Number theory lectures

Transcription

1 Number theory lectures By Dr. Mohammed M. AL-Ashker Associated professor Mathematics Department mail.iugaza.edu Islamic University of Gaza P.O.Box 108, Gaza, Palestine 1

2 Contents 1 Divisibility Theory of integers The Division Algorithm The greatest common divisor The Euclidean algorithm The Diophantine equation ax + by = c Primes and their distributions Fundamental theorem of arithmetics The sieve of Eratosthenes The Goldbach conjecture The theory of congruences Basic properties of congruence Special divisibility tests Linear congruences Fermat s theorem Fermat s little theorem Wilson s theorem

3 Chapter 1 Divisibility Theory of integers 1.1 The Division Algorithm Theorem Given integers a and b, with b > 0, there exist unique integers q and r satisfying, a = qb + r, 0 r < b. The integers q and r are called, respectively, the quotient and remainder in the division of a and b. Proof. We first Show r and q exist. Let S = {a xb : a xb 0, x Z}. We will show that S φ. Since b > 0 and b Z = b 1 = a b a. Let x = a = a xb a + a b a + a 0 = S φ (S not empty), so S contains a least non-negative integer call it r. Then r will be in the form r = a qb, then r exists and also q exists corresponding to r. We will also show that r < b. Suppose r b and r = r + b where r 0 and r + b = a qb = r = a (q + 1)b 0 = r S. Contradiction, (because r < r and r the smallest non-negative element in S) = 0 r < b. We will now prove the uniqueness of r and q. Assume r 1, q 1 satisfying the conditions of the theorem such that r < r 1 and a = q 1 b+r 1 then r 1 = a q 1 b...(1) but a = qb + r then r = a qb...(2), from (1) and (2) by subtraction, we have (r 1 r) = (q q 1 )b = b r 1 r. Contradiction, because r 1 < b and r < b = r 1 r = 0 = r 1 = r and q 1 = q. Example If a = 592, b = 7 then 592 = 84(7) + 4 then q = 84 and r = 4. 3

4 Corollary If a and b are integers, with b 0, then there exist unique integers q and r such that, a = qb + r, 0 r < b. Proof. If b < 0, then b > 0 and by the above theorem q and r a = q b + r, 0 r < b. Since b = b, we may take q = q to arrive a = ( q)( b) + r = qb + r with 0 r < b. Example Let b = 7, a = 1 then 1 = 0( 7) + 1, where 0 r = 1 < b = 7. Let b = 7, a = 2 then 2 = 1( 7) + 5, where 0 r = 5 < b = 7. Let b = 7, a = 61 then 61 = ( 8)( 7) + 5, where 0 r = 5 < b = 7. Let b = 7, a = 59 then 59 = 9( 7) + 4, where 0 r = 4 < b = 7. If a is even integer then a = 2q and a 2 = 4q 2 = 4k where k = q 2. If a is odd then a = 2q+1 and a 2 = 4q 2 +4q+1 = 4(q 2 +q)+1 = 4k +1 where k = q 2 +q. Example The square of any odd integer is of the form 8k + 1 Proof. By division algorithm, any integer is represented as one of the four following forms: 4q, 4q + 1, 4q + 2, 4q + 3. In this classification, only those integers of the forms 4q +1 and 4q +3 are odd. Then 4q +1) 2 = 16q 2 +8q +1 = 8(2q62 + q) + 1 = 8k + 1, where k = 2q 2 + q,and (4q + 3) 2 = 16q q + 9 = 8(2q 2 + 3q + 1) + 1 = 8k + 1, wherek = 2q 2 + 3q + 1. Let a = 7 then a 2 = 49 = The greatest common divisor Definition An integer b is said to be divisible by an integer a 0, in symbols a b, if there exists some integer c such that b = ac. We write a b to indicate that such b is not divisible by a. Example because 12 = 4( 3) but 3 10 c Z 10 = 3c. If a b = a b because if b = ac = b = a( c) = a b. Theorem For integers a, b and c the following hold : 4

5 1) a 0, 1 a, a a. 2) a 1 if and only if a = ±1. 3) If a b and c d, then ac bd. 4) If a b and b c, then a c. 5) a b and b a if and only if a = ±b. 6) If a b and b 0 then a b. 7) If a b and a c then a bx + cy for arbitrary integers x and y. Proof. 1) For a 0, x = 0, 0 = 0a. For 1 a, x = a, a = a1. For a a, x = 1, a = 1a. 2) If a 1, then x Z 1 = ax = x = ±1 or a = ±1 3) If a b and c d,then x, y Z b = xa, d = cy = bd = xyca = ca bd. 4) If a b and b c then x, y Z b = ax, c = by then c = axy = a c. 5) If a b and b a then x, y Z b = ax, a = by = b = byx = 1 = yx = y = ±1, x = ±1 = b = ±a. 6) If a b then there exists c Z such that b = ac, also b 0 then c 0, and b = ac = a c, since c 0 then c 1 and b a c a. 7) If a b and a c then there exists t Z, r Z, b = at and c = ar = bx = atx, cy = ary = bx + cy = a(tx + ry) = a bx + cy. In general if a b k for k = 1, 2,, n then a b 1 x 1 + b 2 x b n x n. Definition The integer d is a common divisor of a and b in case d a, and d b. since 1 a for any integer a then 1 is a common divisor of a and b, so the set of common divisors of any integers a and b is non empty. Remark If a = b = 0, then every integer is a common divisor of a and b, so in this case the set of common divisors of a and b is infinite. If at least one of a or b is not zero then there are only a finite number of positive common divisors. 5

