Systems of Linear Equations

Systems of Linear Equations • Solve the system 4x + 7y = 9, 5x + 2y = 3 in Z13 • Use familiar techniques, realizing that division by k mean multiplying by k-1 mod 13 • 4-1 mod 13 = 10 and 5-1 mod 13 = 8, so multiply first equation by 10 and the second by 8 and reduce modulo 13 to get 40x + 70y = 90 mod 13  x + 5y = 12 40x + 16y = 24 mod 13  x + 3y = 11 • Now subtract second equation from the first to get 2y = 1 mod 13 • Since 2-1 = 7 mod 13, we get y = 7 and then x + 21 = 11 mod 13,x = -10 = 3 mod 13 • Thus we have solved the system of equations to get x = 3, y = 7

Matrix Version of Simultaneous Linear Equations • If we extract the coefficients and right-hand constants, we obtain a matrix version of the equations called an augmented matrix • From our previous example: • We can use elementary row operations (adding a multiple of one row to another) to reduce the first two columns to the 2 by 2 identity matrix • From this it follows that x = 3 and y = 7 • Note that all arithmetic here is done mod 13

Solving in the Rationals • The function from the rationals to Zn given by f(a/b) = ab-1 preserves addition and multiplication. It is only defined if b has an inverse in Zn • Thus, we may solve a system of linear equations in the rationals and then apply the function f. • Example: the system 4x+5y = 9, 5x+2y=3 has rational solution x = -3/17, y = 33/17. • Therefore in Z13, where all nonzero elements have inverses, x 13 -3/17 13 -3/4 13 10/4 13 104-113 1010 = 100 13 9 y 13 33/17 13 7/4 13 74-113 710 = 70 13 5

Solving ax = b in Zn • We have already seen that we can solve ax n b if a is relatively prime to n : just multiply both sides by a-1 • What if a is not relatively prime to n? Say d = gcd(a,n) > 1. • Answer: • If d does not divide b, then there is no solution • Assume d | b. Consider the congruence (a/d)x n/d b/dSince a/d, n/d and b/d are integers and a/d  (n/d), there is a unique solution x0 to this equation. • The solutions to the original congruence ax n b are x0, x0 + 2(n/d), . . . , x0 + (d-1)(n/d) (mod n) • Example: solve 12x = 21 (mod 39) gcd(12,39) = 3 which divides 21, so new congruence is 4x 13 7The solution to the new congruence is x0 = 5, so the solutions of 12x 39 21 are 5, 18, 31 in Z39

Congruences with Different Moduli • It is often necessary to solve two or more congruences with different moduli • Example: x  3 (mod 5), x  5 (mod 7) • This means that x = 3 + 5r and x = 5 + 7s for some integers r,s • Thus x is an element of { …, -12, -7, -2, 3, 8, 13, 18, 23, 28, 33, …}  { …, -16, -9, -2, 5, 12, 19, 26, 33, …} = { …, -37, -2, 33, 68, … } • Notice that the elements in the intersection differ by multiples of 35 = 57, the product of our two moduli • This an example of a general phenomenon

Chinese Remainder Theorem • Lemma If n  m, n | c and m | c, then mn | c • TheoremSuppose gcd(m,n) = 1. For any integers a,b there is exactly one solution mod mn of the congruences x m a, x n b. • Proof • By the generalized euclidean algorithm, there are integers s,t such that gcd(m,n) = 1 = ms + nt. • Thus ms n 1 and nt m 1. • Let x = bms + ant (mod mn). • Then x m ant m a, since nt m 1. • Similarly, x n bms n b, so x is a solution to the two congruences. • Suppose x1 were another solution. Then x m x1 and x n x1. • Thus x-x1 is a multiple of both m and n, hence a multiple of mn (Lemma) • This shows that any two solutions are congruent mod mn

Chinese Remainder Theorem • The proof of the Chinese Remainder Theorem provides an algorithm for finding a solution. • Equivalently, one could proceed as follows: compute the inverse of m mod n, call it z. Then set x = a + mz(b-a). • Chinese Remainder Theorem (General Form) Let m1, …, mr be a pairwise relatively prime set of positive integers and let a1, … , ar be arbitrary integers. Then the systemx m1 a1, x m2 a2, . . . , x mr ar has a unique solution mod m1m2   mr • The proof is a simple induction on r, using the theorem on the previous slide. • CorollaryLet m1, …, mr be a pairwise relatively prime set of positive integers and let a be an arbitrary integer. If x m1 a, x m2 a, . . . and x mr a, then x m1m2…mr a.

Modular Exponentiation • We will have occasion to want to compute xa (mod n) for integers x, a, and n • Example: compute 21234 (mod 789) • Computing 21234 and then reducing mod 789 is not practical, since we would have a number with too many digits before reduction. • Similarly, computing 22 then reducing, then multiplying by 2 and reducing, etc would be much too slow. • There is a better approach: repeatedly square and reduce

Modular Exponentiation • There is a better approach to computing powers mod n than repeated multiplications followed by reducing mod n: • Repeatedly square and reduce • Example: compute 21234 (mod 789) • Compute 22 = 4, then 24 = 42 = 16, 28 = 162 = 256 • 216 = 2562 789 49, 232 789 492 789 34, 264 789 342 789 367 • 2128 789 3672 789 559, 2256 789 5592 789 37, • 2512 789 372 789 580, 21024 789 5802 789 286. • Now 21234 = 21024 + 128 + 64 + 16 + 2 = 21024 2128 264 216 22 789 286(559367)(494) 789 481 • Note that we never had to compute a number bigger than 7882

Modular Exponentiation • The previous method for computing xa (mod n) requires at most log2n multiplications and never generates an intermediate value greater than n2 • The algorithm for computing exponents mod n is shown on the next slide

Modular Exponentiation • For computation of exponents via a computer program, the Modular Exponentiation Algorithm can be used: • pow(x,e,n) { // returns xe mod n prod = 1 while (e > 1) { if ( e mod 2 == 1) prod = prod*x x = x2 mod n e = e/2 } return prod*x mod n

Homework • Page 130: 2, 6 • Page 135-136: : 2, 4, 8 • Compute 4126 mod 127; you may use either the squaring method or by using the Modular Exponentiation algorithm

Systems of Linear Equations