Prime Number Theory

world. The proof is indirect: i.e., the opposite of the desired result is assumed to be correct and it is showed that this assumption leads to a contradiction. Therefore, the assumption must be incorrect and so the result is proved.

Suppose ffiffiffi p2

is rational then it can be put in the form p/q, where p and q are integers andq 6¼0. Therefore, we can choosep,qto be co-prime (i.e., without any common factors) and so

p=q ð Þ²¼2 )p²=q²¼2 )p²¼2q² )2jp² )2jp )p¼2k )p²¼4k² )4k²¼2q² )2k²¼q² )2jq² )2jq

This is a contradiction as we have chosen p and q to be co-prime, and our assumption that there is a rational number that is the square root of two results in a contradiction. Therefore, this assumption must be false and we conclude that there is no rational number whose square is two.

N¼p1p2p3. . .pkþ1:

Clearly, N is not divisible by any of p₁, p₂, p₃, …, p_k since they all leave a remainder of 1. Therefore,N is either a new prime or divisible by a primeq(that is not in the list ofp₁,p₂,p₃,…,p_k.).

This is a contradiction since this was the list of all the prime numbers, and so the assumption that there are a finite number of primes is false, and we deduce that there are an infinite number of primes.

Proof (ii) Consider the integer N=n! + 1. If N is prime then we take p= N. Otherwise, N is composite and has a prime factor p. We will show that p> n.

Suppose, p n then p|n! and since p|N we have p|n! +1 and therefore p|1, which is impossible. Therefore,p> nand the result is proved.

Proof (iii) Letpbe the smallest prime divisor ofn. Sincenis compositen =uvand clearlyp uand p v. Then p² uv= nand sop ffiffiffi

pn .

Proof (iv) Consider the k consecutive integers (k+ 1)! + 2, (k+ 1)! + 3, …, (k+ 1)! +k, (k+ 1)! +k+ 1. Then each of these is composite since j|(k+ 1)! +jwhere 2 j k+ 1.

Algorithm for Determining Primes

The Sieve of Eratosthenes algorithm (Fig. 3.7) is a famous algorithm for deter- mining the prime numbers up to a given number n. It was developed by the Hellenistic mathematician, Eratosthenes.

The algorithm involvesfirst listing all of the numbers from 2 ton. Thefirst step is to remove all multiples of two up ton; the second step is to remove all multiples of three up ton; and so on.

Thekth step involves removing multiples of the kth prime p_k up ton and the steps in the algorithm continue whilep ffiffiffi

pn

. The numbers remaining in the list are the prime numbers from 2 ton.

1. List the integers from 2 ton.

2. For each primep_kup to ffiffiffi pn

remove all multiples ofp_k.

3. The numbers remaining are the prime numbers between 2 andn.

2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Fig. 3.7 Primes between 1 and 50

60 3 Number Theory

The list of primes between 1 and 50 are given in Fig.3.7. They are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47.

Theorem 3.3 (Fundamental Theorem of Arithmetic)Every natural numbern > 1 may be written uniquely as the product of primes

n¼p₁^a1p₂^a2p₃^a3. . .p_k^ak

Proof There are two parts to the proof. The first part shows that there is a fac- torization and the second part shows that the factorization is unique.

Part(a)

Ifnis prime then it is a product with a single prime factor. Otherwise,ncan be factored into the product of two numbersab,wherea> 1 andb > 1. The argument can then be applied to each of a and b each of which is either prime or can be factored as the product of two numbers both of which are greater than one. Con- tinue in this way with the numbers involved decreasing with every step in the process until eventually all of the numbers must be prime. (This argument can be made more rigorous using strong induction).

Part(b)

Suppose the factorization is not unique and letn> 1 be the smallest number that has more than one factorization of primes. Thenn may be expressed as follows:

n¼p₁p₂p₃. . .p_k¼q₁q₂q₃. . .q_r

Clearly,k> 1 andr> 1 andp_i6¼q_jfor (i= 1,…k) and (j= 1,…,r) as otherwise we could construct a number smaller thann(e.g.,n/p_iwherep_i=q_j) that has two distinct factorizations. Next, without loss of generality takep₁<q₁and define the number N by

N¼ðq₁p₁Þq₂q₃. . .q_r

¼p1p2p3. . .pkp1q2q3. . .qr

¼p₁ðp₂p₃. . .p_kq₂q₃. . .q_rÞ

Clearly 1 <N<nand soNis uniquely factorizable into primes. However, clearly p₁is not a divisor of (q₁−p₁), and soNhas two distinct factorizations, which is a contradiction of the choice ofn.