6 The largest common divisor which is called the greatest common divisor which denoted by g.c.d(a, b). Definition Let a and b be given integers with at least one of them different from zero. The g.c.d(a, b) is the positive integer d satisfying: 1) d a and d b. 2) If c a and c b, then c d. Example The positive divisors of 12 are 1, 2, 3, 4, 6, 12 wile those of 30 are 1, 2, 3, 5, 6, 10, 15, 30. the g.c.d(30, 12) = 6. g.c.d( 5, 5) = 5, g.c.d(8, 17) = 1 and g.c.d( 8, 36) = 4. If a = b = 0 there is no greatest common divisor. If we have a 1, a 2,, a n such that at least one of a i 0 for i = 1, 2,, n then g.c.d of a i 0 for i = 1, 2,, n is denoted by g.c.d(a 1, a 2,, a n ). Theorem Given integers a and b not both of which are zero, there exist integers x and y such that g.c.d(a, b) = ax + by. Proof. Let g = g.c.d(a, b), let S = {ax + by : x, y Z and ax + by > 0}. Since a+b, a b, a b, a+b are integers, so at least one of them must be in S = S. Let l = the smallest positive integer in S chose x 0, y 0, So that l = ax 0 + by 0 we will show that l a, and l b. Assume that l a = q and r Z a = lq + r, 0 < r < l = r = a lq = a q(ax 0 + by 0 ) = a(1 qx 0 ) + b( qy 0 ) = r S, Contradiction, because r < l and l is the smallest integer in S = r = 0 and l a...(1). By the same way we can prove that l b...(2) then from (1) and (2) = l is a common divisor of a and b. Since g = g.c.d(a, b) = g a, g b = g ax 0 + by 0 = l = g l, g l, but l is a common divisor of a and b cannot be greater than the g.c.d g = g = l. Corollary If a and b are given integers, not both zeros, then the set T = {ax + by x, y are integers} is precisely the set of all multiples of g = g.c.d(a, b). Proof. Since g a, and g b = g ax + by for all integers x, y Z. Then every member of T is a multiple of g. If g = ax 0 +by 0 for suitable integers x 0 and y 0. So that any multiple ng of g is of the form ng = n(ax 0 + by 0 ) = a(nx 0 ) + 6

7 b(ny 0 ) = ng is a linear combination of a and b and by definition lies in T. Remark If an integer g is expressible in the form g = ax + by then it is not necessary that g is the g.c.d of (a, b). But it follows from such an equation g.c.d (a, b) g. If ax + by = 1 for some integers x and y then g.c.d(a, b) = 1. Definition Two integers a and b, not both of which are zeros, are said to be relatively prime whenever g.c.d(a, b) = 1. Example g.c.d(2, 5) = g.c.d( 9, 16) = g.c.d( 27, 35) = 1 Theorem Let a and b integers not both zeros, then a and b are relatively prime if and only if there exist integers x and y such that 1 = ax + by. Proof. If g.c.d(a, b) = 1, then x, y z 1 = ax + by. Conversely if 1 = ax+by for some x, y Z, and g = g.c.d(a, b) = g a, g b = g ax + by = g 1 = g = 1 = a and b are relatively prime. Corollary If g.c.d(a, b) = d, then g.c.d( a d, b d ) = 1. Proof. d a, d b = a, b are integers since d = g.c.d(a, b) = x, y d d Z d = ax + by = 1 = ax + b y, by previous theorem ( a, b ) = 1. d d d d Example g.c.d( 12, 30) = 6 = g.c.d( 12 6, 30 6 ) = g.c.d( 2, 5 = 1 Remark It is not true that if a c and b c = ab c. For example 6 24, and 8 24 = Corollary If a c and b c with g.c.d(a, b) = 1 = ab c. Proof. If a c, b c, r and s Z c = ar = bs, since g.c.d(a, b) = 1 then there exist x, y Z 1 = ax + by = c = cax + cby = acx + bcy = c = a(bs)x + b(ar)y = ab(ax + ry) = ab c. Theorem If a bc with g.c.d(a, b) = 1, then a c. Proof. Since g.c.d(a, b) = 1 = x, y Z 1 + ax + by = c = acx + bcy since a ac and a bc = a acx+bcy = a c(ax+by) = a c 1 = a c. Remark If g.c.d(a, b) 1 then a c 7

8 Example = 12 9, Theorem Let a, b be integers, no both zero. d, d = g.c.d(a, b) if and only if For a positive integer 1) d a and d b, 2) wherever c a and c b, then c d. Proof. Let d = g.c.d(a, b) = d a, and d a, d b, so (a) holds. Also x, y Z d = ax + by. If c a, c b then c ax + by = c d = condition (2) holds. Conversely, let d be any positive integer satisfying the stated conditions and let c be any common divisor of a and b then c d by condition (2). This implies that d c, so d is the greatest common divisor of a and b. 1.3 The Euclidean algorithm Lemma If a = qb + r, then g.c.d(a, b) = g.c.d(b, r). Proof. If d = g.c.d(a, b) = d a, d b = d a qb = r, = d r = d g.c.d(b, r), let c = g.c.d(b, r) = d c...(1). c b, c r = c bq+r = c a = c g.c.d(b, a) = d = c d...(2). from (1), (2) = c = d. Theorem The Euclidean algorithm. Let a and b be two integers whose greatest common divisor is desired. Assume that a b > 0. Apply the division algorithm to a and b to get, a = q 1 b + r 1, 0 r 1 < b. If r 1 = 0 = b a and g.c.d(a, b) = b. If r 1 0 divide b by r 1 to produce integers q 2 and r 2 satisfying b = q 2 r 1 + r 2, 0 r 2 < r 1. If r 2 = 0, then we stop; otherwise, proceed as before to obtain r 1 = q 3 r 2 + r 3, 0 r 3 < r 2. This division process continues until some zero remainder appears, say at the (n+1)th stage where r n 1 is divided by r n (a zero remainder occurs sooner or 8

9 later, since the decreasing sequence b > r 1 > r 2 > 0 cannot contain more than b integers. The result is the following system of equations. a = q 1 b + r 1, 0 < r 1 < b b = q 2 r 1 + r 2, o < r 2 < r 1 r 1 = q 3 r 2 + r 3, o < r 3 < r 2.. r n 2 = q n r n 1 + r n, 0 < r n < r n 1 r n 1 = q n+1 r n + 0. We argue that r n, the last nonzero remainder which appears in this manner is equal to g.c.d(a, b). Proof. By lemma g.c.d(a, b) = g.c.d(b, r 1 ) = = g.c.d(r n 1, r n ) = g.c.d(r n, 0) = r n. The g.c.d(a, b) can be expressed in the form ax + by. For this, we fall back on Euclidean algorithm. Starting with the next-to-last equation arising from the algorithm, we write r n = r n 2 q n r n 1. Now solve the proceeding equation in the algorithm for r n 1 and substitute to obtain, r n = r n 2 q n (r n 3 q n 1 r n 2 ) = (1 + q n q n 1 )r n 2 + ( q n )r n 3. This represents r n as a linear combination of r n 2 and r n 3. Continuing backwards through the system of equations, we successively eliminate the remainders r n 1, r n 2,, r 2, r 1 until a stage is reached where r n = g.c.d(a, b) is expressed as a linear combination of a and b. Example Use Euclidean algorithm to find the greatest common divisor of 7469 and solution: 7469 = 3(2464) = 32(77) + 0 g.c.d(7469, 2464) = 77. Also 77 = (2464), so x = 1, y = 3. 9

