understand the function notation amod m. This section shows how to translate problems in algebra and number theory into language that will be handled correctly by the computational engine. Note that mod is a function of two variables, with the function written between the two variables. This usage is similar to the common usage of, which is also a function of two variables with the function values expressed as a b, rather than the usual functional notation a, b. Traditionally the congruence notation a b mod m is written with the mod m enclosed inside parentheses since the mod m clarifies the expression a b. In this context, the expression b mod m never appears without the preceding a. On the other hand, the mod function is usually written in the form amod m without parentheses. To evaluate the mod function 1. Leave the insertion point in the expression amod b. 2. Choose Evaluate. 23mod14 9 If a is positive, you can also find the smallest nonnegative residue of a modulo m by applying a Expand to the quotient m. Expand Since , multiplication of by 14 shows that 23mod In terms of the floor function x, the mod function is given by amod m a m a m Multiplication Tables Modulo m You can make tables that display the products modulo m of pairs of integers from the set 0, 1, 2,,m 1. To get a multiplication table modulo m with m 6 1. Define the function g i, j i 1 j From the Matrices submenu, choose Fill Matrix. 3. Select Defined by Function. 4. Enter g in the Enter Function Name box. 5. Select 6 rows and 6 columns. 6. Choose OK. 7. Type mod 6 at the right of the matrix. (Because the insertion point is in mathematics mode; mod automatically turns gray.) 8. Choose Evaluate.

2 mod A more efficient way to generate the same multiplication table is to define g i, j i 1 j 1 mod 6 and follow steps 2-6 above. You can also find this matrix as the product of a column matrix with a row matrix. Evaluate mod Make a copy of this last matrix. From the Edit menu, choose Insert Row(s) and add a new row at the top (position 1); choose Insert Column(s).and add a new column at the left (position 1); fill in the blanks and change the new row and column to Bold font, to get the following multiplication table modulo 6: From the table, we see that 2 4mod6 2 and 3 3mod6 3. A clever approach, that creates this table in essentially one step, is to define g i, j i 2 j 2 mod 6 Choose Fill Matrix from the Matrices submenu, choose Defined by Function from the dialog box, specify g for the function, and set the matrix size to 7 rows and 7 columns. Then replace the digit 1 in the upper left corner by and change the first row and column to Bold font, as before. You can generate an addition table by defining g i, j i j 2mod6. Example If p is a prime, then the integers modulo p form a field, called a Galois field and denoted GF p.fortheprimep 7, you can generate the multiplication table by defining g i, j i 1 j 1 mod 7 and choosing Fill Matrix from the Matrix submenu, then

3 selecting Defined by function from the dialog box. You can generate the addition table in a similar manner using the function f i, j i j 2mod Inverses Modulo m If abmod m 1, then b is called an inverse of a modulo m, and we write a 1 mod m for the least positive residue of b. The computation engine also recognizes both of the forms 1/amod m and 1 a mod m for the inverse modulo m. Note that a 1 mod m exists if and only if a is relatively prime to m; that is, it exists if and only if gcd a, m 1. Thus, modulo 6, only 1 and 5 have inverses. Modulo any prime, every nonzero residue has an inverse. In terms of the multiplication table modulo m, the integer a has an inverse modulo m if and only if 1 appears in row amod m (and 1 appears in column amod m). To compute the inverse of a mod m if gcd a,m 1 1. Enter the inverse in one of the forms a 1 mod m, 1/amod m, or 1 a mod m. 2. Place the insertion point in the expression and choose Evaluate mod 7 3 mod 7 3 1/5mod7 3 5 This calculation satisfies the definition of inverse, because 5 3mod mod mod 6 5 The notations ab 1 mod m, a/bmod m, and a mod m are all interpreted as b a b 1 mod m mod m; that is, find the inverse of b modulo m, multiply the result by a, and then reduce the product modulo m. 3/23mod mod 6 4 Solving Congruences Modulo m To solve a congruence of the form ax b mod m Multiply both sides by a 1 mod m to get x b/amod m. The congruence 17x 23 mod127 has a solution x 91, as the following two evaluations illustrate. 23/17mod Check this result by substitution back into the original congruence.

