Computing Quotient and Remainder. Prime Numbers. Factoring by Trial Division. The Fundamental Theorem of Arithmetic

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1 A Crash Course in Elementary Number Theory L. Felipe Martins Department of Mathematics Cleveland State University Work licensed under a Creative Commons License available at Divisibility, Quotient and Remainder Prime Numbers and Factorization Greatest Common Divisor Modular Arithmetic February 13, 2009 The Theorems of Fermat and Euler 1 / 25 2 / 25 Divisibility a and b integers. b divides a if there is an integer q such that a qb Equivalently: b is a divisor of a b is a factor of a a is a multiple of b 23 divides 3266, because does not divide 2146, because (the remainder of 2146 by 23 is not zero). Every integer a divides 0: 0 0 a 0 does not divide any integer, with the exception of 0 itself. Quotient and remainder a and b integers, b 0. The quotient and remainder of a by b are the only integers q and r characterized by: 1. a qb r 2. br 0 and r b Divisor and remainder always have the same sign. This convention is consistent with Sage and Python. Notation: Quotient: a div b Remainder: a mod b : 343 div 51 6 and 343 mod p 12q p 45q p 8q: 532 div p 45q 12 and 532 mod p 45q 8 3 / 25 4 / 25

2 Computing Quotient and Remainder Prime Numbers Greatest integer function: tx u denotes the largest integer that is not above x q a div b ta{bu r a mod b a qb 20 div 7 t20{7u t u 2 and 20 mod div p 7q t20{p 7qu t u 3 and 20 mod p 7q 20 p 3q p 7q 1 We say that the integer a is prime if: 1. a 1 2. The only positive divisors of a are 1 and a itself. 1 is not prime. Only positive integers can be prime, according to our definition. 2,3,5,7,11,13,...,101,...,2 43,112, If a 1 is not prime, it is said to be composite 0 and 1 are neither prime nor composite. 5 / 25 6 / 25 The Fundamental Theorem of Arithmetic Factoring by Trial Division Every positive integer can be written as product of primes, and this prime factorization is unique, except for the order of the factors. We write a prime factorization as: a p t1 1 pt2 2 ptk k k¹ The prime numbers p1, p2,...,pk are distinct and the exponents t1, t2,...,tk are positive i p ti i Example: factor a a p Can stop when the square of latest prime factor is larger than unfactored part ( ), since every composite b has a nontrivial factor less? b 7 / 25 8 / 25

3 Greatest Common Divisor a and b integers, not both 0. gcdpa, bq largest integer that divides both a and b gcdpa, 0q a for a 0. (gcdp0, 0q is undefined.) Euclidean algorithm: 1. r0 a, r1 b. 2. For i 1: ri 1 ri mod ri Stop when rn 0. Then, gcdpa, bq rn 1. Example: gcdp2420, 1650q i ri gcdp2450, 1650q 110 The Extended Euclidean Algorithm a, b, integers, not both zero 1. x0 1, y0 0, r0 a 2. x1 0, y1 1, r1 b 3. For i 1: 3.1 qi ri 1 div ri 3.2 xi 1 xi 1 qixi 3.3 yi 1 yi 1 qy yi 3.4 ri 1 ri 1 qiri 4. Stop when ri 0. (next row) (previous row) qi (current row) The ri are the same sequence of remainders of the Euclidean algorithm. All rows satisfy the relationship: axi byi ri 9 / / 25 An Example Linear Diophantine Equations a 2420, b 1650 i xi yi ri qi Conclusion: gcdp2420, 1650q 110 and 2420 p 2q a, b integers, not both zero. The equation ax by gcdpx, yq always has integer solutions. A solution can be found by the extended euclidean algorithm. The equation ax by c has integers solutions if and only if gcdpa, bq divides c The equation ax by 1 has integer solutions if and only if gcdpa, bq 1. In this case, a, b are said to be relatively prime or coprime. 11 / / 25