10 Theorem If k > 0, then g.c.d (ka, kb) = k g.c.d (a, b). Proof. Let g = g.c.d(a, b) then g = the least positive value of ax + by. So for k > 0, g.c.d(ka, kb) = least positive value of (kax + kby) = k (least positive value of ax + by) = k g.c.d(a, b) = kg. Corollary For any integer k 0, g.c.d(ka, kb) = k g.c.d(a, b). Proof. If k < 0 = k = k = g.c.d(ak, bk) = g.c.d( ak, bk) = g.c.d(a k, b k ) = k g.c.d(a, b). Example g.c.d(12, 30 = 3g.c.d(4, 10) = (3) 2g.c.d(2, 5) = 6. Definition An integer c is said to be a common multiple of two nonzero integers a and b whenever a c and b c. Evidently 0 is a common multiple of a and b. Example For a = 12, b = 30, all of the integers 0, 60, 120, 180, are common multiple of 12 and 30. l.c.m( 12, 30) = 60. The least positive common multiple Definition The Least common multiple of two nonzero integers a and b, denoted by l.c.m(a, b), is the positive integer m satisfying: (1) a m and b m. (2) if a c and b c, with c > 0, then m c. Remark If a, b 0 then l.c.m(a, b) exists and l.c.m(a, b) ab. Theorem For positive integers a and b, g.c.d(a, b)l.c.m(a, b) = ab. Proof. Let d = g.c.d(a, b) = d a, d b = r, s Z a = dr, b = ds. If m = ab drds, then m = as = br because m = = as = br = a m, b m = d d m is a positive common multiple of a and b. Now let c be any positive integer that is a common multiple of a and b then a c and b c = c = au = bv for some u, v Z = u = c a, v = c b. Since d = g.c.d(a, b) = x, y Z d = ax + by, then c m = cd c x + c b a m = ab = ab d g.c.d(a,b) = c(ax+by) = ab ab y = vx + uy Z = m c = m c = m = l.c.m(a, b) and = ab = md = l.c.m(a, b)g.c.d(a, b). Corollary Given positive integers a and b, l.c.m(a, b) = ab if and only if g.c.d(a, b) = 1. 10

11 Proof. l.c.m(a, b)g.c.d(a, b) = ab if g.c.d.(a, b) = 1 then l.c.m(a, b) 1 = ab. Conversely, let l.c.m(a, b)g.c.d(a, b) = ab and l.c.m(a, b) = ab then abg.c.d(a, b) = ab = g.c.d(a, b) = 1. Example Find l.c.m(7469, 2464). solution: g.c.d(7469, 2464) by Euclidean algorithm is equal 77 So l.c.m(7469, 2464) = (7469)(2464) g.c.d(7469,2464) = Definition If a, b, c Z not all zero then the g.c.d(a, b, c) is defined to be the positive integer d having the properties: (1) d is a divisor of each a, b, c (2) If e divides the integers a, b, c then e d. Example g.c.d(39, 42, 54) = 3. Definition Three integers are said to be relatively prime as a triple if g.c.d(a, b, c) = 1, yet not relatively primes in pairs. 1.4 The Diophantine equation ax + by = c Theorem The linear diophantine equation ax + by = c has a solution if and only if d c, where d = g.c.d(a, b). If x 0, y 0 is any particular solution of this equation, then all other solutions are given by x = x 0 + b d t, y = y 0 a t for varying integer t. d Proof. Let d = g.c.d(a, b) = d a, d b = r, s Z a = dr, b = ds. If ax + by = c has a solution then there exists x 0, y 0 such that ax 0 + by 0 = c = drx 0 + dsy 0 = c = d(rx 0 + sy 0 ) = c = d c. Conversely if d c = t Z, c = dt, but d = g.c.d(a, b) = x 0, y 0 Z ax 0 + by 0 = d = atx 0 + bty 0 = td = a(tx 0 ) + b(ty 0 ) = c. If x = tx 0, y = ty 0 then the diophantine equation ax+by = c has a solution x = tx 0, and y = ty 0. To proof the second assertion of the theorem, let x 0, y 0 be a solution of the diophantine equation ax+by = c = ax 0 +by 0 = c. Let x, y be any other solution, then ax + by = c = ax 0 + by 0 = ax + by = c = a(x x 0 ) = b(y 0 y ). 11

12 But d a, and d b = r, s Z, a = rd, b = ds = d = g.c.d(a, b) = g.c.d(dr, ds) = dg.c.d(r, s) = g.c.d(r, s) = 1 = rd(x x 0 ) = sd(y 0 y ) = r s(y 0 y ) but g.c.d(r, s) = 1 = r (y 0 y ) = t Z, y 0 y = tr. By substituting we have, r(x x 0 ) = str = x x 0 = st = x = x 0 + st = x 0 + ( b )t and d y = y 0 rt = y 0 ( a)t. d It is only to show that these values satisfy the diophantine equation ax+by = c. ax + by = a[x 0 + b d t] + b[y 0 a d t] = ax 0 + by 0 + ( ab d ab d )t = ax 0 + by 0 = c. Therefore there are infinite number of solutions of the given equation, one for each value of t. Example Determine all solutions in integers of the following Diophantine equation 56x + 72y = 40. Solution:we apply Euclidian algorithm to find g.c.d(56, 72). 72 = = = Then g.c.d(56, 72) = 8, since 8 40, then the Diophantine equation has solution. 8 = = 56 (72 56)3 = 56(1 + 3) 3(72) = 4(56) (3)72. Then 40 = 20(56) 15(72). Then x 0 = 20, y 0 = 15 are the particular solutions and all solutions are x = t, y = 15 7t, where t Z. Corollary If g.c.d(a, b) = 1 and if x 0, y 0 is a particular solution of the linear diophantine equation ax + by = c, then all solutions are given by x = x 0 + bt, y = y 0 at for integral values of t. 12

