Proofs from THE BOOK

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Martin Aigner · Günter M. Ziegler

Proofs from THE BOOK

Sixth Edition

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Martin Aigner Günter M. Ziegler

Proofs from THE BOOK

Sixth Edition

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Martin Aigner Günter M. Ziegler

Proofs from THE BOOK

Including Illustrations by Karl H. Hofmann

123 Sixth Edition

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Martin Aigner Institut für Mathematik Freie Universität Berlin Berlin, Germany

Günter M. Ziegler Institut für Mathematik Freie Universität Berlin Berlin, Germany

ISBN 978-3-662-57264-1 ISBN 978-3-662-57265-8 (eBook) https://doi.org/10.1007/978-3-662-57265-8

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Preface

Paul Erd˝os Paul Erd˝os liked to talk about The Book, in which God maintains the perfect

proofs for mathematical theorems, following the dictum of G. H. Hardy that there is no permanent place for ugly mathematics. Erd˝os also said that you need not believe in God but, as a mathematician, you should believe in The Book. A few years ago, we suggested to him to write up a ﬁrst (and very modest) approximation to The Book. He was enthusiastic about the idea and, characteristically, went to work immediately, ﬁlling page after page with his suggestions. Our book was supposed to appear in March 1998 as a present to Erd˝os’ 85th birthday. With Paul’s unfortunate death in the summer of 1996, he is not listed as a co-author. Instead this book is dedicated to his memory.

“The Book”

We have no definition or characterization of what constitutes a proof from The Book: all we offer here is the examples that we have selected, hop- ing that our readers will share our enthusiasm about brilliant ideas, clever insights and wonderful observations. We also hope that our readers will enjoy this despite the imperfections of our exposition. The selection is to a great extent influenced by Paul Erd˝os himself. A large number of the topics were suggested by him, and many of the proofs trace directly back to him, or were initiated by his supreme insight in asking the right question or in making the right conjecture. So to a large extent this book reflects the views of Paul Erd˝os as to what should be considered a proof from The Book.

A limiting factor for our selection of topics was that everything in this book is supposed to be accessible to readers whose backgrounds include only a modest amount of technique from undergraduate mathematics. A little linear algebra, some basic analysis and number theory, and a healthy dollop of elementary concepts and reasonings from discrete mathematics should be sufﬁcient to understand and enjoy everything in this book.

We are extremely grateful to the many people who helped and supported us with this project — among them the students of a seminar where we discussed a preliminary version, to Benno Artmann, Stephan Brandt, Stefan Felsner, Eli Goodman, Torsten Heldmann, and Hans Mielke. We thank Margrit Barrett, Christian Bressler, Ewgenij Gawrilow, Michael Joswig, Elke Pose, and Jörg Rambau for their technical help in composing this book. We are in great debt to Tom Trotter who read the manuscript from ﬁrst to last page, to Karl H. Hofmann for his wonderful drawings, and most of all to the late great Paul Erd˝os himself.

Berlin, March 1998 Martin Aigner · Günter M. Ziegler

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VI

Preface to the Sixth Edition

The idea to this project was born during some leisurely discussions at the Mathematisches Forschungsinstitut in Oberwolfach with the incomparable Paul Erd˝os in the mid-1990s. It is now nearly twenty years ago that we presented the ﬁrst edition of our book on occasion of the International Congress of Mathematicians in Berlin 1998. At that time we could not possibly imagine the wonderful and lasting response our book about The Book would have, with all the warm letters, interesting comments and suggestions, new editions, and as of now thirteen translations. It is no exagger- ation to say that it has become a part of our lives.

In addition to numerous improvements and smaller changes, many of them suggested by our readers, for the present sixth edition we wrote an entirely new chapter with Gurvits’s proof of Van der Waerden’s permanent conjecture, used this to derive asymptotics for the number of Latin squares, added a new, fourth proof for the Euler theorem

n≥1 1

n² =π²/6, and present a new geometric explanation for Heath-Brown’s involution proof for the Fermat two squares theorem.