3.3.1 Greatest Common Divisors (GCD)

Letaandbbe integers not both zero. Thegreatest common divisor dofaandbis a divisor ofaandb(i.e.,d|aandd|b), and it is the largest such divisor (i.e., ifk|aand k|b thenk|d). It is denoted by gcd (a,b).

Properties of greatest common divisors

(i) Leta andb be integers not both zero then exists integersxandysuch that:

d¼gcdða;bÞ ¼axþby

(ii) Letaand bbe integers not both zero then the set S = {ax+ bywherex,y2 ℤ} is the set of all multiples ofd = gcd (a,b).

Proof (of i) Consider the set of all linear combinations ofaandb forming the set {ka+nb: k, n 2 ℤ}. Clearly, this set includes positive and negative numbers.

Choosexand ysuch thatm= ax+by is the smallest positive integer in the set.

Then we shall show thatmis the greatest common divisor.

We know from the division algorithm thata= mq+ rwhere 0 r< m.Thus r¼amq¼aðaxþbyÞq¼ ð1qxÞaþ yqð Þb

ris a linear combination ofaand b and sormust be 0 from the definition ofm.

Therefore,m|aand similarly m|band so mis a common divisor ofa andb.Since, the greatest common divisord is such thatd|aand d|b andd mwe must have d= m.

Proof (of ii) This follows sinced|aandd|b =>d|ax+ byfor all integersxandyand so every element in the set S = {ax+ bywherex,y2ℤ} is a multiple ofd.

Relatively Prime

Two integersa,b are relatively prime if gcd(a,b) = 1 Properties

Ifp is a prime andp|abthenp|aorp|b.

Proof Supposep athen from the results on the greatest common divisor we have gcd (a,p) = 1. That is,

raþsp¼1 )rabþspb¼b

)p bðsincej pjrabandp spbj and sopjrabþspbÞ

3.3.2 Least Common Multiple (LCM)

Ifm is a multiple of a andm is a multiple of b then it is said to be a common multiple of a and b. The least common multiple is the smallest of the common multiples ofa andb and it is denoted by lcm (a,b).

62 3 Number Theory

Properties

Ifxis a common multiple ofaandbthenm|x.That is, every common multiple of aand bis a multiple of the least common multiple m.

Proof We assume that both aand b are nonzero as otherwise the result is trivial (since all common multiples are 0). Clearly, by the division algorithm we have

x¼mqþr where 0r\m

Sincexis a common multiple ofaandbwe havea|xandb|xand also thata|mandb|

m. Therefore,a|randb|r.and soris a common multiple ofaandband sincemis the least common multiple we haver is 0. Therefore,xis a multiple of the least common multiplemas required,

3.3.3 Euclid’s Algorithm

Euclid’s⁴algorithm is one of the oldest known algorithms and it provides a pro- cedure for finding the greatest common divisor of two numbers. It appears in Book VII of Euclid’s Elements, and the algorithm was known prior to Euclid (Fig.3.8).

Lemma Let a, b, q, and r be integers with b> 0 and 0 r< b such that a= bq+r.Then gcd(a,b) = gcd(b,r).

Proof LetK= gcd(a,b) and letL= gcd(b,r) and we therefore need to show that K=L. Supposemis a divisor ofaandbthen asa=bq+ rwe havemwhich is a divisor ofrand so any common divisor ofaand bis a divisor ofr.

Similarly, any common divisornofbandris a divisor ofa. Therefore, the greatest common divisor ofaandbis equal to the greatest common divisor ofbandr.

Fig. 3.8 Euclid of Alexandria

4Euclid was a third century B.C. Hellenistic mathematician and is considered the father of geometry.

Theorem 3.4 (Euclid’s Algorithm) Euclid’s algorithm for finding the greatest common divisor of two positive integers a and b involves applying the division algorithm repeatedly as follows:

a¼bq0þr1 0\r1\b b¼r1q1þr2 0\r2\r1

r1¼r2q2þr3 0\r3\r2

. . .. . .. . .. . . . . .. . .. . .. . .

r_n2¼r_n1q_n1þr_n 0\r_n\r_n1 r_n1¼r_nq_n

Then r_n(i.e., the last nonzero remainder) is the greatest common divisor of a and b:

i.e., gcd(a,b) = r_n.

Proof t is clear from the construction thatrnis a divisor ofrn−1,rn−2,…,r3,r2,r1

and ofaandb.Clearly, any common divisor ofaandbwill also divider_n. Using the results from the lemma above we have

gcdða;bÞ

¼gcdðb;r₁Þ

¼gcdðr1r2Þ

¼. . .