13 Example Find all solutions of the diophantine equation, 30x + 17y = 300. Solution 30 = = = = Then the g.c.d(30, 17) = 1. Then 1 = 13 4(3) = 13 (17 13)(3) = 13(4) 3(17) = (30 17)4 3(17) 1 = 4(30) 7(17). Then 300 = 1200(30) 2100(17). Then the particular solutions are x 0 = 1200, y 0 = 2100 and all solutions are x = t, y = t. 13

14 Chapter 2 Primes and their distributions 2.1 Fundamental theorem of arithmetics Definition An integer p > 1 is called a prime number, or simply a prime, if its only positive divisors are 1 and p. An integer greater than 1 which is not a prime termed composite. Example ,3,5,7 are primes but 4,6,8,9,10 are composite. 2 is the only even prime, and all other primes are odd. Theorem If p is a prime and p ab, then p a or p b. Proof. Assume that p a = g.c.d(a, p) = 1 = if p ab = p b. if g.c.d(a, p) = p = p a. Corollary If p is a prime and p a 1 a 2 a n, then p a k for some k, where 1 k n. Proof. Assume that p a k for 1 k n = g.c.d(p, a k ) = 1 for 1 k n = g.c.d(p, a 1, a 2, a n ) = 1 = p a 1 a 2 a n contradiction. So there exists some k, 1 k n such that p a k. Corollary If p, q 1, q 2,, q n are all primes and p q 1 q 2 q n then p = q k for some k, where 1 k n. Proof. By previous corollary (2.1.2) k, 1 k n p q k, being a prime, q k is not divisible by any positive integer other than 1 or q itself, since p > 1 then p = q k 14

15 Theorem (Fundamental theorem of arithmetic) Every positive integer n > 1 can expressed as a product of primes, this representation is unique, a part from the order in which the factors occur. Proof. If n is prime then done. If n is composite then there exists a prime p 1 such that, n = p 1 n 1, 1 < n 1 < n. If n 1 is prime done, if not there exists p 2 and n 2 such that n 1 = p 2 n 2, 1 < n 2 < n 1 < n = n = p 1 p 2 n 2. After k 1 times we have, n = p 1 p 2 p k 1 n k 1 and 1 < n k < < n 2 < n 1 < n, the sequence cannot be infinite, so after n k 1 steps, n k 1 must be prime say n k 1 = p k = n = p 1 p 2 p k. To establish the uniqueness, suppose that n has two representations as a product of primes, say n = p 1 p 2 p k = q 1 q 2 q s, where k s and p i, q j are primes and p 1 p 2 p k, q 1 q 2 q s. Since p 1 q 1 q 2 q s, there exists i, 1 i s such that p 1 = q i q 1, similarly q 1 p 1 then q 1 = p 1. If s = k after repeated the same process k times we have unique representation of n. If k < s then after k times we have p 1 p 2 p k = p 1 p 2 p k q k+1 q k+2 q s and by cancelling the common factors we obtain 1 = q k+1 q k+2 q s which is a contradiction. Since q i > 1 then k = s and p 1 = q 1, p k = q k. So n has unique representation. Corollary Any positive integer n > 1 can be written uniquely in a canonical form n = p k 1 1 p k 2 2 p kr r, where, for i = 1, 2,, r, each k i is positive integer and each p i is a prime, with p 1 < p 2 < < p r. Example = , 4725 = , and = Theorem The number 2 is irrational number. Proof. Suppose that 2 is rational number say 2 = a, where a and b are both integers with b g.c.d(a, b) = 1. 15

16 Then a 2 = 2b 2 = b a 2. If b > 1, then by the fundamental theorem of arithmetics, there exists a prime p p b = p a 2 = p a = g.c.d(a, p) p, contradiction unless b = 1. If b = 1 = a 2 = 2 which is impossible because no integer can be multiplied by itself to give 2. Theorem Suppose a and b are positive integers. Let the distinct primes dividing a or b (or both) be p 1, p 2, p n. suppose a = p j 1 1 p j 2 2 p j n n and b = p k 1 1 p k 2 2 p k n n (some of the j s and the k s may be zero). Let m i be the smaller and M i be the larger of j i and k i for i = 1, 2,, n. (a) a b if and only if j i k i for i = 1, 2,, n. (b) g.c.d(a, b) = p m 1 1 p m 2 2 p mn n. (c) l.c.m(a, b) = p M 1 1 p M 2 2 p M n n. 2.2 The sieve of Eratosthenes Theorem If n is composite it must have a prime factor p n. Proof. Let n be composite, then n = d 1 d 2 where d 1 and d 2 > 1. If d 1 and d 2 > n then n = d 1 d 2 > ( n) 2 = n which impossible. Suppose d n, then d 1 is either prime or else has a prime divisor p n. Corollary If the integer n > 1 has no prime divisor n, then n is prime. Example Determine wether n = 1999 is prime number or composite number. Solution: n = 1999 < 45, then all of the primes 2, 3, 5,, 43 < 45 are not divisors of 1999 so n = 1999 is prime. Theorem (Euclid)There are an infinite number of primes. Proof. Suppose that p 1, p 2,, p r are a finite number of primes. Let n = 1 + p 1 p 2 p r. Since n > 1, then if p is a prime number and p n, then p is one of the primes p 1, p 2,, p r because = p p 1 p 2 p r = p n p 1 p 2 p r = p 1 = p = 1, a contradiction because p > 1. So p dose not belong to set of primes p 1, p 2,, p r, so the number of primes is infinite. 16

17 Remark If p n is the nth prime number in the natural order then p n+1 p 1 p 2 p n + 1 for n 1. For example if n = 3 then p 4 = 7 p 1 p 2 p = (2) (3) (5) + 1 = 31. The above estimation is wide; a sharper limitation to this size of p n is given in the following theorem. Theorem If p n is the nth prime number, then p n 2 2n 1. Proof. By induction on n. If n = 1, then p 1 = 2 20 = 2 1 = 2 = p 1 2. Assume that the result is true for n > 1,then p n 2 2n 1. We will show the result is true for n + 1. Since p n+1 p 1 p 2 p n n = n And since n 1 = 2 n 1, then p n+1 2 2n But 1 2 2n 1 for all n > 1, then, p n+1 2 2n n 1 = 2 2 2n 1 = 2 2n. So the result is true for n + 1. Corollary For n 1, there are at least n + 1 primes less than 2 2n. Proof. p 1, p 2,, p n+1 are all less than 2 2n. 2.3 The Goldbach conjecture Lemma The product of two or more integers of the form 4n + 1 is of the same form. Proof. Let k = 4n + 1 and k = 4m + 1 then k k = (4n + 1)(4m + 1) = 16nm + 4n + 4m + 1 = 4(4nm + n + m) + 1 which is of the desired form. Theorem There is infinite number of primes of the form 4n

