PDF 1 Factorisation of Integers

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1 Factorisation of Integers

Definition N={1, 2, 3, . . .} are the natural numbers.

Definition Z={. . . , −2, −1, 0, 1, 2, . . .} are theintegers.

Closed under the binary operations +, ×, −

Definition α∈R then bαc is the greatest integer which is less than or equal to α.

Ex b3c= 3, √ 2

= 1, b−πc=−4 Then bαc6α <bαc+ 1

Proposition 1 If a and b are two integers with b > 0 then there are integers q and r with 06r < b and a=qb+r.

Proof. Let α= ^a_b.

⇒ 0 6 ^a_b −_a

b

< 1

⇒ 0 6 a−b_a

b

< b so if r=a−b_a

b

then a =qb+r with q=_a

b

.

Definition If a=cb (a, b, c∈Z) we say a is amultiple of b, or b divides a and write b|a.

Proposition 2 If b6= 0, c6= 0 then

(a) b|a and c|b ⇒ c|a (b) b|a ⇒ bc|ac

(c) c|d and c|e ⇒ ∀m, n∈Z, c|dm+en.

Proposition 3 Let a, b >0. If b|a and b6=a then b < a.

Definition If b|a and b 6= 1 or a then we sayb is aproper divisor of a. Ifb doesnot divide a write b -a.

Definition P ={p ∈ N :p > 1 and the only divisors of p are 1 and p} are the prime numbers. Then N\(P∪ {1}) are the composite numbers.

P={2, 3, 5, 7, 11, 13, 17, 19, 23, . . .}.

Theorem 1 Every n >1, n∈N, is a product of prime numbers.

Proof. If n ∈ P we are done. If n is not prime, let q₁ be the least proper divisor of n.

Thenq₁ is prime (since otherwise, by Prop 3, it would have a smaller proper divisor). Let

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n = q₁n₁, 1< n₁ < n. If n₁ is prime we are done. If not n₁ = q₂n₂, 1 < n₂ < n₁ < n.

This process must terminate in less than n steps. Hence n=q₁q₂. . . q_s with s < n.

Ex 10725 = 3·5·5·11·13

In a prime factorization ofnarrange the primes so thatp₁ < p₂ <· · ·< p_kand exponents αi ∈N, 16i6k so

n = p^α₁¹p^α₂²· · ·p^α_k^k

=

k

Y

j=1

p^α_j^j

is the standard factorisation of n.

Prime Numbers

We can use the sieve of Eratosthenesto list the primes 26p6N. If n6 N and n is not prime, then n must be divisible by a prime p6√

N (if p₁ >√ N and p₂ >√

N ⇒ p₁p₂ > N).

List all of the integers between 2 and N

2, 3, 4,5, . . . , N successively remove

(i) 4, 6, 8, 10, . . . even integers from 2² on (ii) 9, 15, 21,27, . . . multiples of 3 from 3² on (iii) 25,35, 55, 65, . . . multiples of 5 from 5² on etc.

i.e. remove all integers which are multiples of a prime p < √

N. We are left with all primes up to N.

Ex N = 16, √ N = 4

{2, 3, 64, 5,66, 7, 68,69, 610,11, 612, 13, 614, 615,16}6

Theorem 2 |P|=∞, i.e. there are an infinite number of primes.

Proof. Let P={p₁, p₂, . . . , p_n} with p₁ < p₂ <· · · < p_n and let q =Qn

j=1p_j+ 1. Then q > pj ∀j ⇒ q 6∈ P so q is composite. But pi|q ⇒ pi|q−Qn

j=1pj = 1 ⇒ pi = 1 which is false. Hence |P|=∞.

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How many primes are there ?

Note:

∞

X

n=1

1

n = ∞

∞

X

n=1

1

n² = π² 6 <∞.

We can show

∞

X

j=1

1 pj

=∞ so the primes are denser than the squares.