We thank everyone who helped and encouraged us over all these years. For the second edition this included Stephan Brandt, Christian Elsholtz, Jürgen Elstrodt, Daniel Grieser, Roger Heath-Brown, Lee L. Keener, Christian Lebœuf, Hanfried Lenz, Nicolas Puech, John Scholes, Bernulf Weißbach, and many others. The third edition beneﬁtted especially from input by David Bevan, Anders Björner, Dietrich Braess, John Cosgrave, Hubert Kalf, Günter Pickert, Alistair Sinclair, and Herb Wilf. For the fourth edition, we were particularly indebted to Oliver Deiser, Anton Dochtermann, Michael Harbeck, Stefan Hougardy, Hendrik W. Lenstra, Günter Rote, Moritz W. Schmitt, and Carsten Schultz for their contributions. For the ﬁfth edition, we gratefully acknowledged ideas and suggestions by Ian Agol, France Dacar, Christopher Deninger, Michael D. Hirschhorn, Franz Lemmermeyer, Raimund Seidel, Tord Sjödin, and John M. Sullivan, as well as help from Marie-Sophie Litz, Miriam Schlöter, and Jan Schneider.

For the present sixth edition, very valuable hints were provided by France Dacar again, as well as by David Benko, Jan Peter Schäfermeyer, and Yuliya Semikina.

Moreover, we thank Ruth Allewelt at Springer in Heidelberg and Christoph Eyrich, Torsten Heldmann, and Elke Pose in Berlin for their continuing sup- port throughout these years. And ﬁnally, this book would certainly not look the same without the original design suggested by Karl-Friedrich Koch, and the superb new drawings provided again and again by Karl H. Hofmann.

Berlin, March 2018 Martin Aigner · Günter M. Ziegler

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Number Theory 1

1. Six proofs of the inﬁnity of primes . . . 3

2. Bertrand’s postulate . . . 9

3. Binomial coefﬁcients are (almost) never powers . . . 15

4. Representing numbers as sums of two squares . . . 19

5. The law of quadratic reciprocity . . . 27

6. Every ﬁnite division ring is a ﬁeld . . . 35

7. The spectral theorem and Hadamard’s determinant problem . . . 39

8. Some irrational numbers . . . 47

9. Four timesπ²/6 . . . 55

Geometry 65

10. Hilbert’s third problem: decomposing polyhedra . . . 67

11. Lines in the plane and decompositions of graphs . . . 77

12. The slope problem . . . 83

13. Three applications of Euler’s formula . . . 89

14. Cauchy’s rigidity theorem . . . 95

15. The Borromean rings don’t exist . . . 99

16. Touching simplices . . . 107

17. Every large point set has an obtuse angle . . . 111

18. Borsuk’s conjecture . . . 117

Analysis 125

19. Sets, functions, and the continuum hypothesis . . . 127

20. In praise of inequalities . . . 143

21. The fundamental theorem of algebra . . . 151

22. One square and an odd number of triangles . . . 155

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VIII Table of Contents

23. A theorem of Pólya on polynomials . . . 163

24. Van der Waerden’s permanent conjecture . . . 169

25. On a lemma of Littlewood and Offord . . . 179

26. Cotangent and the Herglotz trick . . . 183

27. Buffon’s needle problem . . . 189

Combinatorics 193

28. Pigeon-hole and double counting . . . 195

29. Tiling rectangles . . . 207

30. Three famous theorems on ﬁnite sets . . . 213

31. Shufﬂing cards . . . 219

32. Lattice paths and determinants . . . 229

33. Cayley’s formula for the number of trees . . . 235

34. Identities versus bijections . . . 241

35. The ﬁnite Kakeya problem . . . 247

36. Completing Latin squares . . . 253

Graph Theory 259

37. Permanents and the power of entropy . . . 261

38. The Dinitz problem . . . 271

39. Five-coloring plane graphs . . . 277

40. How to guard a museum . . . 281

41. Turán’s graph theorem . . . 285

42. Communicating without errors . . . 291

43. The chromatic number of Kneser graphs . . . 301

44. Of friends and politicians . . . 307

45. Probability makes counting (sometimes) easy . . . 311

About the Illustrations 321

Index 323

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Number Theory

1

Six proofs

of the inﬁnity of primes 3 2

Bertrand’s postulate 9 3

Binomial coefﬁcients

are (almost) never powers 15 4

Representing numbers as sums of two squares 19 5

The law of

quadratic reciprocity 27 6

Every ﬁnite division ring is a ﬁeld 35

7

The spectral theorem and Hadamard’s

determinant problem 39 8

Some irrational numbers 47 9

Four timesπ²/6 55

“Irrationality andπ”

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Six proofs

of the inﬁnity of primes

Chapter 1

It is only natural that we start these notes with probably the oldest Book Proof, usually attributed to Euclid (ElementsIX, 20). It shows that the sequence of primes does not end.