18 Proof. Assume that there exist only finitely many primes of the form 4n + 3, call them q 1, q 2, q s. Consider the positive integer N = 4q 1 q 2 q s 1 = 4(q 1 q 2 q s 1) + 3 and let N = r 1 r 2 r t be its prime factorization. Since N is odd integer, we have r k 2 for all k, so that each r k is either of the form 4n + 1 or 4n + 3. By previous lemma, the product of any number of primes of the form 4n + 1 is again an integer of this type. For N to take the form 4n + 3, as it clearly does, N must contain at leat one prime factor r i of the form 4n+3. But r i cannot be found among the listing q 1, q 2, q s for this would lead to the contradiction that r i 1. The only possible conclusion is that there are infinitely many primes of this form 4n + 3. Theorem If a and b are relatively prime positive integers then the arithmetic progression a, a + b, a + 2b, contains infinitely many primes. Proof. Suppose that p = a + nb, where p is prime. If n k = n + kp, for k = 1, 2, 3, then the n k th term in the progression is a + n k b = a + (n + kp)b = (a + nb) + kpb = p + kpb = p a + n k b. So the arithmetic progression must contain infinitely many composite numbers which are divisible by infinitely prime numbers. 18

19 Chapter 3 The theory of congruences 3.1 Basic properties of congruence Definition Let n be a fixed positive integer. Two integers a and b are said to be congruent modulo n, symbolized by a b (mod n) if n divides the difference a b; that is provided that a b = kn for some integer k. Example If n = 7 then 3 24( mod 7), and 31 11( mod 7). If n (a b), then we say that a is not incongruent to b modulo n and in this case we write a b (mod n). For example (mod 7) since 7 (25 12) = 13. Remark (1) Any two integers are congruent modulo 1. (2) Two integers are congruent modulo 2 when they are both even or both odd. (3) Given integer a, let q and r be its quotient and remainder upon division by n, so that a = nq + r, 0 r < n = a r = nq = n a r = a r (mod n). There are n choices of r, r = 0, 1,, n 1. So every integer a is congruent modulo n to exactly one of the numbers 0, 1, 2,, n 1. and a 0 (mod n) n a. The set of n integers 0, 1, 2, n 1 is called the set of least positive residues modulo n. 19

20 Definition The set of n integers a 1, a 2, a 3, a n is said to form a complete residues set (or a complete residue system) modulo n if every integer y is congruent modulo n to one and only one of the a i for 1 i n i;e the set {a 1, a 2, a 3, a n } is called complete residue set modulo n if y Z, one and only one a i, 1 i n y a i (mod) n. Example Any integers a 1, a 2, a n are congruent modulo to 0, 1, 2, n 1, taken in some order. Example , 4, 11, 13, 22, 82, 91 constitute a complete set of residues modulo 7, because 12 2, 4 3, 11 4, 13 6, 22 1, 82 5 and 91 0 all modulo 7. Remark Any n integers from a complete residue set modulo n if and only if no two of the integers are congruent modulo n. Theorem For arbitrary integers a and b, a b (mod n) if and only if a and b leave the same nonnegative remainder when divided by n. Proof. Let a b (mod n) = n a b, so k Z a b = nk = a = b + nk by division algorithm b = qn + r, 0 r < n so n leaves the remainder r, then a = b+kn = (qn+r)+kn = (q +k)n+r = a has the same remainder as b. conversely, suppose that a = q 1 n + r, b = q 2 n + r 0 r < n. Then a b = (q 1 n + r) (q 2 n + r) = (q 1 q 2 )n = n a b = a b (mod n). Example ( 7)9 + 7 then by the above theorem (3.1.1) 11 ( 2)9 + 7, (mod 9). 20

21 Theorem Let n > 0 be fixed and a, b, c, d be arbitrary integers. then the following properties hold: (1) a a (mod n). (2) If a b (mod n), then b a (mod n) (3) If a b (mod n) and b c (mod n), then a c(mod n). (4) If a b (mod n) and c d (mod n), then a+c b+c (mod n) and ac bd (mod n). (5) If a b (mod n), then a + c b + c (mod n) and ac bc (mod n). (6) If a b (mod n), then a k b k (mod n) for any positive integer k. Proof. (1) a a = 0 = n a a = a a (mod n). (2) n a b = k Z a b = nk = b a = kn = n b a = b a (mod n). (3) Suppose that a b (mod n) and b c (mod n) then there exist integers h and k satisfying a b = nk and b c = kn. It follows that a c = (a b) + (b c) = hn + kn = (h + k)n, = n a c = a c (mod n). (4) If a b (mod n) and c d (mod n), then k 1, k 2 Z a b = k 1 n, c d = k 2 n = (a + c) (b + d) = k 1 n + k 2 n = (k 1 + k 2 )n = a + b (b + d) (mod n). For the second assertion of (4) ac = (b+k 1 n)(d+k 2 n) = bd+(bk 2 +dk 1 +k 1 k 2 n)n = ac bd (mod n). (5) a b (mod n), then n a b = n (a + c) (b + c) also n ca cb c Z = a + c b + c (mod n) and ca bc (mod n). (6) For k = 1 the result holds. Assume it is true for n = k = a k b k (mod n), then by (4) since a b (mod n) = a a k b b k (mod n) = a k+1 b k+1 (mod n). So it is true for n = k + 1. So the induction step is complete. 21

