# A Guide to Elementary Number Theory

Underwood Dudley
Volume: 41
Edition: 1
Pages: 152
https://www.jstor.org/stable/10.4169/j.ctt6wpw94

1. Front Matter
(pp. I-VI)
2. Introduction
(pp. VII-VIII)
Underwood Dudley

ThisGuideis short, so this introduction will be short as well.

It contains most of the topics that are typically part of a first course in elementary number theory as well as some others that could be. It is intended for two kinds of readers. The first are those who once knew but have forgotten, for instance, which integers are the sum of two squares and why they have that property, and who want their memories refreshed. The second are those who were never told about sums of two squares and want to find out about them, quickly and...

(pp. IX-X)
4. CHAPTER 1 Greatest Common Divisors
(pp. 1-6)

The main concern of elementary number theory is the integers, …, −2, −1, 0, 1, 2 …, especially those that are positive.

We say thatddividesaifa/dis an integer:

Definitiond|a(“ddividesa”) if there is an integerksuch thata=kd.

Thus 3 | 24 and 5 | − 100. Zero does not divide any non-zero integer.

Ifddivides each of a set of integers, it divides any linear combination of them. Here is a proof for a set of size two—the general case is the same except...

5. CHAPTER 2 Unique Factorization
(pp. 7-10)

Though we are not born with it, the knowledge that every positive integer is the product of primes in only one way (except for permutations of the factors) may seem to be so natural as to not be worth commenting on. Nevertheless, we will prove it, if only for completeness’ sake.

Definition An integer greater than 1 that has no positive divisors other that itself and 1 is calledprime. An integer greater than 1 that is not prime iscomposite. The integer 1 is neither prime nor composite: it is aunit.

Theorem (Euclid)There are infinitely many primes....

6. CHAPTER 3 Linear Diophantine Equations
(pp. 11-12)

Adiophantine equation(named after Diophantus of Alexandria (c. third century) who was, as far as we know, the first to consider equations with restricted solutions) will be for us one for which we are to find solutions in integers. Some diophantine equations, such asx⁴ +y⁴ = 2 have solutions (x=y= 1) and some, such asx⁴ +y⁴ = 3, do not.

A simple diophantine equation is thelinear diophantine equation,ax+by=c. If (a,b) = 1 it has infinitely many solutions.

Theorem If (a,b) = 1, then all...

7. CHAPTER 4 Congruences
(pp. 13-16)

The invaluable congruence notation was devised by Gauss.

Definitionab(modm) (“ais congruent tobmodulom”) if and only ifm| (ab).

We will always suppose thatmis positive.

For example, 7 ≡ 17 (mod 10), and another way of saying “nis even” is “n≡ 0 (mod 2)”.

The definition has some immediate consequences that, for want of a better word, we will label as theorems.

Theoremab(modm)if and only if there is an integer k such that a=b+km.

Proof...

8. CHAPTER 5 Linear Congruences
(pp. 17-20)

When confronted with the linear equationax=bwe say “x=b/a” and pass on. The linear congruenceaxb(modm) is a little more complicated.

For one thing, if we have one integer that satisfies it we have infinitely many, because we can add to or subtract from it any multiple ofm. We single one number out:

Definition Asolution to axb(modm) is an integer that satisfies the congruence that is a least residue (modm). (That is, one of 0, 1, …,m− 1.)

For example, 2x≡...

9. CHAPTER 6 The Chinese Remainder Theorem
(pp. 21-24)

Some ancient Chinese manuscripts posed the problem of determining an integer given its remainders on division by other integers, whence the name of the theorem. Its statement is

TheoremThe system of congruences xai(modmi),i= 1, 2, …,k,where the moduli are pairwise relatively prime (i.e., (mi,mj) = 1if ij) has a unique solution modulo m1m2mk.

There are two standard proofs, the first of which generalizes what is done when the system is solved in the obvious manner, as in the

Example Let us solve$x\equiv 1\ (\bmod \ 3),\quad x\equiv 2\ (\bmod \ 5),\quad x\equiv 3\ (\bmod \ 7).$

The first...