If x >0, let S(x) = #{n∈N:n² 6x}. Then S(x) =b√

xc. We can show π(x) = #{p∈P:p6x}

∼ x

log(x)

Definition A modulusis a set of integers closed under ±. The zero modulusis just {0}. If a∈Z then M ={na:n∈Z} is a modulus.

Proposition 4 If M is a modulus with a, b∈M and m, n∈Z then ma+nb∈M. Proof. a ∈M ⇒ a+a = 2a ∈M ⇒ 2a+a = 3a ∈M etc. so ma∈M and so is nb, thus ma+nb∈M.

Proposition 5 If M 6= {0} is a modulus, it is the set of multiples of a fixed positive integer.

Proof. Let d be the least positive integer in M with 0< d.

Claim: every element of M is a multiple of d. If not (???) let n ∈ M have d -n. Then n=dq+r with 16r < d. But r=n−dq ∈M (!!!).

Definition Leta, b∈Zand let M ={ma+nb:m, n∈Z} then M is generated by din that M ={nd:n ∈Z}. We calld the greatest common divisor orGCD of a and b, and write (a, b) = d.

Proposition 6

(i) ∃x, y ∈Z so ax+by = (a, b) (ii) ∀x, y ∈Z, (a, b)|ax+by (iii) If e|a and e|b then e|(a, b)

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Definition If (a, b) = 1 we say a and b are coprime.

Ex The GCD (greatest common divisor) is normally computed using the Euclidean Algorithm. From Proposition 5: (a= 323, b = 221)

323 = 221·1 + 102 so 102∈M 221 = 102·2 + 17 so 17∈M 102 = 17·6 + 0

so 17 is the least positive integer inM ⇒ (323, 221) = 17. Reading back:

17 = 221−2·102

= 221−2·(323−221)

= 3·221−2·323 so (a, b) =xa+yb ⇒ x=−2, y = 3.

Proposition 7 If p∈P and p|ab then p|a or p|b.

Proof. If p-a then (a, p) = 1. By Prop 6(i)∃x, y ∈Z so xa+yp = 1

⇒xab+ybp = b But p|abso ab=qp. Hence (xq+yb)p=b so p|b.

Proposition 8 If c >0 and (a, b) =d then (ac, bc) =dc.

Proof. ∃x, y ∈Z so

xa+yb = d

⇒ x(ac) +y(bc) = dc

Theorem 3 (Fundamental Theorem of Arithmetic) The standard factorisation of a number n∈N is unique.

Proof. If p | ab· · ·m, by Proposition 7, p must divide one of the factors. If each of these is prime, then p must be one of them. If n = p^α₁¹· · ·p^α_iⁱ = q₁^β¹· · ·q_j^β^j are two standard factorizations of n, each p must be a q and each q a p. Hence i = j. Since p1 < p2 <· · ·< pk and q1 < q2 <· · ·< qk, p` =q` for 16`6k. If β1 < α1, divide n by p^β₁¹ to getp^α₁¹^−β¹p^α₂²· · ·=p^β₂²· · · ⇒ α₁ =β₁ etc.

Proposition 9 Let a, b∈N have non-standard factorisations a=

m

Y

j=1

p^α_j^j

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and

b=

m

Y

j=1

p^β_j^j with α_j >0, β_j >0 then

(a, b) =

m

Y

j=1

p^{min (α}_j ^j^{, β}^j⁾.

Ex

a = 2²3⁴5¹ b = 2¹3⁰5¹

⇒ (a, b) = 2¹3⁰5¹

Definition Leta, b∈ Z⁺ ={0, 1, 2, . . .} =N∪ {0}. Theleast common multiple or LCM of a and b is the smallest common multiple of a and b and is written {a, b}.

Ex {3, 4}= 12

Proposition 10 With the same notation as for Proposition 9, {a, b}=

m

Y

j=1

p^{max (α}_j ^j^{, β}^j⁾.

Proposition 11 Any common multiple of a and b is a multiple of the least common multiple.

Proposition 12 {a, b}(a, b) = ab Proof.

LHS =

m

Y

j=1

p^{max (α}_j ^j^{, β}^j^{)+min (α}^j^{, β}^j⁾.