4 17 91 mod Note that, since 91 is a solution to the congruence 17x 23 mod127, additional solutions are given by n, where n is any integer. In fact, x 91 mod127 is just another way of writing x n for some integer n. Pairs of Linear Congruences Since linear congruences of the form ax b mod m can be reduced to simple congruences of the form x c mod m, we consider systems of congruences in this latter form. To solve a pair of linear congruences x c mod m and x d mod n 1. Check that gcd m, n 1 so that a solution exists. 2. Rewrite the congruences as equations x km c, x rn d, whence km c rn d. 3. Rewrite this equation as the congruence km d c mod n and divide both sides by m to solve for k. 4. Place the insertion point in the congruence k d c /m mod n and choose Evaluate. 5. Using the computed value for k, place the insertion point in the equation x km c and choose Evaluate. 6. The complete set of solutions are the solutions of x km c mod mn, with k d c /m mod n. Example Consider the system of two congruences x 45 mod237 x 19 mod419 Checking, gcd 237, 419 1, so237 and 419 are relatively prime. The first congruence canberewrittenintheformx k for some integer k. Substituting this value into the second congruence, we see that k r for some integer r. This last equation can be rewritten in the form 237k 19 45mod419, which has the solution k /237mod Hence, x Checking, 14265mod and 14265mod Example The complete set of solutions is given by x s mod99303 Thus, the original pair of congruences has been reduced to a single congruence, x mod99303 In general, if m and n are relatively prime, then one solution to the pair x a mod m x b mod n is given by x a m b a /m mod n A complete set of solutions is given by x a m b a /m mod n rmn where r is an arbitrary integer. Systems of Linear Congruences You can reduce systems of any number of congruences to a single congruence by solving systems of congruences two at a time. The Chinese remainder theorem states that, if the moduli are

5 relatively prime in pairs, then there is a unique solution modulo the product of all the moduli. To solve a system of linear congruences x c i mod m i, i 1,2,...,t 1. Check that gcd m i, m j 1 for every i j so that a solution exists. 2. Solve the congruences one pair at a time to obtain a complete solution. Example Consider the system of three linear congruences x 45 mod237 x 19 mod419 x 57 mod523 Checking, gcd , and gcd 237, 419 1; hence this system has a solution. The first two congruences can be replaced by the single congruence x mod99303 ; hence the three congruences can be replaced by the pair x mod99303 x 57 mod523 As before, k r for some integers k and r. Thus, k /99303mod ; hence x This system of three congruences can thus be reduced to the single congruence x mod Extended Precision Arithmetic Computer algebra systems support exact sums and products of integers that are hundreds of digits long. To do extended precision arithmetic 1. Generate a set of mutually relatively prime bases, and do modular arithmetic modulo all of these bases. 2. Solve the resulting system of linear congruences. For example, consider the vector 997, 999, 1000, 1001, 1003, 1007, 1009 of bases. Factorization shows that the entries are pairwise relatively prime. Factor Consider the two numbers and You can represent these numbers by reducing the numbers modulo each of the bases. Thus,

6 mod mod mod mod mod mod mod mod mod mod mod mod mod mod Thus, the product is represented by the vector mod mod mod mod mod mod mod The product is now a solution to the system x 739 mod997 x 909 mod999 x 864 mod1000 x 652 mod1001 x 671 mod1003 x 207 mod1007 x 1003 mod1009 Powers Modulo m To calculate large powers modulo m Place the insertion point in an expression of the form a n mod m and choose Evaluate. Example Define a , n , and m Applying the command Evaluate to a n mod m yields the following: a n mod m Fermat s Little Theorem states that, if p is prime and 0 a p, then a p 1 mod p 1 The integer 1009 is prime, and the following is no surprise.

7 Evaluate mod Generating Large Primes There is not a built-in function to generate large primes, but the underlying computational system does have such a function. The following is an example of how to define functions that correspond to existing functions in the underlying computational system. (Click here for a general discussion of how to access such functions.) In this example, p x is defined as the Scientific WorkPlace (Notebook) Name for the MuPAD function, nextprime(x), which generates the first prime greater than or equal to x. To define p x as the next-prime function 1. From the Definitions submenu, choose Define MuPAD Name. 2. Enter nextprime(x) as the MuPAD Name. 3. Enter p x as the Scientific Notebook (WorkPlace) Name. 4. Under The MuPAD Name is a Procedure, check That is Built In to MuPAD or is Automatically Loaded. 5. Choose OK. Test the function using Evaluate. p 5 5 p p p Example The Rivest-Shamir-Adleman (RSA) cipher system is based directly on Euler s theorem and requires a pair of large primes. First, generate a pair of large primes say, q p and r p (In practice, larger primes are used; such as, q and r ) Then n qr and the number of positive integers n and relatively prime to n is given by n q 1 r Let x be plaintext (suitably generated by a short section of English text). Long messages must be broken up into small enough chunks that each plaintext integer x is smaller than the modulus n. Choose E to be a moderately large positive integer that is relatively prime to n, for example, E The ciphertext is given by y x E mod n Let D mod n Then friendly colleagues can recover the plaintext by calculating

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