10. CHAPTER 7 Fermat’s Theorem
(pp. 25-26)

Fermat’s theorem, sometimes called “Fermat’s Little Theorem” to distinguish it from Fermat’s Last Theorem about solutions ofxn+yn=zn, is useful in many places.

TheoremIf p is prime and(a,p) = 1,then ap−1≡ 1 (modp).

Though hardly deserving the name of lemma—“observation” would be more appropriate—we need the

LemmaIf(a,m) = 1,then the least residues of a, 2a, …, (m− 1)a(modm)are a permutation of1, 2, …,m− 1.

For example, form= 8 anda= 3 the two sets...

11. CHAPTER 8 Wilson’s Theorem
(pp. 27-28)

Wilson’s theorem was not proved by Wilson, but by Lagrange in 1770.

TheoremAn integer p,p≥ 2,is prime if and only if(p− 1)! ≡ −1 (modp).

For example, 5 is prime and 4! = 24 ≡ −1 (mod 5) whereas 6 is composite and 5! = 120 ≡ 0 (mod 6).

The theorem gives a necessary and sufficient condition for an integer to be a prime, but it does not provide a practical primality test. It is, however, useful. For example, a polynomial has been constructed whose positive integer values are, for integer values...

12. CHAPTER 9 The Number of Divisors of an Integer
(pp. 29-30)

To keep statements uncomplicated, in this chapter and the next three all integers will be positive.

Letd(n) denote the number of positive divisors ofn. For example, because 60 has divisors 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60, we haved(60) = 12.

We can calculated(n) from the prime-power decomposition ofn.

TheoremIf$n=p_{1}^{{{e}_{1}}}p_{2}^{{{e}_{2}}}\cdots p_{k}^{{{e}_{k}}}$then$d(n)=({{e}_{1}}+1)({{e}_{2}}+1)\cdots ({{e}_{k}}+1).$

ProofAny divisor ofnhas the form$n=p_{1}^{{{a}_{1}}}p_{2}^{{{a}_{2}}}\cdots p_{k}^{{{a}_{k}}}$where 0 ≤aieifori= 0, 1, …,ei. Eachaimay be chosen inei+ 1 ways, so the total number...

13. CHAPTER 10 The Sum of the Divisors of an Integer
(pp. 31-32)

Letσ(n) denote the sum of the divisors ofn. For example,$\sigma (30)=1+2+3+5+6+10+15+30=72.$

As we will see shortly, this can be obtained more quickly from$\sigma (30)=\sigma (2\cdot 3\cdot 5)=\sigma (2)\sigma (3)\sigma (5)=3\cdot 4\cdot 6=72.$

For a primep,$\sigma ({{p}^{n}})=1+p+{{p}^{2}}+\cdots +{{p}^{n}}.$

This is enough for us to determineσ(n) for allnas is shown by the following theorem.

TheoremIf$p_{1}^{{{e}_{1}}}p_{2}^{{{e}_{2}}}\cdots p_{k}^{{{e}_{k}}}$is the prime-power decomposition of n, then$\sigma (n)=\sigma (p_{1}^{{{e}_{1}}})\sigma (p_{2}^{{{e}_{2}}})\cdots \sigma (p_{k}^{{{e}_{k}}}).$

ProofWe will use mathematical induction. The theorem is true fork= 1. Suppose that it is true fork=r. We want to show that this implies its truth fork=r+ 1. Let$n=p_{1}^{{{e}_{1}}}p_{2}^{{{e}_{2}}}\cdots p_{r}^{{{e}_{r}}}p_{r+1}^{{{e}_{r+1}}}=Np_{r+1}^{{{e}_{r+1}}}.$

To...

