Introduction to Number Theory

Introduction to Number Theory Integers: Z={…-3, -2, -1, 0, 1, 2, 3, …} Operations: addition, multiplication, subtraction. Given any two integers a, bZ we can define a + b Z a  b Z a  b Z Integers Z are closed under operations ‘+’, ‘’, ‘’. Closure properties under +, ,  If a, b are integers, then a+b, a b, ab are integers.

Commutative Law If a, b are integers, thena+b=b+a; ab=ba Associative Law If a, b are integers, then a+(b+c) = (a+b)+c; a( bc ) = (ab)c Distributive Law If a, b are integers, then a(b+c) = ab+ ac Identity elements for addition and multiplication For all integer a, a+0=a; a 1= a Additive inverse For all integer a, a + (a) = (a)+a = 0

divisor of a multiple of b If a, b Z, b 0, then it may be that (a / b) Z, so integers are not closed under division. Example. If we divide 10 by 5, the result is an integer again. So, we say, that 10 is divisible by 5 because there exists an integer n (2 in this case) such that 10=5 n. But 10 is not divisible by 3 (within integer domain). Definition. Given two integers a, b Z, b 0, we say that a is divisible by b and denote it b | a,if there exists an integer n Z, such that a = b  n .

For any a, b, c 6) If a | b and b | c, then a | c. Proof. a | b  b = ax for some xZ (by defn of divisibility) b | c  c = by for some yZ (by defn of divisibility) By substitution we have c = (ax)y. By associative law, c = a(xy). xy=k is an integer by the closure property of integers under multiplication. c = ak means that a | c .

7) If a | b then a | bc. Proof. By the definition of divisibility a | b implies that there is some integer x such that b = ax. For any integer c, bc = (ax)c = a(xc) by associative property of multiplication. By the closure property integers under multiplication xc = k, an integer, so bc =ak, i. e. a | bc 8) If a | b and a | c then a | (bx + cy) for any x, y Z Proof is left as an exercise. Does the linear equation 21x + 3y= 133 has integer solution ? No, because for any integer x, y, 3| (21x + 3y), but 133 is not divisible by 3.

Example. Prove that the expression n 3 4n is divisible by 3 for all integers n 0 Proof by induction on n 0. To prove n  0 [P (n): 3|(n 3 4n)] We use the pattern: 1) (Basis) check P (0): 03  40=0, 3|0. 2) n  0 [P(n)  P (n +1)] Induction Hypothesis. Assume P(n)for n = k , for some k  0, i. e. assume 3|(k 3  4k) Induction Step. We need to show that P(n=k+1) holds, i. e. 3|((k+1)3  4(k+1)). Here you need to use your creativity and some algebra skills… By definition of divisibility we have from IH: k 3  4k =3m for some integer m, i. e. k 3  4k =3m (k+1)3  4(k+1) = k3+3k2+3k+14(k+1)= (k 3  4k )+ 3k2+3k+14 = (k 3  4k )+ 3k2+3k 3 = 3m + 3k2+3k 3 =3(m + k2+k 1)

remainder dividend divisor quotient b 3a 4a 2a a 0 a 2a r What is the outcome when any integer b is divided by any positive integer a >0 in general case? For example, if 25 is divided by 7, the quotient is 3 and the remainder is 4. These numbers are related by 25=37+4. Division Algorithm. Given any integers a and b, with a>0, there existuniqueintegers q and r such that b = aq + r, with 0 r <a. Proof. Consider the multiples of a on the real line: We can always find such a multiple aq, that the remainder r=baq satisfies 0 r <a.

b 3a 4a 2a a 0 a 2a b  3a b  4a b  2a b  a b b + a b +2a b= … =a+(b a)=2a+(b2a)= 3a+(b3a)=4a+(b4a)=… Consider set R ={… b+2a,b+a,b, b  a, b2a , … } Among nonnegative terms of R there exists the smallest element (by Well-Ordering Principle). Denote this smallest nonnegative r, then we have that for some q and r: r = b  qa  0 and r a < 0 . If not (i. e. if r a0), b = qa +r = (q+1)a +ra and r is not the smallest positive term (contradiction). So, 0 r <a .

To prove the uniqueness of q and r, suppose thereis another pair q1 and r1 satisfying the same conditions, i. e. we have b=aq+r and b=aq1+r1, with 0 r, r1 <a. Assume r < r1, so that 0 < r1  r <a. Then by subtracting we have r1  r = a(q  q1), so a | (r1  r ) by definition of divisibility. Since r1  r >0 and a >0 it means that r1  r  a in contradiction to 0 r, r1 <a. Hence r = r1 and q =q1, i.e. the quotient and remainder are unique.