¼gcdðrn2rn1Þ

¼gcdðr_n1;r_nÞ

¼rn

Lemma Let n be a positive integer greater than one then the positive divisors of n are precisely those integers of the form:

d¼p^b₁¹p^b₂²p^b₃³. . .p^b_k^k ðwhere 0b_ia_iÞ;

where the unique factorization of n is given by n¼p^a₁¹p^a₂²p^a₃³. . .p^a_k^k

Proof Supposedis a divisor ofnthenn= dq. By the unique factorization theorem the prime factorization ofnis unique, and so the prime numbers in the factorization ofd must appear in the prime factorsp₁,p₂,p₃,…,p_kof n.

Clearly, the power bi of p_i must be less than or equal to ai: i.e., bi ai. Conversely, wheneverbi aithen clearlyddividesn.

64 3 Number Theory

3.3.4 Distribution of Primes

We already have shown that there are an infinite number of primes. However, most integer numbers are composite and a reasonable question to ask is how many primes are there less than a certain number. The number of primes less than or equal toxis known as the prime distribution function (denoted byp(x)) and it is defined by

pð Þ ¼x R

px1 ðwherepis primeÞ

The prime distribution function satisfies the following properties:

(i) lim

x!1 pðxÞx ¼0 (ii) lim

x!1pðxÞ ¼ 1

Thefirst property expresses the fact that most integer numbers are composite, and the second property expresses the fact that there are an infinite number of prime numbers.

There is an approximation of the prime distribution function in terms of the logarithmic functionðx=lnxÞas follows:

x!1lim pðxÞ

x=lnx¼1 ðPrime Number TheoremÞ

The approximationx/lnxtop(x) gives an easy way to determine the approximate value of p(x) for a given value of x. This result is known as thePrime Number Theorem, and Gauss originally conjectured this theorem.

Palindromic Primes

A palindromic prime is a prime number that is also a palindrome (i.e., it reads the same left to right as right to left). For example, 11, 101, 353 are all palindromic primes.

All palindromic primes (apart from 11) have an odd number of digits. It is an open question as to whether there are an infinite number of palindromic primes.

Letr(m) denote the sum of all the positive divisors of m(includingm):

rð Þ ¼m R_djmd

Lets(m) denote the sum of all the positive divisors ofm(excludingm):

s mð Þ ¼rð Þ m m:

Clearly,s(m) = mand r(m) = 2mwhenmis a perfect number.

Theorem 3.5 (Euclid–Euler Theorem) The positive integer n is an even perfect number if and only ifn = 2^p−1(2^p− 1), where 2^p− 1 is a Mersenne prime.

Proof Suppose n= 2^p−1(2^p−1), where 2^p−1 is a Mersenne prime then rð Þ ¼rn 2^p1ð2^p1Þ

¼r2^p1

rð2^p1Þ

¼r2^p1

2^p ð2^p1 is prime with 2 divisors: 1 and itselfÞ

¼ð2^p1Þ2^p ðSum of arithmetic seriesÞ

¼ð2^p1Þ2:2^p1

¼2:2^p1ð2^p1Þ

¼2n

Therefore,nis a perfect number sincer(n) = 2n.

The next part of the proof is to show that any even perfect number must be of the form above. Letnbe an arbitrary even perfect number (n= 2^p−1q) withqodd and so the gcd (2^p−1,q) = 1 and so

rðnÞ

¼r2^p1q

¼r2^p1 rð Þq

¼ð2^p1Þrð Þq rð Þn

¼2n ðsincenis perfectÞ

¼2:2^p1q

¼2^pq Therefore,

2^pq

¼ð2^p1Þrð Þq

¼ð2^p1Þðs qð Þ þqÞ

¼ð2^p1Þs qð Þ þð2^p1Þq

¼ð2^p1Þs qð Þ þ2^pqq Therefore, (2^p−1)s(q) = q

Therefore,d =s(q) is a proper divisor ofq.However,s(q) is the sum of all the proper divisors ofqincludingd, and sodis the only proper divisor ofqandd = 1.

Therefore,q= (2^p−1) is a Mersenne prime.

66 3 Number Theory

EuleruFunction

The Euler⁵u function (also known as thetotient function) is defined for a given positive integer n to be the number of positive integers k less than n that are relatively prime to n (Fig.3.9). Two integers a, b are relatively prime if gcd(a, b) = 1.

uð Þ ¼n _1k\nR 1 wheregcdðk;nÞ ¼1