14. CHAPTER 11 Amicable Numbers
(pp. 33-34)

Two integers areamicableif the sum of the divisors of one, excluding the number itself, is the other, and vice versa. That is,mandnare amicable ifσ(m) −m=nandσ(n) −n=mor if$\sigma (m)=\sigma (n)=m+n.$

For example, the smallest pair of amicable numbers, (220, 284), is such because$\sigma (220)=\sigma ({{2}^{2}}\cdot 5\cdot 11)=\sigma ({{2}^{2}})\sigma (5)\sigma (11)=7\cdot 6\cdot 12=504$and$\sigma (284)=\sigma ({{2}^{2}}\cdot 71)=\sigma ({{2}^{2}})\sigma (71)=7\cdot 72=504=220+284.$

The name dates back to ancient Greek number mysticism. (The ancient Greeks, lacking theσfunction, would have verified the amicability of 220 and 284 by calculating$1+2+4+5+10+11+20+22+44+55+110=284$and$1+2+4+71+142=220.)$

Amicable numbers belong more to recreational mathematics than to number...

15. CHAPTER 12 Perfect Numbers
(pp. 35-36)

An integer is called aperfect numberif is equal to the sum of its positive divisors, not including itself. Thus 6 = 1 + 2 + 3 and 28 = 1 + 2 + 4 + 7 + 14 are perfect numbers. The name dates back to ancient Greek number mysticism, in which, for example, even numbers were female (because they were easily divided into two parts, so they were weak and hence female—that is the way that ancient Greek number mystics thought). Regardless of their name, perfect numbers are of mathematical interest.

The first recorded theorem on...

16. CHAPTER 13 Euler’s Theorem and Function
(pp. 37-40)

After Fermat’s Theorem thatap−1≡ 1 (modp) ifpis a prime and (a,p) = 1, it is natural to ask if there is a similar theorem when the modulus is not prime. There is.

Definition (Euler’sϕ-function) Letϕ(n) denote the number of positive integers that are less than or equal tonand relatively prime ton.

For example,ϕ(12) = 4 because the integers from 1 to 12 that are relatively prime to 12 are 1, 5, 7, and 11.

If (a,m) ≠ 1 thenar≡ 1 (modm) is impossible because...

17. CHAPTER 14 Primitive Roots and Orders
(pp. 41-48)

We know that if (a,m) = 1 thenaϕ(m)≡ 1 (modm). However, there may be smaller positive integer exponents with that property. For example,ϕ(15) = 8 so if (a, 15) = 1 thena⁸ ≡ 1 (mod 15). Buta⁴ ≡ 1 (mod 15) for allarelatively prime to 15.

Definition If (a,m) = 1, theorderofa(modm) is the smallest positive integertsuch thatat≡ 1 (modm).

The table (mod 15)$\begin{array}{lcccccccc} a & 1 & 2 & 4 & 7 & 8 & 11 & 13 & 14 \\ {{a}^{2}} & & 4 & 1 & 4 & 4 & 1 & 4 & 1 \\ {{a}^{4}} & & 1 & & 1 & 1 & & 1 & \\ \end{array}$shows that the order of 1 (mod 15) is 1, the order of 4, 11,...

18. CHAPTER 15 Decimals
(pp. 49-50)

Some decimal expansions of fractions terminate, such as$\frac{3}{8}=.375$, and others, such as$\frac{2}{9}=.222\ldots$, do not. Which occurs is determined by the denominator of the fraction (if (a,n) = 1, the expansion ofa/nterminates or fails to terminate if and only if the expansion of 1/nterminates or fails to terminate) so the question as to which is which is answered by the

TheoremThe decimal expansion of1/n terminates if and only if n= 2a5bfor some nonnegative integers a and b.

ProofIf 1/n= 1/(2a5b) then multiplying the numerator and denominator by 5ab...

(pp. 51-56)

After solving linear congruencesaxb(modm) it would be natural to try to solve quadratic congruencesax² +bx+c≡ 0 (modm). If$m=p_{1}^{{{e}_{1}}}p_{2}^{{{e}_{2}}}\cdots p_{k}^{{{e}_{k}}}$the congruence is equivalent to the system of congruences$a{{x}^{2}}+bx+c\equiv 0\ (\bmod\ p_{i}^{{{e}_{i}}})$,i= 1, 2, …,k, and if we can solve those then the Chinese Remainder Theorem gives us a solution (modm). It is the case that forr> 1, solutions toax² +bx+c≡ 0 (modpr) can be gotten from solutions toax² +bx+c≡ 0 (modpr−1), so we...