But ∀x, y max (x, y) + min (x, y) =x+y. Hence LHS =

m

Y

j=1

p^α_j^j^+β^j =

m

Y

j=1

p^α_j^j ·

m

Y

j=1

p^β_j^j =ab.

Alternative Characterisation of the GCD

By Proposition 6 (ii), (a, b)|ax+by.

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Let x= 1, y = 0 ⇒ (a, b)|a.

Let x= 0, y = 1 ⇒ (a, b)|b.

Sog = (a, b) is a common divisor of a and b. By Proposition 6 (iii), ife|a and e|b then e|g i.e. g is divisible by every common divisor. Hence it is the greatest. This property:

“being a common divisor divisible by every common divisor” characterises the GCD up to sign.

Proof. If g₁ and g₂ satisfy this property, then g₁ and g₂ are both common divisors with g₁|g₂ and g₂|g₁. Hence g₂ = αg₁ = αβg₂ ⇒ αβ = 1 if g₂ 6= 0. Hence α = ±1. So g₁ =±g₂. The GCD, so defined by the above property, ismade unique by fixing the sign, g >0.

Ex

divisors of 12 = {±1, ±2, ±3, ±4, ±6, ±12}=D₁₂ divisors of 18 = {±1, ±2, ±3, ±6, ±9, ±18}=D₁₈ common divisors = {±1, ±2, ±3, ±6}=D₁₂∩D₁₈ So±6 satisfies the property. Hence, fixing the sign, 6 = (12, 18).

Linear Equations in Z

Proposition 13 Given a, b, n ∈ Z, the equation ax+by = n has an integer solution x, y ⇔ (a, b)|n.

Proof. (⇐) By Proposition 6 (i) ∃x, y such thatax+by= (a, b). Since (a, b)|n, ∃csuch that (a, b)c=n Hence a(xc) +b(yc) = (a, b)c=n and xc, yc is the solution.

(⇒) By Proposition 6 (ii), (a, b)|ax+by =n.

Proposition 14 Let (a, b) = 1 and let x₀, y₀ be a solution to ax+by =n (a solution exists by Proposition 13). Then all solutions are given by

x = x0+bt

y = y₀−at , t∈Z.

Proof.

a(x₀+bt) +b(y₀−at) = ax₀+abt+by₀−bat

= n

so each such x and y is a solution. If ax0 + by0 = n and ax +by = n also, then a(x−x₀) +b(y−y₀) = n−n = 0. But (a, b) = 1. Hence b|x−x₀ ⇒ bt=x−x₀ so x=x₀+bt ⇒ abt+b(y−y₀) = 0 ⇒ y−y₀ =−atif b 6= 0.

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Theorem 4 If(a, b) = 1, a >0, b >0then every integer n > ab−a−bis representable as n =ax+by, x>0, y>0 and ab−a−b is not.

Proof. By Proposition 14,

x = x₀+bt y = y₀−at

Choose t so that 0 6y₀−at < a ⇒ 06y₀−at6a−1. But

(x₀+bt)a=n−(y₀−at)b > ab−a−b−(a−1)b =−a

⇒ (x₀+bt) > −1

⇒ (x₀+bt) > 0.

Hence n is representable. Finally suppose ax+by=ab−a−b (???) x>0, y >0.

⇒ a(x+ 1) +b(y+ 1) =ab.

But (a, b) = 1, hencea|y+ 1 (a(x+ 1−b) =b(−y−1)) and b|x+ 1. ⇒ a 6y+ 1 and b6x+ 1 so ab= (x+ 1)a+ (y+ 1)b>ba+ab= 2ab (!!!).

Definition n ∈N

σ(n) = sum of the divisors of n

= X

d|n

d

Ex σ(12) = 1 + 2 + 3 + 4 + 6 + 12 = 28 σ(6) = 1 + 2 + 3 + 6 = 12 = 2(6).