4 Definition of Congruence An Example a, b are congruent modulo m if and only if m divides a b Notation: a b pmod mq a b pmod mq if and only if a and b leave the same remainder when divided by m. a 0 pmod mq if and only if m divides a. For a given modulo m, the congruence relation is an equivalence relation. Cogruence is compatible with addition and multiplication: a b pmod mq and c d pmod mq imply a c b d pmod mq and ac bd pmod mq Compute mod 19 Start computing: pmod 19q pmod 19q pmod 19q pmod 19q pmod 19q Division of 2009 by 6 gives pmod 19q 13 / / 25 Linear Congruences Solving Linear Congruences A linear congruence is an equation: ax b pmod mq ax b The linear congruence is equivalent to: pmod mq m ax b, that is, ax b my for some y which is in turn equivalent to: ax my b The linear congruence has solutions if and only if gcdpa, mq divides b. 1. Use the extended euclidean algorithm to solve: au mv g where g gcdpa, mq 2. If g gcdpa, mq b, the equation has the solution: x0 ub{g 3. A maximal set of noncongruent solutions is: tx0, x0 d, x0 2d,..., x0 pg 1qdu where d m{g. The number of noncongruent solutions is g gcdpa, mq 15 / / 25

5 Example Multiplicative Inverses Modulo m 35x 10 pmod 240q a is invertible modulo m if there is a x such that: The extended euclidean algorithm gives gcdp35, 240q 5 and 35 p 41q Multiplying by 10{5 2 we get one solution: x0 2 p 41q pmod 240q The stepsize is 240{5 48, and we get the solutions: {158, , , , u, which reduced modulo 240 give: t158, 206, 14, 62, 110u ax 1 pmod mq, in which case x is said to be an inverse of a modulo m. a is invertible modulo m if and only if gcdpa, mq 1 If it exists, the inverse of a is unique modulo m, that is, if x1 and x2 both satify the equation above, then x1 x2 pmod mq a 1 mod m denotes the inverse x of a such that 0 x m. 17 / / 25 Examples 1. Find the inverse (if it exists) of 65 modulo 321. Solution: The extended euclidean algorithm gives: Congruence Classes The congruence class of a modulo m is the set: 65 p 79q ras tx P Z x a pmod mqu Since gcdp65, 321q 1, the inverse exists, and one inverse of 65 modulo 321 is 79. Thus: 65 1 mod mod Find the inverse (if it exists) of 214 modulo 321. Solution: The extended euclidean algorithm gives: p 160q 107, and since gcdp214, 321q 107 1, 214 is not invertible modulo 321. if m 6: r0s t..., 12, 6, 0, 6, 12,...u r1s t..., 11, 5, 1, 7, 13,...u r4s t..., 18, 2, 4, 10, 16,...u r10s r4s because 10 4 pmod 6q Any element b in ras is said to be a representative of the congruence class ras, and rbs ras. The set of congruence classes modulo m is: Z{mZ tr0s, r1s,..., rm 1su 19 / / 25

6 Operations in Z{mZ Fermat s Theorem ras rbs ra bs and rasrbs rabs Definition is consistent, that is, it does not depend on the representatives chosen for the congruence classes. in Z{6Z, we have: r2s r3s r5s, r2sr3s r6s r0s, r5sr5s r25s r1s, r35sr1000s r 1sr4s r 4s r2s pz{mz,, q is a commutative ring: Addition and multiplication are defined, and have the usual properties : commutative, associative, is distributive with respect to, 1 is the identity of multiplication, etc. Elements are not required to have a multiplicative inverse: ras is invertible if and only if gcdpa, mq 1, and then ras 1 ra 1 mod ms. From now on, the modulus is a prime number p. Every a that is not a multiple of p is invertible modulo p Cancellation law: if ab ac and a is not a multiple of p, then b c pmod pq If p a, the two lists of integers: 1, 2,..., p 1 and 1a, 2a, pp 1qa consist of the same integers, modulo p Thus: 1 2 pp 1q 1a 2a pp 1qa pmod pq and, rearranging terms: 1 2 pp 1q 1 2 pp 1q a p 1 pmod pq. Using cancellation again: a p 1 1 pmod pq. Example 21 / 25 Euler s φ function 22 / 25 Compute mod is prime and 12 is not a multiple of 19. Fermat s theorem gives: pmod 19q Using division: pmod 19q Computing pmod 19q: pmod 19q pmod 19q pmod 19q pmod 19q φpaq number of integers in t1, 2,..., au that are relatively prime to a If the prime factorization of a is p t1 1 pt2 2 ptk k, then: φpaq p t1 1 1 pp1 1qp t2 1 tk 1 2 pp2 1q pk ppk 1q a p1 p2 pk Example: Compute φp44982q φp44982q 2 0 p2 1q 3 2 p3 1q 7 1 p7 1q 17 0 p17 1q / / 25

7 Euler s Theorem Let a and m be integers. If gcdpa, mq 1, then: a φpmq 1 pmod mq Example: compute mod φp44982q and gcdp13, 44982q 1. so, by Euler s thoerem: pmod 44982q pmod 44982q 25 / 25