20. CHAPTER 17 Gauss’s Lemma
(pp. 57-60)

Gauss’s Lemma is needed to prove the Quadratic Reciprocity Theorem, that for odd primespandq, (p/q) = (q/p) unlesspq≡ 3 (mod 4), in which case (p/q) = −(q/p), but it also has other uses.

Theorem (Gauss’s Lemma)Suppose that p is an odd prime,pa,and that among the least residues(modp)of$a,2a,\ldots ,\left( (p-1)/2 \right)a$exactly g are greater than(p− 1)/2.Then(a/p) = (−1)g.

ProofDivide the least residues (modp) of$a,2a,\ldots ,\left( (p-1)/2 \right)a$into two classes:r1,r2, …,rkthat are less than or equal to (p− 1)/2...

21. CHAPTER 18 The Quadratic Reciprocity Theorem
(pp. 61-66)

We can now prove the Quadratic Reciprocity Theorem that for distinct odd primespandq, (p/q) = (q/p) unlesspq≡ 3 (mod 4), in which case (p/q) = −(q/p). It was first proved by Gauss, who was very fond of it, probably because of its elegance (there seems to be no reason why square roots (modp) should be related to square roots (modq)) and its applications to quadratic forms. He gave eight proofs and many have been found since. We give the details of Gauss’s third proof.

We need a lemma:

LemmaIf p...

22. CHAPTER 19 The Jacobi Symbol
(pp. 67-70)

It is natural to ask if the Legendre symbol (a/p) could be generalized to (a/b) wherebis not an odd prime. The Jacobi symbol does this.

Definition Ifais an integer, (a,b) = 1, andbis an odd integer with prime factorization$b={{p}_{1}}{{p}_{2}}\cdots {{p}_{k}}$, where the primes are not necessarily distinct, then the Jacobi symbol is defined by$(a/b)=\prod\limits_{i=1}^{k}{(a/{{p}_{i}})}$where the (a/pi) are Legendre symbols.

Suppose thatais a quadratic residue ofb, so there is an integerrwithr² ≡a(modb). That implies thatr² ≡a(modpi) and so...

23. CHAPTER 20 Pythagorean Triangles
(pp. 71-74)

Pythagorean triangles are right triangles whose sides have integer lengths, as in the 3, 4, 5 and 5, 12, 13 triangles. Their sides and hypotenuse satisfya² +b² =c², and it is natural to ask what triples (a,b,c) satisfy that equation. It is also natural to write the equation as${{a}^{2}}={{c}^{2}}-{{b}^{2}}=(c+b)(c-b),$say that the product of two numbers is a square when both factors are squares, set$c+b={{r}^{2}},\quad c-b={{s}^{2}},\quad {{a}^{2}}={{r}^{2}}{{s}^{2}},$solve fora,b, andc, and triumphantly think, “Done!” The idea is a good one, and works, but we have to be careful, as we are in...

24. CHAPTER 21 x⁴ + y⁴ ≠ z⁴
(pp. 75-78)

We will show thatx⁴ +y⁴ =z⁴ has no solutions in positive integers by using Fermat’s method of infinite descent. Fermat assumed that the equation had a solution and then showed that that implied there was a solution in smaller positive integers. That solution would then give a smaller solution in positive integers, and so on. Since an infinitely descending sequence of positive integers does not exist, the assumption is incorrect and the equation has no solutions.

It is possible that Fermat thought that his method would apply to powers other than 4 and this led him to...

25. CHAPTER 22 Sums of Two Squares
(pp. 79-82)

Some positive integers can be written as a sum of two squares of integers, such as 5 = 2² + 1², and others, such as 7, cannot. Here is how to tell them apart:

TheoremA positive integer can be written as a sum of two squares if and only if its prime-power decomposition does not contain a prime congruent to3 (mod 4)raised to an odd power.

Thus 3² · 5 · 7² · 13 is a sum of two squares and 3² · 5² · 7³ · 13² is not.

If we factor out all possible even...

26. CHAPTER 23 Sums of Three Squares
(pp. 83-84)

TheoremA positive integer is a sum of three integer squares if and only if it is not of the form4a(8k+ 7).