Perfect Numbers

Definition A perfect number is equal to the sum of its proper divisors

n = X

d|n 16d < n

d

orσ(n) = 2n.

Ex 6, 28

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Proposition 15 If n=Qm

j=1p^α_j^j then σ(n) =

m

Y

j=1

p^α_j^j⁺¹−1 p_j −1

Proof. All divisors ofn have the form d=p^x₁¹· · ·p^x_m^m with 06xj 6αj. Hence σ(n) =

α1

X

x1=0

· · ·

αm

X

xm=0

p^x₁¹· · ·p^x_m^m

=

α1

X

x1=0

p^x₁¹

!

· · ·

αm

X

xm=0

p^x_m^m

!

= RHS above.

Definition A functionf :N→Nis calledmultiplicative ifa, b∈Nand (a, b) = 1 ⇒ f(ab) = f(a)f(b)

Proposition 16 (a, b) = 1 ⇒ σ(ab) =σ(a)σ(b) i.e. σ is a multiplicative function.

Proof. This follows from Proposition 15.

Theorem 5 Let p = 2ⁿ−1 be prime. Then m = ¹₂p(p+ 1) = 2ⁿ⁻¹(2ⁿ−1) is perfect.

Every even perfect number has this form.

Proof. m = ¹₂p(p+ 1) = 2ⁿ⁻¹p¹ and p is odd. By Proposition 15 σ(m) = 2ⁿ−1

2−1 · p²−1 p−1

= (2ⁿ−1)(p+ 1)

= p(p+ 1)

= 2m

so m is perfect. Let a be an even perfect number. a = 2ⁿ⁻¹u, u > 1,2 - u. (Note that σ(2^α) = 2^α+1−16= 2·2^α, so no power of 2 is perfect.) Since σ is multiplicative,

σ(a) = 2ⁿ−1

2−1σ(u) = 2a= 2ⁿu since a is perfect. Hence

σ(u) = 2ⁿu

2ⁿ−1 =u+ u 2ⁿ−1.

Butu|u and ₂n^u−1|usou has just two divisors henceu∈Pand ₂n^u−1 = 1 ⇒ u= 2ⁿ−1.

Conjecture There are no odd perfect numbers.

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Definition If p= 2ⁿ−1∈P we say p is a Mersenne Prime.

Theorem 6 If n > 1 and aⁿ−1 is prime then a= 2 and n is prime.

Proof. If a >2 then a−1|aⁿ−1 (aⁿ−1 = (a−1)(aⁿ⁻¹+aⁿ⁻²+· · ·+ 1)) soaⁿ−16∈P. If a= 2 and n=j`, wherej is a proper divisor of n, then 2ⁿ−1 = (2^j)^`−1 is divisible by 2^j −1 (a= 2^j in the equation above). Hence n ∈P.

web: http://www.utm.edu/research/primes/mersenne.shtml Theorem 7 If 2^m+ 1 ∈P then m= 2ⁿ.

Proof. If m=qr, whereq is odd, then

2^qr+ 1 = (2^r)^q+ 1 = (2^r+ 1)(2^r(q−1)−2^r(q−2)+· · ·+ 1) and 1<2^r+ 1<2^qr+ 1 so 2^qr+ 1 cannot be prime. Hence m has no odd prime factor. Hence m= 2ⁿ, n∈N.

Note The factorization

aⁿ−bⁿ = (a−b)(aⁿ⁻¹+aⁿ⁻²b+aⁿ⁻³b²+· · ·+bⁿ⁻¹) works here for odd n since

aⁿ+ 1 = aⁿ−(−1)ⁿ

= (a−(−1))(aⁿ⁻¹+aⁿ⁻²(−1) +aⁿ⁻³(−1)²+· · ·(−1)ⁿ⁻¹)

= (a+ 1)(a+ 1)(aⁿ⁻¹−aⁿ⁻²+aⁿ⁻³− · · ·+ 1)

Fermat Numbers

Definition The nth Fermat number,F_n= 2²ⁿ+ 1 F₀ = 3, F₁ = 5, F₂ = 17, F₃ = 257, F₄ = 65537.