Euclid’s Proof. For any ﬁnite set{p₁, . . . , p_r} of primes, consider the numbern = p₁p₂· · ·p_r+ 1. Thisnhas a prime divisorp. But pis not one of thep_i: otherwisepwould be a divisor ofnand of the product p₁p₂· · ·p_r, and thus also of the differencen−p₁p₂· · ·p_r = 1, which is impossible. So a ﬁnite set{p₁, . . . , p_r}cannot be the collection ofallprime

numbers.

Before we continue let us ﬁx some notation. N ={1,2,3, . . .}is the set of natural numbers,Z={. . . ,−2,−1,0,1,2, . . .}the set of integers, and P={2,3,5,7, . . .}the set of primes.

In the following, we will exhibit various other proofs (out of a much longer list) which we hope the reader will like as much as we do. Although they use different view-points, the following basic idea is common to all of them:

The natural numbers grow beyond all bounds, and every natural number n ≥ 2has a prime divisor. These two facts taken together forcePto be inﬁnite. The next proof is due to Christian Goldbach (from a letter to Leon- hard Euler 1730), the third proof is apparently folklore, the fourth one is by Euler himself, the ﬁfth proof was proposed by Harry Fürstenberg, while the last proof is due to Paul Erd˝os.

Second Proof. Let us ﬁrst look at theFermat numbersFn= 2²ⁿ+1for n= 0,1,2, . . .. We will show that any two Fermat numbers are relatively prime; hence there must be inﬁnitely many primes. To this end, we verify the recursion ⁿ−1

k=0

Fk = Fn−2 (n≥1),

from which our assertion follows immediately. Indeed, ifmis a divisor of,

F₀ = 3 F₁ = 5 F₂ = 17 F₃ = 257 F₄ = 65537 F₅ = 641·6700417 The ﬁrst few Fermat numbers

say,Fk andFn (k < n), thenmdivides 2, and hence m = 1or2. But m= 2is impossible since all Fermat numbers are odd.

To prove the recursion we use induction onn. Forn= 1we haveF₀= 3 andF₁−2 = 3. With induction we now conclude

n k=0

F_k = ⁿ⁻¹

k=0

F_k

F_n = (F_n−2)F_n =

= (2²ⁿ−1)(2²ⁿ+ 1) = 2²ⁿ⁺¹−1 = F_n+1−2.

M. Aigner, G. M. Ziegler, Proofs from THE BOOK, https://doi.org/10.1007/978-3-662-57265-8_1

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4 Six proofs of the inﬁnity of primes Third Proof. SupposePis ﬁnite andpis the largest prime. We consider the so-called Mersenne number2^p−1 and show that any prime factorq of2^p−1is bigger thanp, which will yield the desired conclusion. Letqbe a prime dividing2^p−1, so we have2^p≡1 (modq). Sincepis prime, this Lagrange’s theorem

IfGis a ﬁnite (multiplicative) group and U is a subgroup, then |U| divides|G|.

Proof. Consider the binary relation

a∼b:⇐⇒ ba⁻¹∈U.

It follows from the group axioms that ∼ is an equivalence relation.

The equivalence class containing an elementais precisely the coset

U a={xa:x∈U}. Since clearly|U a| = |U|, we ﬁnd thatGdecomposes into equivalence classes, all of size |U|, and hence that|U|divides|G|. In the special case whenUis a cyclic subgroup {a, a², . . . , a^m} we ﬁnd that m (the smallest positive integer such thata^m = 1, called the orderofa) divides the size|G|of the group. In particular, we have a^|G|= 1.

means that the element2has orderpin the multiplicative groupZq\{0}of the ﬁeldZq. This group hasq−1elements. By Lagrange’s theorem (see the box) we know that the order of every element divides the size of the group, that is, we havep|q−1, and hencep < q.

Now let us look at a proof that uses elementary calculus.

Fourth Proof. Letπ(x) := #{p≤x:p∈P}be the number of primes that are less than or equal to the real number x. We number the primes P = {p₁, p₂, p₃, . . .}in increasing order. Consider the natural logarithm logx, deﬁned aslogx=x

1 1 tdt.

2 1 1

n n+1 Steps above the functionf(t) =¹_t

Now we compare the area below the graph off(t) = ¹_twith an upper step function. (See also the appendix on page 12 for this method.) Thus for n≤x < n+ 1we have

logx ≤ 1 + 1 2+1

3 +· · ·+ 1 n−1 +1

n

≤ 1

m, where the sum extends over allm∈Nwhich have only prime divisorsp≤x.