Proof(“only if”) We want to show that any integer of the formn= 4a(8k+ 7) is not a sum of three squares. Suppose that$n={{4}^{a}}(8k+7)={{x}^{2}}+{{y}^{2}}+{{z}^{2}}$for some integersx,y, andzwitha> 1. Thenx² +y² +z² ≡ 0 (mod 4). Since squares (mod 4) are either 0 or 1, this is possible only ifx² ≡y² ≡z² ≡ 0 (mod 4). That is,x,yand...

27. CHAPTER 24 Sums of Four Squares
(pp. 85-88)

Euler tried for years to show that every positive integer is a sum of four squares, but the first proof was given by Lagrange in 1770. It uses Euler’s identity that shows that the product of two sums of four squares is a sum of four squares:

Lemma$({{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}})({{r}^{2}}+{{s}^{2}}+{{t}^{2}}+{{u}^{2}})={{(ar+bs+ct+du)}^{2}}+{{(as-br+cu-dt)}^{2}}+{{(at-bu-cr+ds)}^{2}}+{{(au+bt-cs-dr)}^{2}}.$

ProofMultiply it out.

The identity becomes obvious when viewed as a statement about quaternions, but quaternions were not discovered until more than a century after Euler.

TheoremEvery positive integer is a sum of four integer squares.

ProofIf we can show that every prime is a sum of four...

28. CHAPTER 25 Waring’s Problem
(pp. 89-90)

In 1770, Waring asserted that every positive integer is a sum of four squares, nine cubes, nineteen fourth powers, and so on.

Letg(k) denote the smallest number ofkth powers needed to represent every positive integer as a sum.

Theorem$g(k)\ge {{2}^{k}}+\left[ {{(3/2)}^{k}} \right]-2$.

ProofLet$n={{2}^{k}}\left[ {{(3/2)}^{k}} \right]-1$. We show that to writenas a sum ofkth powers takes at least${{2}^{k}}+\left[ {{(3/2)}^{k}} \right]-2$terms.

Because$n\le{{2}^{k}}{{(3/2)}^{k}}-1={{3}^{k}}-1$, writingnas a sum ofkth powers will use only 2ks and 1ks. To use as few terms as possible, we want to use as many 2ks as possible. This number will...

29. CHAPTER 26 Pell’s Equation
(pp. 91-94)

Pell’s equation antedates John Pell (1611–1685) by several centuries and Pell never solved it, but the vagaries of mathematical nomenclature are such that his name is irrevocably linked to it.

The problem is to solvex² −Ny² = 1 in positive integers,Nnot a perfect square. (IfNis a perfect square, the left-hand side can be factored and the equation is no longer interesting.) The equation arises in many places. As we will show, if we have the smallest solution in positive integers, we have them all. For example, the smallest solution ofx² − 2y²...

30. CHAPTER 27 Continued Fractions
(pp. 95-100)

We can write a rational number, for example 23/17, as acontinued fraction:$\begin{array}{lll} \frac{23}{17} & = & 1+\frac{6}{17}=1+\frac{1}{17/6}=1+\frac{1}{2+5/6} \\ & = & 1+\frac{1}{2+\frac{1}{6/5}} \\ & = & 1+\frac{1}{2+\frac{1}{1+\frac{1}{5}}}. \\ \end{array}$

Any rational number can be put in this form, which we will write as [a0,a1, …,ak], so 23/17 = [1, 2, 1, 5]. Any finite continued fraction represents a rational number, which can be determined by evaluating the fraction from the bottom up. The representation is unique, except that$[{{a}_{0}},{{a}_{1}},\ldots ,{{a}_{k}}]=[{{a}_{0}},{{a}_{1}},\ldots ,{{a}_{k}}-1,1].$

The process used to generate the continued fraction,$\begin{array}{rcl} a & = & [a]+{{r}_{1}},\ {{a}_{0}}=[a], \\ 1/{{r}_{1}} & = & [1/{{r}_{1}}]+{{r}_{2}},\ {{a}_{1}}=[1/{{r}_{1}}], \\ 1/{{r}_{2}} & = & [1/{{r}_{2}}]+{{r}_{3}},\ {{a}_{2}}=[1/{{r}_{2}}], \\ & \vdots & \\ \end{array}$can be applied to irrational numbers as well to give infinite continued fraction representations. (If the continued fraction were finite, it would represent a rational...