F_i ∈P for 06i64. No other Fermat prime is known.

F₅ 6∈P.

(Euler, 1732): 641|2²⁵ + 1 = 641·6700417.

Proof. Let

a = 2⁷ b = 5 a−b³ = 3

1 +ab−b⁴ = 1 + 5·3 = 2⁴

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Therefore

2²⁵ + 1 = (2⁸)⁴+ 1

= (2a)⁴+ 1

= 2⁴a⁴+ 1

= (1 +ab−b⁴)a⁴ + 1

= (1 +ab)a⁴+ 1−a⁴b⁴

= (1 +ab)a⁴+ (1−a²b²)(1 +a²b²)

= (1 +ab)[a⁴+ (1−ab)(1 +a²b²)]

and 1 +ab= 641.

Theorem 8 (Lagrange) If p∈P, the exact power α of p dividing n! (p^αkn!) is α=

n p

+

n p²

+

n p³

+· · ·

Proof.

n! = 1·2· · ·(p−1)

·p(p+ 1)· · ·2p· · ·(p−1)p

·p²

· · ·

There are j

n p

k

multiples of p,j

n p²

k

multiples of p², etc.

Each multiple of p contributes 1 to α. Each multiple of p² has already contributed 1, being a multiple of p, so contributes 1 more toα leading to

n p

+ n

p²

etc. Hence

α = n

p

+ n

p²

+ n

p³

+· · ·+ n

p^r

where r is the first N such that p^r+1 > n. So j

n p^β

k

= 0 ∀β >r+ 1.

Ex n= 12, p= 3 so

α =

12 3

+

12 9

+

12 27

= 4 + 1 + 0

= 5.

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12! = 12 · 11 · 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1

↓ ↓ ↓ ↓

1 2 1 1

and 3⁵k12! .

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2 Congruences

Definition a≡b (mod m) if m|a−b, m6= 0, a, b, m∈Z. If so we say aiscongruent tob modulo m. We call m the modulus.

Proposition 17 ≡ is an equivalence relation on Z and the set of equivalence classes forms a ring (Z^m,+, ·, [ 1 ]_m) where

[a]_m+ [b]_m = [a+b]_m [a]_m·[b]_m = [a·b]_m

Proposition 18

a₁ ≡ b₁ (mod m) a₂ ≡ b₂ (mod m)

⇒ a₁·a₂ ≡ b₁·b₂ (mod m) a₂+a₂ ≡ b₁+b₂ (mod m)

Proposition 19

ac ≡ bd (mod m) c ≡ d (mod m) (c, m) = 1







⇒ a≡b (mod m)

Proof. (a−b)c+b(c−d) =ac−bd≡0 (mod m) ⇒ m|(a−b)c ⇒ m|a−b soa ≡b (mod m).

If m ∈ P then (c, m) = 1 ∀c ∈ Z with m - c, c 6= 0 and ∃x, y ∈ Z so that cx+my = (c, m) = 1 socx ≡1 (mod m). Hence [c]_m has a multiplicative inverse class [x]_m and (Z^m, +, ·, [ 1 ]_m) is a field GF(m) called a Galois field.

Note [c]_m is called a residue class with representative c. Each class has a smallest non-negative representative.

Ex m= 5

GF(5) ={[ 0 ]₅, [ 1 ]₅, [ 2 ]₅, [ 3 ]₅,[ 4 ]₅}

Proof. If c∈Zand m >0, ∃q, r so that c=mq+r, 06r < m and c≡r (mod m) ⇒ [c]_m = [r]_m

Euler’s Phi Function φ

Definition φ(n) = #{i 6 n : 1 6 i and (i, n) = 1} is the number of natural numbers less than n and coprime to n.

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Ex φ(1) = 1, φ(2) = 1, φ(4) = 2 since (1, 4) = 1, (2, 4) = 2, (3, 4) = 1, (4, 4) = 4.