Since every such mcan be written in auniqueway as a product of the form

p≤x

p^k^p, we see that the last sum is equal to

p∈P p≤x

k≥0

1 p^k

.

The inner sum is a geometric series with ratio ¹_p, hence

logx ≤

p∈P p≤x

1

1−¹_p =

p∈P p≤x

p p−1 =

π(x)

k=1

p_k p_k−1.

Now clearlyp_k≥k+ 1, and thus p_k

p_k−1 = 1 + 1

p_k−1 ≤ 1 + 1

k = k+ 1 k , and therefore

logx ≤

π(x)

k=1

k+ 1

k = π(x) + 1.

Everybody knows that logxis not bounded, so we conclude thatπ(x)is unbounded as well, and so there are inﬁnitely many primes.

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Six proofs of the inﬁnity of primes 5 Fifth Proof. After analysis it’s topology now! Consider the following

curious topology on the setZof integers. Fora, b∈Z,b >0, we set N_a,b={a+nb:n∈Z}.

Each setN_a,bis a two-way infinite arithmetic progression. Now call a set O ⊆ Zopenif eitherO is empty, or if to everya∈ O there exists some b > 0 withN_a,b ⊆O. Clearly, the union of open sets is open again. If O₁, O₂are open, anda ∈ O₁∩O₂ withN_a,b₁ ⊆ O₁ andN_a,b₂ ⊆ O₂, thena∈Na,b₁b₂ ⊆O₁∩O₂. So we conclude that any finite intersection of open sets is again open. So, this family of open sets induces a bona fide topology onZ.

Let us note two facts:

(A) Any nonempty open set is inﬁnite.

(B) Any setNa,bis closed as well.

Indeed, the ﬁrst fact follows from the deﬁnition. For the second we observe N_a,b = Z\

b−1 i=1

N_a+i,b,

which proves thatN_a,bis the complement of an open set and hence closed.

“Pitching ﬂat rocks, inﬁnitely”

So far the primes have not yet entered the picture — but here they come.

Since any numbern= 1,−1has a prime divisorp, and hence is contained inN_0,p, we conclude

Z\ {1,−1} =

p∈P

N_0,p. Now ifPwere ﬁnite, then

p∈PN_0,pwould be a ﬁnite union of closed sets (by (B)), and hence closed. Consequently,{1,−1}would be an open set,

in violation of (A).

Sixth Proof. Our ﬁnal proof goes a considerable step further and demonstrates not only that there are inﬁnitely many primes, but also that the series

p∈P1

p diverges. The ﬁrst proof of this important result was given by Euler (and is interesting in its own right), but our proof, devised by Erd˝os, is of compelling beauty.

Let p₁, p₂, p₃, . . . be the sequence of primes in increasing order, and assume that

p∈P 1

p converges. Then there must be a natural number k such that

i≥k+1 1

p_i < ¹₂. Let us callp₁, . . . , p_k thesmall primes, and p_k+1, p_k+2, . . .thebigprimes. For an arbitrary natural numberNwe there-

fore ﬁnd

i≥k+1

N

p_i < N

2. (1)

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6 Six proofs of the inﬁnity of primes LetN_bbe the number of positive integersn≤N which are divisible by at least one big prime, andN_sthe number of positive integersn≤Nwhich have only small prime divisors. We are going to show that for a suitableN

Nb+Ns < N,

which will be our desired contradiction, since by deﬁnitionNb+Nswould have to be equal toN.

To estimateN_b note that ^N_p

icounts the positive integersn ≤ N which are multiples ofp_i. Hence by (1) we obtain

N_b ≤

i≥k+1

N pi

< N

2. (2)

Let us now look atN_s. We write everyn≤N which has only small prime divisors in the formn=a_nb²_n, wherea_nis the square-free part. Everya_n is thus a product ofdifferentsmall primes, and we conclude that there are precisely2^kdifferent square-free parts. Furthermore, asb_n ≤√

n≤√ N, we ﬁnd that there are at most√

Ndifferent square parts, and so N_s ≤ 2^k√

N .

Since (2) holds foranyN, it remains to ﬁnd a numberNwith2^k√ N ≤^N₂ or2^k+1≤√

N, and for thisN = 2^2k+2will do.