(pp. 101-102)

Multigrades consist of integers whose sums are the same for more than one power, such as${{1}^{k}}+{{5}^{k}}+{{8}^{k}}+{{12}^{k}}={{2}^{k}}+{{3}^{k}}+{{10}^{k}}+{{11}^{k}}$fork= 1, 2, 3. We can write this as$[1,5,8,12]{{=}^{3}}[2,3,10,11]$.

It can be shown that to have agreement of the firstkpowers, a multigrade must have at leastk+ 1 elements on each side.

Multigrades are mostly of recreational interest. Two theorems that can be verified by algebra are:

TheoremIf$[{{a}_{1}},{{a}_{2}},\ldots ,{{a}_{r}}]{{=}^{k}}[{{b}_{1}},{{b}_{2}},,\ldots ,{{b}_{r}}]$then$[{{a}_{1}}+c,{{a}_{2}}+c,\ldots ,{{a}_{r}}+c]{{=}^{k}}[{{b}_{1}}+c,{{b}_{2}}+c,\ldots ,{{b}_{r}}+c].$

This allows multigrades to be normalized with smallest entry 0 or 1.

TheoremIf$[{{a}_{1}},{{a}_{2}},\ldots ,{{a}_{r}}]{{=}^{k}}[{{b}_{1}},{{b}_{2}},,\ldots ,{{b}_{r}}]$then$[{{a}_{1}},{{a}_{2}},\ldots ,{{a}_{r}},{{b}_{1}}+c,{{b}_{2}}+c,\ldots ,{{b}_{r}}+c]{{=}^{k+1}}[{{b}_{1}},{{b}_{2}},\ldots ,{{b}_{r}},{{a}_{1}}+c,{{a}_{2}}+c,\ldots ,{{a}_{r}}+c].$

This allows multigrades of higher order to be constructed...

32. CHAPTER 29 Carmichael Numbers
(pp. 103-104)

Though it is sufficient thatpbe prime forapa(modp) to be true for alla, it is not necessary. There are composite numbers with that property.

Definition Ifnis composite andana(modn) for alla, thennis aCarmichael number.

The first few Carmichael numbers are 561, 1105, 1729, 2465, and 2821.

The next theorem was proved by Korselt in 1899 but it was not until 1910 that the first Carmichael number was found.

Theoremn is a Carmichael number if and only if n is not divisible by...

33. CHAPTER 30 Sophie Germain Primes
(pp. 105-106)

In the 1820s, Sophie Germain proved that ifpis a prime such that 2p+ 1 is also prime thenxp+yp=zphad no solutions in positive integers relatively prime top. Though her result has been superseded by Andrew Wiles’ proof of Fermat’s Last Theorem, her name has been given to such primes, which are calledSophie Germain primes. The first few are 2, 3, 5, 11, and 23. It is almost certainly true that there are infinitely many, though there is no proof of this. Exceedingly large ones continue to be discovered.

A sufficient...

34. CHAPTER 31 The Group of Multiplicative Functions
(pp. 107-110)

A functionfon the positive integers is multiplicative if (m,n) = 1 impliesf(mn) =f(m)f(n). We have seen some:$\sigma (n)=\sum\limits_{d|n}{d},\quad d(n)=\sum\limits_{d|n}{1,}\quad \phi (n)=n\prod\limits_{p|n}{\left( 1-\frac{1}{p} \right).}$

Here are some more:$i(n)=n,\quad s(n)=1,\quad e(n)=\left\{ \begin{array}{ll} 1, & n=1 \\ 0, & n\ne 1. \\ \end{array} \right.$

That they are multiplicative is not hard to verify.

We introduce an operation, theDirichlet convolution, on multiplicative functions.