Ex p∈P ⇒ φ(p) =p−1 since (p, 1) = 1, (p, p) = pand (p, j) = 1, 1< j < p.

Consider m >1. In Zm, [c]_m will have an inverse class ⇔ (c, m) = 1.

(⇐)cx+my= (c, m) = 1 ⇒ cx≡1 (mod m).

Hence thenumber of classes which have inverses is φ(m).

Definition A reduced residue system is a complete set of representatives for those classes with inverses.

Ex {1,3} is such a system for Z4.

Proposition 20 If a₁, . . . , a_φ(m) is a reduced residue system and (m, k) = 1 then ka₁, . . . , ka_φ(m) is also a reduced residue system.

Proof. (a_i, m) = 1 ⇒ (ka_i, m) = 1. If ka_i ≡ ka_j (mod m) ⇒ a_i ≡ a_j (mod m) ⇒ i=j. Hence the kai represent distinct residue classes, and each is coprime with m.

Theorem 9 (Euler) (a, m) = 1 ⇒ a^φ(m) ≡1 (mod m).

Proof. The {aa_i : 1 6 i 6 φ(m)} and {a_i : 1 6 i 6 φ(m)} represent the same classes (albeit in a different order). Hence

φ(m)

Y

j=1

(aa_j) ≡

φ(m)

Y

j=1

a_j (mod m)

⇒ a^φ(m)





φ(m)

Y

j=1

aj



 ≡





φ(m)

Y

j=1

aj



 (mod m) and so a^φ(m)≡1 (mod m) since (a_j, m) = 1 means we can cancel.

Corollary (Fermat’s Little Theorem) (a, p) = 1 ⇒ a^p ≡a (mod p).

Proof. φ(p) =p−1 so a^p−1 ≡1 (mod p) ⇒ a^p ≡a (mod p).

Note Simple probabilistic primality test: Check q ∈ N through considering a^q ≡ a (mod q) for random a with (a, q) = 1.

Note Euler’s a^φ(m) ≡1 (mod m) is the basis of RSA public key cryptography.

Proposition 21 Let (m, m⁰) = 1, let x run over a complete residue system (mod m) and x⁰ over a complete system (mod m⁰). Then mx⁰+m⁰x runs over a complete system (mod mm⁰).

Proof. Consider the mm⁰ numbers mx⁰+m⁰x. If mx⁰+m⁰x ≡ my⁰ +m⁰y (mod mm⁰)

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then

mx⁰ ≡ my⁰ (mod m⁰) m⁰x ≡ m⁰y (mod m)

⇒ x⁰ ≡ y⁰ (mod m⁰) x ≡ y (mod m)

since (m, m⁰) = 1. So each class is distinct. The result follows since there aremm⁰ classes (mod mm⁰).

Proposition 22 Same as before but ‘complete’ → ‘reduced’.

Proof. Claim: (mx⁰+m⁰x, mm⁰) = 1. If not (???) Let p∈P havep|(mx⁰+m⁰x, mm⁰).

If p|m then p|m⁰x. But (m, m⁰) = 1 so p - m⁰ hence p|x and p|(m, x) which is false (!!!). This proves the claim.

Claim: Every a ∈ Z, (a, mm⁰) = 1 satisfies a ≡ mx⁰ + m⁰x (mod mm⁰) for x, x⁰ with (x, m) = (x⁰, m⁰) = 1. By the above ∃x, x⁰ so a ≡ mx⁰ +m⁰x (mod mm⁰). If (x, m) = d 6= 1 then (a, m) = (mx⁰+m⁰x, m) = (m⁰x, m) = (x, m) = d 6= 1 which is false. Similarly (x⁰, m⁰) = 1.

By the above, the numbersmx⁰+m⁰xare incongruent. hence we have a reduced residue system of this form.

Theorem 10 φ is a multiplicative function.

Proof. If (m, m⁰) = 1,

φ(mm⁰) = #{RRS(mm⁰)}

= #{RRS(m)} ·#{RRS(m⁰)}

= φ(m)·φ(m⁰)