22 Example Show that (mod 41). Solution: (mod 41) = 2 20 = (2 5 ) 4 ( 9) 4 (mod 41) (81)(81) ( 1)( 1) 1 (mod 41). Example Find the remainder obtained upon dividing the sum Solution: 1! + 2! + 3! + 4! 99! + 100! by 12. fork 4, 4! 0 (mod 12) = k! 5 6 k 0(mod 12) 1! + 2! + 3! + 4! + 100! (mod 12). So the sum leaves a remainder of 9 when divided by 12. Remark If a b (mod n) = ac bc (mod n) but the converse need not true. for example (mod 6) but 4 1 (mod 6). Theorem If ca cb (mod n), then a b (mod n ) where d = gcd(c, n). d Proof. If ca cb (mod n), then n ca cb = k Z c(a b) = kn but gcd(c, n) = d = gcd( c, n) = 1 = c (a b) = k n = n c (a b). d d d d d d But gcd( c, n) = 1, so by previous theorem n (a b) = a b (mod n). d d d d Corollary If ca cb (mod n) and gcd(c, n) = 1 then a b (mod n). Proof. If ca cb (mod n) then n ca cb = c(a b) but gcd (n, c) = 1 = n a b = a b (mod n). Corollary If ca cb (mod p), where p is prime number and p c then a b (mod p). Proof. p c = gcd(p, c) = 1 and p ca cb = c(a b) = p a b = a b (mod p). Example (mod 9) = (mod 9) and gcd(3, 9) = 3 = 11 5 (mod 3). Example (mod 8) then 5( 7) gcd(5, 8) = 1 then 7 9 (mod 8). 5(9) (mod 8), since 22

23 Remark (1) If a b 0 (mod n) then it is not necessarily true that a 0 (mod n) and b 0 (mod n). For example (mod 12) but 3 0 (mod 12) and 4 0 (mod 12). (2) If a b 0 (mod n) and gcd(a, n) = 1 then, b 0 (mod n). (3) If p is prime and a b 0 (mod p) then either a 0 (mod p) or b 0 (mod p). 3.2 Special divisibility tests Given an integer b > 1, any positive integer N can be written uniquely in terms of powers of b as N = a m b m + a m 1 b m a 2 b 2 + a 1 b + a 0. Where the coefficient a k can take on the b different values 0, 1, 2,, b 1. By division algorithm N = q 1 b + a 0, 0 a 0 < b. If q 1 b, we can divide once more, obtaining q 1 = q 2 b + a 1, 0 a 1 < b 1. Now substitute for q 1 in the earlier equation to get N = (q 2 b + a 1 )b + a 0 = q 2 b 2 + a 1 b + a 0. As long as q 2 b, we can continue in the same fashion, going one more step: q 2 = q 3 b + a 2, where 0 a 2 < b, we hence N = q 3 b 3 + a 2 b 2 + a 1 b + a 0. Since N > q 1 > q 2 > 0 is a strictly decreasing sequence of integers, this process must eventually terminate say, at the (m-1)th stage, where q m 1 = q m b + a m 1, 0 a m 1 < b and0 q m < b. Setting a m = q m, we reach the representation N = a m b m + a m 1 b m a 1 b + a 0. 23

24 Which was our aim. The number N = a m b m + a m 1 b m a 1 b + a 0 may be replaced by the simpler symbol N = (a m a m 1 a 2 a 1 a 0 ) b. We call this the base b place value notation for N. If b = 2 the resulting system of enumeration is called the binary number system. Example a = 105 can be written in binary system as 105 = = Or in abbreviated form 105 = ( ) 2. And ( ) 2 translates into = = 79. decimal system: If the base b = 10 then we can represent any integer with this base. Example a = 1492 = The integers 1, 4, 9, 2 are called the digits of the given number. 1 is called the thousands digit, 4 the hundreds digits, 9 the tens digit, and 2 the unit digit. m Theorem Let P (x) = c k x k be a polynomial function of x with k=0 integral coefficients c k. If a b (mod n), then P (a) P (b) (mod n). Proof. Since a b (mod n) then a k b k (mod n) for k = 0, 1, 2,, m therefore c k a k c k b k (mod n) for all such k. Adding these congruences, we conclude that m k c k a k c k b k (mod n). k=0 k=0 or P (a) P (b) (mod n). Definition If P (x) is a polynomial with integral coefficients, one says that a is a solution of the congruence P (x) 0 (mod n) if P (a) 0 (mod n). 24

25 Corollary If a is a solution of P (x) 0 (mod n) and a b (mod n), then b is also a solution. Proof. From previous theorem P (a) P (b) (mod n). Hence if a is a solution of P (x) 0 (mod n), then P (b) P (a) (mod n), making b a solution. Theorem Let N = a m 10 m + a m 1 10 m 1 + a a 0, be the decimal expansion of the positive integer N, 0 a k < 10, and let S = a 0 +a 1 + +a m. Then 9 N if and only if 9 S. m Proof. Consider P (x) = a k x k, a polynomial with integral coefficients. 10 k=0 1 (mod 9) = P (10) P (1) (mod 9), but P (10) = N and P (1) = a 0 + a a m = S = N S (mod 9). It follows that N 0 (mod 9) if and only if S 0 (mod 9). Which is the wanted prove. Theorem Let N = a m 10 m + a m 1 10 m 1 + a a 0, be the decimal expansion of the positive integer N, 0 a k < 10, and let T = a 0 a 1 + a 2 (±1) m a m. Then 11 N if and only if 11 T. m Proof. Put P (x) = a k x k. Since 10 1 (mod 11), we get P (10) P ( 1) (mod 11). k=0 But N = P (10), T = P ( 1) = a 0 a 1 +a 2 +(±1) m a m. So N T (mod 11), it follows that N 0(mod 11) if and only if T 0 (mod 11). Example Let N = then = 27 is divisible by 9 = 9 N. And it divisible by 11 because = 11 is divisible by 11 = 11 N. 3.3 Linear congruences Definition An equation ax b (mod n) is called a linear congruence and x 0 is a solution of this congruence if ax 0 b (mod n). Theorem The linear congruence ax b (mod n) has a solution if and only if d b, where d = gcd(a, n). If d b, then it has d mutually incongruent solutions modulo n. 25