Definition Iffandgare multiplicative functions, then$(f*g)(n)=\sum\limits_{d|n}{f(d)g(n/d).}$

For example,$(i*s)(n)=\sum\limits_{d|n}{i(d)s(n/d)=\sum\limits_{d|n}{i(d)}=\sum\limits_{d|n}{d=\sigma (n).}}$

TheoremIf f and g are multiplicative then fg is multiplicative.

ProofIf (m,n) = 1 andd|mnthend=d1d2whered1|mandd2|n. Then$\begin{array}{lll} (f*g)(mn) & = & \sum\limits_{d|mn}{f(d)g(mn/d)} \\ & = & \sum\limits_{{{d}_{1}}{{d}_{2}}|mn}{f({{d}_{1}}{{d}_{2}})g\left( (m/{{d}_{1}})(n/{{d}_{2}}) \right).} \\ \end{array}$

Because (d1,d2) = 1 and...

35. CHAPTER 32 Bounds for π(x)
(pp. 111-116)

Letπ(x) denote the number of primes less than or equal tox. The Prime Number Theorem, thatπ(x) is asymptotic tox/lnxin the sense that$\underset{x\to \infty }{\mathop{\lim }}\,\frac{\pi (x)}{x/\ln x}=1$(actually,$\int_{2}^{x}{dt/\ln t}$is better thanx/lnx, as was noticed by Gauss in 1792), was not proved until 1896. More details can be found in Chapter 35. The proof of the theorem, as might be expected of something that eluded Gauss, is difficult. Here we will get a weaker result, namely the

TheoremFor x≥ 2,$\frac{1}{4}\ln 2<\frac{\pi (x)}{x/\ln x}<32\ln 2.$

We will be following work by Tchebyshev, who established the inequality with upper...

36. CHAPTER 33 The Sum of the Reciprocals of the Primes
(pp. 117-120)

We know that$\sum\nolimits_{n=1}^{\infty }{1/n}$diverges so, because the primes are so much rarer than integers—the gaps between successive primes can be arbitrarily large—we might think that$\sum\nolimits_{p\ \text{prime}}{1/p}$has a chance of converging.

The Prime Number Theorem indicates that this would be mistaken.

Theorem$\sum\limits_{p}{\frac{1}{p}}$diverges.

Plausibility Argument The Prime Number Theorem says that$\pi (x)\sim \frac{x}{\ln x}.$

Because$\ln \pi (x)\sim \ln \frac{x}{\ln x}=\ln x-\ln (\ln x)$we have$\frac{\ln \pi (x)}{\ln x}\sim 1-\frac{\ln (\ln x)}{\ln x},$which goes to 1 asx→ ∞. Thus lnx~ lnπ(x) and so$\pi (x)\sim \frac{x}{\ln \pi (x)}$.

That is,$x\sim \pi (x)\ln \pi (x).$

Letx=pn, thenth prime. Thenπ(x) =π(pn) =nso this becomes${{p}_{n}}\sim n\ln n$and...

37. CHAPTER 34 The Riemann Hypothesis
(pp. 121-122)

Definition If the real part of the complex variables=ς+itis greater than 1, theRiemann zeta function,ς(s), is$\sum\limits_{n=1}^{\infty }{\frac{1}{{{n}^{s}}}}.$

When$\Re (s)>1$the series converges. Its connection with the primes comes from Euler’s

TheoremIf$\Re (s)>1$,$\begin{array}{lll} \varsigma (s) & = & \sum\limits_{n=1}^{\infty }{\frac{1}{{{n}^{s}}}} \\ {} & = & \prod \limits_{p}{\frac{1}{1-{{p}^{-s}}}}. \\ \end{array}$

Semi-proofThe proof uses the same idea as was used in the semi-proof of the corresponding statement in Chapter 33 where 1 appeared instead ofs. Questions of convergence must be dealt with, which we do not do here.