Definition. If an integer n>1 divides the difference a b, we say that a  b (mod n): a is congruent to b modulo n. Examples of valid congruences: 63  0 (mod 3) ……….meaning 3 | 63 7  1 (mod 8) ……….meaning 8 | (7 (1)) 25   1(mod 13)…….meaning 13 | (25 (1)) What are the properties of relation Rn = {(a, b) | a  b (mod n)} on Z? a  b (mod n)  b  a (mod n) (symmetric!) because n | (a b)  n | (b a) Reflexive? Transitive?

Note, that relation a  b (mod n) means that a and b differ by a multiple of n, i. e. a = sn+r , b =tn+r (for some s,t Z), Or a and b have the same remainder r when divided by n. So that a b =(st)n, or n | (a b). Example. Take n =3. Then any a Zcan be represented as a = 3q+r, with 0 r <3. So, with respect to division by 3 a remainder may have the values: r = 0 (a is divisible by 3), 1 or 2 (a is not divisible by 3). In accordance to this we have 3 equivalence classes of integers with the same remainder (residue).

…3  03  6  9  12(mod 3)… …2  1 4  7  10  13(mod 3)… …12 5  8  11  14(mod 3)… ‘congruence’ classes So, for any n we have n congruence classes (for residue r=0, 1, 2, … n). r =0: …, 3, 0, 3, 6, 9, 12, … r =1: …, 2, 1, 4, 7, 10, 13, … r =2: …, 1, 2, 5, 8, 11, 14, … What are congruence classes with respect to relation a  b (mod 2)? r =0: n = 2q, q Z(even numbers) r =1: n = 2q+1, q Z(odd numbers)

The properties that follow from the definition: 1. a  b (mod n), b  a (mod n) are equivalent statements (symmetry) 2. If a  b(mod n) and b  c(mod n), then a  c (mod n) (transitivity) 3. If a  b(mod n) and c  d(mod n), then for any integer x, y ax + cy  bx+ dy (mod n) 4. If a  b(mod n) and c  d(mod n), then ac  bd (mod n). Special cases: If a  b(mod n), then a2b2(mod n), a3b3(mod n), etc. If a  b(mod n), then if ac  bc (mod n). 5. If a  b(mod n), then am bm (mod n) for anym1 (can be proved by induction on m1). Example (4): 6  15 (mod 3)  (6 15)=3k  6x 15x =3kx 10  4 (mod 3)  (10 4)=3s  10y 4y = 3s y 6x + 10y (15x+ 4y) =3(kx+ s y)  6x+10y 15x+4y (mod 3)

The last property can be used in solving following problem. Example. Find the remainder when 320 is divided by 25. Try to find a power of 3 that equals a multiple of 25 with small ‘residue’, for instance 33=27 320 = 336+2 = (33)69 = (27)6 9  (2)6 9(mod 25) ………because 27 2(mod 25) = 25 2 9 = 32 2 9  7 2 9 (mod 25)………. because 32 7(mod 25) = 63 2  13 2 (mod 25)…………. because 6313 (mod 25) =26  1 (mod 25)

Definition. Let a, b Z. A positive integer d is called a common divisor of a and b if d | a and d | b . For any two integers a and b the greatest common divisor is always defined (except the case a = b =0) and is denoted by gcd (a, b). Note, that gcd (a, b) 1. • if gcd (a, b) =1, integers a and b are called relatively prime. • gcd (a, b) = gcd (b, a) = gcd (a, b) = gcd (a, b) = gcd (a, b)

Proof. Consider set of integers {ax+by|x, y Z}. This set includes positive, negative values and 0. Choose x0and y0so that l =ax0+by0 is the least positive integer l in the set. Prove that l | a and l | b. Prove l | a by contradiction. Assume l | a , i. e. a = ql+r , where 0<r<l. Hence r = a  ql = a  q(ax0+by0 )= a(1 qx0)+b(qy0). So, r {ax+by|x, y Z}, but it smaller than l in contradiction with assumption, that l is the least positive element in the set. l | b is proved analogously. / Theorem. The gcd(a, b) is the least positive value of ax+by, where x and y range over all integers. Let d =gcd(a, b). It means that a = d n and b = d m. Then, l = ax0+by0 = d(nx0+my0), thus d | l and so l d . But l > d is impossible, since d is the greatest common divisor. So, l =d.

Introduction to Number Theory