26 Proof. Let ax b (mod n) has a solution the there exists x 0 Z ax 0 b (mod n) = n ax 0 b = y Z ax 0 b = ny = ax 0 ny = b, but this equation has a solution, then gcd(a, n) = d b. Conversely, let d b and let ax b (mod n) = n ax b = y Z ax b = ny = ax ny = b, since gcd(a, n) = d b = ax ny = b has a solution, say x 0, y 0 and all solutions are x = x 0 + nt, y = y d 0+ a t, for some choice of t. Then the congruence d ax b (mod n) has a solution if and only if gcd(a, n) = d b. Let now x = x 0 + n t be all solutions for the congruence, where 0 t d 1. d We will show that the solutions are not congruent modulo n. x 0, x 0 + n d,, x 0 + (d 1)n d Suppose that x 0 + n d t 1 x 0 + n d t 2 (mod n), where 0 t 1 < t 2 d 1. We have nt d 1 nt d 2 (mod n), but gcd( n, n) = n = t d d 1 t 2 (mod n ) = t n/d 1 t 2 (mod d) = d t 2 t 1 = d < t 2 t 1, but 0 t 2 t 1 < d, contradiction. Then x 0 + n d t 1 x 0 + n d t 2 (mod n). It remains to show that any other solution x 0 + ( n )t is congruent modulo n to d one of the d integers listed above. By division algorithm t = qd + r where 0 r d 1. Hence x 0 + n d t = x 0 + n d (qd + r) = x 0 + nq + n d r x 0 + n r (mod n), d whith x 0 + n r being one of our d selected solutions. d Corollary If gcd(a, n) = 1, then the linear congruence ax b (mod n) has a unique solution modulo n. Example Solve the congruence 18x 30 (mod 42). Solution gcd(18, 42) = 6 and 6 30 = the congruence has exactly six solutions which are incongruent modulo 42. Then 18x 30 (mod 42) 3x 5 (mod 7) 6x 10 (mod 7) 26

27 x 3 (mod 7) = x 3 (mod 7) x 4 (mod 7) = x 0 = 4. All solutions of the original congruence are x = x t = 4 + 7t, 0 t < 6, which are x = 4, 11, 18, 25, 32, 39 (mod 42). Example Solve the congruence 9x 21 (mod 30). Solution gcd(9, 30) = 3, 3 21, so the congruence is solvable, then 3x 7 (mod 10) 21x 49 (mod 10) x 9 (mod 0) = x 0 = 9. All solutions are x = x 0 + t 30 3 = t for 0 t < 3. Theorem (The Chinese Remainder theorem) Let n 1, n 2,, n r be positive integers such that gcd(n i, n j ) = 1 for i j. Then the system of linear congruences x a 1 (mod n 1 ) x a 2 (mod n 2 ). x a r (mod n r ) has a common solution which is unique modulo n 1 n 2 n r. r Proof. Let n = n i = n 1 n 2 n r and N k = n for 1 k r. n k i=1 Since gcd(n i, n j ) = 1 for i j, then gcd(n k, n k ) = 1 = a unique solution x k for the congruence N k x 1 (mod n k ) k, 1 k r. = a k N k x k a k (mod n k ) k, 1 k r, and a k N k x k 0 (mod n j ) for j k. Let r x = a k N k x k a k N k x k (mod n k ). k=1 27

28 the x is a common solution for the original system of congruences. But x k was chosen to satisfy the congruence N k x 1 (mod n k ), which forces x a k 1 a k (mod n k ). This implies that the solution of the given system of congruences exists. To show uniqueness, Suppose that x is another common solution of the congruences x a k (mod n k ), for 1 k r. Then x x (mod n k ) for 1 k r. then lcm(n 1, n 2, n r ) x x, but gcd(n i, n j ) = 1 for i j, then n x x then x x (mod n). Example Find the common solution of the system of congruences x 2 (mod 3), x 3 (mod 5), x 2 (mod 7). Solution a 1 = 2, a 2 = 3, a 3 = 2 and n 1 = 3, n 2 = 5, n 3 = 7 n = = 105. N 1 = n = 35, N 2 = n = 21, N 3 = n = 15. n 1 n 2 n 3 The linear congruences 35x 1 (mod 3), 21x 1 (mod 5), 15x 1 (mod 7), has solutions x 2 (mod 3), x 1 (mod 5), x 1 (mod 7). So x 1 = 2, x 2 = 1, x 3 = 1 respectively. Thus the common solution of the original system is given by 3 x = a k N k x k = = (mod 105). k=1 The solution of the congruence ax + by c (mod n) The congruence ax+by c (mod n) has a solution if and only gcd(a, b, n) c.if gcd(a, b, n) = 1, then either gcd(a, n) = 1 or gcd(b, n) = 1. Let gcd(a, b) = 1. The congruence can be written in the form ax c by (mod n). 28

29 then the congruence has a unique solution x for each of the n incongruent values of y. Example Solve the congruence 7x + 4y 5 (mod 12), then 7x 5 4y (mod 12). Substitution of y 5 (mod 12) then 7x 15(mod 12). Then 5x 15 (mod 12) = x 3(mod 12) It follows that x 3 (mod 12) and x 5(mod 12) is one of the 12 incongruent solutions of 7x + 4y 5(mod 12). Theorem The system of linear congruences ax + by r(mod n)...(1) cx + dy s(mod n)...(2) has a unique solution modulo n whenever gcd(ad bc, n) = 1. Proof. multiply the congruence (1) by d and the congruence (2) by b we get adx + bdy rd(mod n) cbx + dby sb(mod n). Now subtract the last equations we get (ad cb)x rd bs (mod n)...(3). Since gcd(ad bc, n) = 1 then the congruence (ad bc)z 1 (mod n) has a unique solution; denote the solution by t. Now multiply the congruence (3) by t we obtain (ad bc)tx (dr bs)t (mod n). Then x (dr bs)t (mod n). 29

30 now if we multiply the congruence (1) by c and the congruence (2) by a and subtract we obtain (ad bc)y (as cr)(mod n). now multiply the congruence by t we get y t(as cr)(mod n). So we have get the solution of the congruence. Example Solve the system of congruences 7x + 3y 10 (mod 16), 2x + 5y 9 (mod 16) Solution: Since gcd( , 16) = 1 so there is a solution of the congruences. Multiply the first congruence by 5, the second by 3 and subtract we get 29x 23(mod 16) then 13x 7(mod 16), now multiply by 5 we get 65x 35 (mod 16) = x 3(mod 16). Now multiply the first congruence by 2 and the second by 7 we get 14x + 6y 20(mod 16) 14x + 35y 63(mod 16). now by subtraction we get 29y 43(mod 16) = 13y 11(mod 16), multiply by 5 we get 65y 55(mod 16) = y 7(mod 16). 30