Euler proved the

TheoremIf$\Re (s)>0$,$\varsigma (s)=\frac{1}{1-{{2}^{1-s}}}\sum\limits_{n=1}^{\infty }{\frac{{{(-1)}^{n-1}}}{{{n}^{s}}}}.$

Partial ProofIf$\Re (s)>1$,$\begin{array}{lll} \varsigma (s)-\sum\limits_{{n=1}}^{\infty }{\frac{{{(-1)}^{n-1}}}{{{n}^{s}}}} & = & \sum\limits_{n=1}^{\infty }{\frac{1-{{(-1)}^{n-1}}}{{{n}^{s}}}} \\ & = & \sum\limits_{n\ even}{\frac{2}{{{n}^{s}}}} \\ & = & \sum\limits_{m=1}^{\infty }{\frac{2}{{{(2m)}^{s}}}={{2}^{1-s}}\varsigma (s).} \\ \end{array}$

When this is extended to$\Re (s)>0$...

38. CHAPTER 35 The Prime Number Theorem
(pp. 123-124)

The Prime Number Theorem is thatπ(x), the number of primes less than or equal tox, is asymptotically equal tox/lnx(so that$\underset{x\to \infty }{\mathop{\lim }}\,\frac{\pi (x)}{x/\ln x}=1)$or, as Gauss observed, to$\int_{2}^{\infty }{\frac{dt}{\ln t}}$. Riemann and Tchebyshev laid the groundwork for the proof, which was finally achieved in 1896 by Hadamard and de la Vallée Poussin. Here is an extremely brief sketch of the ideas in the proof. Pages of details have been omitted.

TheMangoldt functionis defined by$\Lambda (n)=\left\{ \begin{array}{ll} \text{In }p, & n={{p}^{k}},k\ge 1 \\ 0, & \text{otherwise}\text{.} \\ \end{array} \right.$

It has a connection with the Riemann zeta function, which does not involve the primes in its definition:$-\frac{{\varsigma }'(s)}{\varsigma (s)}=\sum\limits_{n=1}^{\infty }{\frac{\Lambda (n)}{{{n}^{s}}}.}$

Let...

39. CHAPTER 36 The abc Conjecture
(pp. 125-126)

Ifaandbare relatively prime andc=a+b, it is usually the case thatcis smaller than the product of the primes that dividea,b, andc, as 8+25 = 33 and 33 < 2·5·3·11. This does not always happen, however, as is shown by 5 + 27 = 32 and 32 > 5·3·2. The exceptions are sufficiently rare that in 1985 Masser and Oesterlé made theabc conjecture:

ConjectureGiven ε> 0there is a number k, which depends on ε, such that\[cis true for all a, b, c with a+b...

40. CHAPTER 37 Factorization and Testing for Primes
(pp. 127-130)

To see if an integerNis prime, it is always possible to try divisors 2, 3, 5, … up to the last prime less than or equal to$\sqrt{N}$. If none of them dividesN, thenNis prime.

The test can take as many as$\sqrt{N}$steps, which is large ifNis large. A test devised by Lucas is the basis for some prime-testing algorithms that are more efficient.

TheoremIf an−1≡ 1 (modn)and if, for each prime p that divides n− 1,a(n−1)/p≢ 1 (modn),then n is prime....

41. CHAPTER 38 Algebraic and Transcendental Numbers
(pp. 131-134)

Though almost all numbers are transcendental, transcendental numbers can be hard to identify.

Definition A number isalgebraicif it is the root of a polynomial equation with integer coefficients.

So$\sqrt{2}$andiare algebraic, being roots ofx² − 2 = 0 andx² + 1 = 0, as are all the roots ofx⁵ +x+ 1 = 0.

The roots of a polynomial equation with rational coefficients are also algebraic, because multiplication by a common denominator can make the coefficients integers. It is a fact (that we will not prove) that if the coefficients are...

42. CHAPTER 39 Unsolved Problems
(pp. 135-136)

Number theory abounds with problems that are easy to state but hard to solve. The Riemann Hypothesis and theabcConjecture have already been mentioned. There follow some more.

TheGoldbach Conjecture, made by Goldbach in 1742, is that every even number greater than 2 is a sum of two primes. This was not a conjecture made for conjecture’s sake: Goldbach was trying to help Euler find a proof of the theorem that every integer is a sum of four squares. The conjecture, though undoubtedly true, has turned out to be harder than the foursquares theorem. In 1973 it was...

43. Index
(pp. 137-140)