2.7
Theorems and Conjectures involving prime num-
bers
We have proved that there are infinitely many primes. We have also proved that there are arbitrary large gaps between primes. The question that arises naturally here is the following: Can we estimate how many primes are there less than a given number? The theorem that answers this question is the prime number theorem. We denote byπ(x)the number of primes less than a given positive numberx. Many mathematicians worked on this theorem and conjectured many estimates before Chebyshev finally stated that the estimate isx/logx. The prime number theorem was finally proved in 1896 when Hadamard and Poussin produced independent proofs. Before stating the prime number theorem, we state and prove a lemma involving primes that will be used in the coming chapters.
Lemma 9. Letpbe a prime and letm ∈Z+. Then the highest power ofpdividing
m!is ∞ X i=1 m pi
Proof. Among all the integers from 1 till m, there are exactly hmpi integers that are divisible byp. These arep,2p, ...,hmpip. Similarly we see that there arehmpi
i
integers that are divisible bypi. As a result, the highest power ofpdividingm!is
X i≥1 i m pi − m pi+1 =X i≥1 m pi
Theorem 20. The Prime Number TheoremLetx >0then
So this theorem says that you do not need to find all the primes less thanxto find out their number, it will be enough to evaluatex/logxfor largexto find an estimate for the number of primes. Notice that I mentioned thatxhas to be large enough to be able to use this estimate.
Several other theorems were proved concerning prime numbers. many great mathematicians approached problems that are related to primes. There are still many open problems of which we will mention some.
Conjecture 1. Twin Prime ConjectureThere are infinitely many pairs primesp
andp+ 2.
Conjecture 2. Goldbach’s ConjectureEvery even positive integer greater than 2 can be written as the sum of two primes.
Conjecture 3. The n2 + 1 Conjecture There are infinitely many primes of the
formn2 + 1, wherenis a positive integer.
Conjecture 4. Polignac ConjectureFor every even number2nare there infinitely many pairs of consecutive primes which differ by2n.
Conjecture 5. Opperman Conjecture Is there always a prime between n2 and
Chapter 3
Congruences
A congruence is nothing more than a statement about divisibility. The theory of congruences was introduced by Carl Friedreich Gauss. Gauss contributed to the basic ideas of congruences and proved several theorems related to this theory. We start by introducing congruences and their properties. We proceed to prove theo- rems about the residue system in connection with the Euler φ-function. We then present solutions to linear congruences which will serve as an introduction to the Chinese remainder theorem. We present finally important congruence theorems derived by Wilson, Fermat and Euler.
3.1
Introduction to congruences
As we mentioned in the introduction, the theory of congruences was developed by Gauss at the beginning of the nineteenth century.
Definition 12. Let m be a positive integer. We say thatais congruent tobmodulo m ifm |(a−b)whereaandbare integers, i.e. ifa=b+kmwherek ∈Z.
Ifais congruent tobmodulom, we writea≡b(mod m). 51
Example 24. 19 ≡ 5(mod7). Similarly2k+ 1 ≡ 1(mod2)which means every odd number is congruent to 1 modulo 2.
There are many common properties between equations and congruences. Some properties are listed in the following theorem.
Theorem 21. Leta, b, candddenote integers. Letmbe a positive integers. Then:
1. Ifa≡b(mod m), thenb ≡a(mod m).
2. Ifa≡b(mod m)andb≡c(mod m), thena≡c(mod m). 3. Ifa≡b(mod m), thena+c≡b+c(mod m).
4. Ifa≡b(mod m), thena−c≡b−c(mod m). 5. Ifa≡b(mod m), thenac≡bc(mod m).
6. Ifa≡b(mod m), thenac≡bc(mod mc), forc > 0.
7. Ifa≡b(mod m)andc≡d(mod m)thena+c≡(b+d)(mod m). 8. Ifa≡b(mod m)andc≡d(mod m)thena−c≡(b−d)(mod m). 9. Ifa≡b(mod m)andc≡d(mod m)thenac≡bd(mod m).
Proof. 1. If a ≡ b(mod m), then m | (a −b). Thus there exists integer k
such thata−b = mk, this impliesb−a =m(−k)and thusm | (b−a). Consequentlyb≡a(mod m).
2. Since a ≡ b(mod m), then m | (a − b). Also, b ≡ c(mod m), then
m|(b−c). As a result, there exit two integerskandlsuch thata=b+mk
3.1. INTRODUCTION TO CONGRUENCES 53 3. Sincea≡b(mod m), thenm|(a−b). So if we add and subtractcwe get
m|((a+c)−(b+c))
and as a result
a+c≡b+c(mod m).