31 Chapter 4 Fermat s theorem 4.1 Fermat s little theorem Theorem (Fermat s theorem) If p is a prime and p a then a p 1 1(mod p). Proof. Consider the first p 1 positive multiples of a; that is the integers a, 2a, 3a,, (p 1)a. None of these numbers is congruent modulo p to any other, nor is any congruent to zero, because if ra sa(mod p), for 1 r s p 1 and gcd(a, p) = 1 = r s(mod p). A contradiction. Therefore, the above set of integers must be congruent modulo p to 1, 2, 3,, p 1, taken in some order. multiplying all these congruences together, we find that a 2a 3a (p 1)a (p 1)(mod p). Then a p 1 (p 1)! (p 1)!(mod p) but p (p 1)! then gcd((p 1)!, p) = 1 = a p 1 1(mod p). Corollary If p is a prime, then a p a (mod p) for any integer a. Proof. when p a then p a p = p a p a = a p a(mod p). If p a, then a p 1 1(mod p) = a p a(mod p). Theorem (a + 1) p a p + 1 (mod p) a + 1(mod p). 31

32 Proof. (a+1) p = a p + ( p 1 ) a p 1 + ( p p 1 ) a+1 a p (mod p) a+1 (mod p). Example Verify that (mod 11). Solution: 5 38 = = (5 10 ) 3 (5 2 ) 4 since gcd(5, 11) = 1 then (mod 11) = (mod 11) = (5 2 ) (mod 11) 4 (mod 11). Note:(1) If n is composite then it is not necessarily true a n a(mod n). For Example = = (2 7 ) (mod 117) and 2 7 = (mod 117), we have (mod 117) (121) (mod 117) (mod 117) 2 21 (2 7 ) 3 (11) (mod 117) 2 (mod 117). The number 117 = 13 9 is a composite number. (2) If a n 1 1 (mod n) = n is not necessarily prime. For example (mod 341) but 341 = is not prime. Lemma If p and q are distinct primes such that a p a(mod q) and a q a(mod p), then a pq a(mod pq). Proof. (a q ) p a q (mod p) a Z. While a q a (modp) = a pq a (mod p) or p a pq a, also q a pq a. Since lcm(p, q) = pq a pq a = a pq a (mod pq). Example Show that (mod 341), where 341 = 11(31). Solution:2 10 = 1024 = = (mod31) and 2 31 = 2(2 10 ) (mod 11) = (2 11 ) 31 2 (mod 11 31) = (mod 341) Since gcd(2, 341) = 1 = (mod 341). 4.2 Wilson s theorem Theorem If p is a prime then (p 1)! 1(mod p). 32

33 Proof. If p = 2 or p = 3 then the proof is trivial. Let p > 3, Suppose that a is any one of the p 1 positive integers 1, 2, 3,, p 1 and consider the linear congruence ax 1 (mod p), then gcd(a, p) = 1. So the congruence ax 1(mod p) has a unique solution modulo p, a Z with 1 a p 1, satisfying aa 1(mod p). Since p is prime a = a if and only if a = 1 or p 1, because 1 1 1(mod p) and (p 1) 2 1(mod p). Or the congruence a 2 1(mod p) is equivalent to (a 1)(a + 1) 0(mod p), then either (a 1) 0(mod p) or a + 1 0(mod p) then a 1(mod p) = a = 1 or a + 1 = p = a = p 1. If we omit the numbers 1 and p 1, the effect is to group the remaining integers 2, 3,, p 2 into pairs a, a, where a a ; Such that aa 1(mod p). When these p 3 2 congruences are multiplied together and the factors rearranged we get 2 3 (p 2) 1(mod p) or then (p 2)! 1(mod p) (p 1)(p 2)! (p 1) 1(mod p) which implies that (p 1)! 1(mod p). Example Verify Wilson s theorem for p = 13. Solution:We divide the integers 2, 3, 11 into p 3 2 = 5 pairs each of whose products is congruent to 1 modulo 13. Now write out these congruences explicitly we get 2 7 1(mod 13) 3 9 1(mod 13) (mod 13) 5 8 1(mod 13) (mod 13) Multiply the above congruence gives the result 11! = (2 7)(3 9)(4 10)(5 8)(6 11) 1(mod13). = 12! 12 1(mod 13) = (p 1)! 1(mod p) with p = 13. Note:The converse of Wilson s theorem is also true If (n 1)! 1(mod n), then n must be prime. 33.

34 To prove this: let n be not prime, then n has a divisor d, with 1 < d < n. Furthermore, since d n 1, d occurs as one of the factors in (n 1)!, whence d (n 1)!, Now we are assuming that n (n 1)! + 1 and so d (n 1)! + 1 too, then d 1 which is a contradiction. So n is prime. Theorem The quadratic congruence x (mod p), where p is an odd prime, has a solution if and only if p 1(mod 4). Proof. Let a be any solution of x (mod p), so that a 2 1(mod p). Since p a, the outcome of applying Fermat s Theorem is 1 = a p 1 (a 2 ) p 1 2 ( 1) p 1 2 (mod p). The possibility that p = 4k + 3 for k Z is not arise because ( 1) p 1 2 = ( 1) 2k+1 = 1 = 1 1(mod p) = p 2 which false. Therefore p = 4k + 1 because ( 1) p 1 2 = ( 1) 2k = 1 = 1 1(mod p), so p 1(mod 4). Conversely, Let p 1(mod 4). By Wilson s theorem, we have (p 1)! = 1 2 ( p 1 ) ( p+1 ) (p 2)(p 1), also 2 2 p 1 1(mod p), p 2 2(mod p). p + 1 ( p 1 )(mod p). 2 2 Rearranging the factor produces, then (p 1)! 1 ( 1) (2) ( 2) ( p 1 2 )( p 1 )(mod p), 2 (p 1)! ( 1) p 1 2 (1 2 ( p 1 2 ))2 (mod p). If p 1(mod 4) = p = 4k + 1, then 1 (( p 1 2 )!)2 (mod p). The conclusion ( p 1 2 )! satisfies the quadratic congruence x2 +1 0(mod p). Example Solve the congruence x (nod 13). Solution: Since p = 13 and 13 1(mod 4) then the congruence has a solution, so x = ( p 1 )! = 6! = 720 5(mod 13) is a solution. 2 34