4. Sincea ≡b(mod m), thenm |(a−b)so we can subtract and addcand we get
m |((a−c)−(b−c))
and as a result
a−c≡b−c(mod m).
5. If a ≡ b(mod m), thenm | (a−b). Thus there exists integerk such that
a−b =mkand as a resultac−bc=m(kc). Thus
m|(ac−bc)
and hence
ac≡bc(mod m).
6. If a ≡ b(mod m), thenm | (a−b). Thus there exists integerk such that
a−b =mkand as a result ac−bc=mc(k). Thus mc|(ac−bc) and hence ac≡bc(mod mc).
7. Since a ≡ b(mod m), then m | (a − b). Also, c ≡ d(mod m), then
m|(c−d). As a result, there exits two integerskandlsuch thata−b =mk
andc−d =ml. Note that
(a−b) + (c−d) = (a+c)−(b+d) =m(k+l).
As a result,
m|((a+c)−(b+d)),
hence
a+c≡b+d(mod m).
8. Ifa=b+mkandc=d+mlwherekandlare integers, then
(a−b)−(c−d) = (a−c)−(b−d) =m(k−l).
As a result,
m |((a−c)−(b−d)),
hence
a−c≡b−d(mod m).
9. There exit two integersk andl such thata−b =mk andc−d = mland thusca−cb=m(ck)andbc−bd=m(bl). Note that
(ca−cb) + (bc−bd) =ac−bd=m(kc−lb).
As a result,
m|(ac−bd),
hence
3.1. INTRODUCTION TO CONGRUENCES 55 Examples 1. 1. Because14≡8(mod6)then8≡14(mod6).
2. Because22≡10(mod6)and10≡4(mod6). Notice that22≡4(mod6). 3. Because50≡20(mod15), then50 + 5 = 55≡20 + 5 = 25(mod15). 4. Because50≡20(mod15), then50−5 = 45≡20−5 = 15(mod15). 5. Because19≡16(mod3), then2(19) = 38≡2(16) = 32(mod3).
6. Because19≡16(mod3), then2(19) = 38≡2(16) = 32(mod2(3) = 6).
7. Because19≡3(mod8)and17≡9(mod8), then19 + 17 = 36≡3 + 9 = 12(mod8).
8. Because19≡ 3(mod8)and17≡9(mod8), then19−17 = 2 ≡3−9 =
−6(mod8).
9. Because19≡3(mod8)and17≡9(mod8), then19(17) = 323≡3(9) = 27(mod8).
We now present a theorem that will show one difference between equations and congruences. In equations, if we divide both sides of the equation by a non- zero number, equality holds. While in congruences, it is not necessarily true. In other words, dividing both sides of the congruence by the same integer doesn’t preserve the congruence.
Theorem 22. 1. Ifa, b, candmare integers such thatm >0,d= (m, c)and
ac≡bc(mod m), thena≡b(mod m/d).
2. If(m, c) = 1thena =b(mod m)ifac≡bc(mod m).
Proof. Part 2 follows immediately from Part 1. For Part 1, if ac ≡ bc(mod m), then
Hence there existsk such thatc(a−b) = mk. Dividing both sides by d, we get
(c/d)(a−b) = k(m/d). Since (m/d, c/d) = 1, it follows thatm/d | (a−b). Hencea≡b(mod m/d).
Example 25. 38≡10(mod7). Since(2,7) = 1then19≡5(mod7).
The following theorem combines several congruences of two numbers with different moduli.
Theorem 23. If
a≡b(mod m1), a≡b(mod m2), ..., a≡b(mod mt)
wherea, b, m1, m2, ..., mtare integers andm1, m2, ..., mtare positive, then
a≡b(modhm1, m2, ...mti)
Proof. Sincea≡b(mod mi)for all1≤i≤t. Thusmi |(a−b). As a result,
hm1, m2, ..., mti |(a−b)
(prove this as an exercise). Thus
a≡b(modhm1, m2, ...mti).
Exercises
1. Determine whether 3 and 99 are congruent modulo 7 or not. 2. Show that ifxis an odd integer, thenx2 ≡1(mod8)
3. Show that ifa, b, mandnare integers such thatmandnare positive,n|m
anda≡b(mod m), thena≡b(mod n).
4. Show that ifai ≡bi(mod m)fori= 1,2, ..., n, wheremis a positive integer
andai, biare integers forj = 1,2, ..., n, thenPin=1ai ≡Pni=1bi(mod m)