Appendix: Inﬁnitely many more proofs

Our collection of proofs for the infinitude of primes contains several other old and new treasures, but there is one of very recent vintage that is quite different and deserves special mention. Let us try to identify sequencesS of integers such that the set of primesPS that divide some member ofS is infinite. Every such sequence would then provide its own proof for the infinity of primes. The Fermat numbers F_n studied in the second proof form such a sequence, while the powers of2don’t. Many more examples

Issai Schur

are provided by a theorem of Issai Schur, who showed in 1912 that for every nonconstant polynomialp(x)with integer coefﬁcients the set of all nonzero values{p(n)= 0 :n∈N}is such a sequence. For the polynomial p(x) = x, Schur’s result gives us Euclid’s theorem. As another example, for p(x) = x²+ 1we get that the “squares plus one” contain inﬁnitely many different prime factors.

The following result due to Christian Elsholtz is a real gem: It generalizes Schur’s theorem, the proof is just clever counting, and it is in a certain sense best possible.

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Six proofs of the inﬁnity of primes 7 LetS= (s₁, s₂, s₃, . . .)be a sequence of integers. We say that

• Sisalmost injectiveif every value occurs at mostctimes for some constantc,

• Sisof subexponential growthif|sn| ≤2²^f(n)for alln, wheref:N→R+

is a function with _log^f(n)

2n →0.

In place of 2 we could take any other base larger than 1; for example,

|sn| ≤ e^e^f(n) leads to the same class of sequences.

Theorem. If the sequenceS = (s₁, s₂, s₃, . . .)is almost injective and of subexponential growth, then the setPSof primes that divide some member ofSis inﬁnite.

Proof. We may assume thatf(n)is monotonely increasing. Otherwise, replacef(n)byF(n) = max_i_≤_nf(i); you can easily check that with this F(n)the sequenceSagain satisﬁes the subexponential growth condition.

Let us suppose for a contradiction thatPS = {p₁, . . . , p_k}is ﬁnite. For n∈N, let

sn =εnp^α₁¹· · ·p^α_k^k, withεn∈ {1,0,−1}, αi≥0,

where theα_i =α_i(n)depend onn. (Fors_n = 0we can putα_i = 0for alli.) Then

2^α¹⁺^···^+α^k ≤ |s_n| ≤2²^f(n) fors_n= 0, and thus by taking the binary logarithm

0≤α_i≤α₁+· · ·+α_k≤2^f(n) for1≤i≤k.

Hence there are not more than2^f(n)+ 1different possible values for each α_i=α_i(n). Sincef is monotone, this gives a ﬁrst estimate

#{distinct|s_n| = 0forn≤N} ≤ (2^f(N⁾+ 1)^k ≤ 2^(f(N^)+1)k. On the other hand, sinceSis almost injective onlycterms in the sequence can be equal to0, and each nonzero absolute value can occur at most2c times, so we get the lower estimate

#{distinct|s_n| = 0forn≤N} ≥ N−c 2c . Altogether, this gives

N−c

2c ≤ 2^k(f(N)+1).

Taking again the logarithm with base2on both sides, we obtain log₂(N−c)−log₂(2c) ≤ k(f(N) + 1) for allN.

This, however, is plainly false for large N, as k andc are constants, so

log₂(N−c)

log₂N goes to1forN→ ∞, while _log^f(N)

2N goes to0.

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8 Six proofs of the inﬁnity of primes Can one relax the conditions? At least neither of them is superﬂuous.

That we need the “almost injective” condition can be seen from sequences S like (2,2,2, . . .)or (1,2,2,4,4,4,4,8, . . .), which satisfy the growth condition, whilePS ={2}is ﬁnite.

As for the subexponential growth condition, let us remark that it cannot be weakened to a requirement of the form _log^f(n)

2n ≤ ε for a ﬁxedε > 0. To see this, one analyzes the sequence of all numbers of the formp^α₁¹· · ·p^α_k^k arranged in increasing order, where p₁, . . . , p_k are ﬁxed primes andk is large. This sequenceSgrows roughly like2²^f(n)with_log^f(n)

2n ≈ ¹_k, whilePS

is ﬁnite by construction.

References

[1] B. ARTMANN:Euclid — The Creation of Mathematics,Springer-Verlag, New York 1999.

[2] C. ELSHOLTZ:Prime divisors of thin sequences,Amer. Math. Monthly119 (2012), 331-333.

[3] P. ERD ˝OS:Über die Reihe₁

p,Mathematica, Zutphen B7(1938), 1-2.

[4] L. EULER: Introductio in Analysin Inﬁnitorum, Tomus Primus, Lausanne 1748; Opera Omnia, Ser. 1, Vol. 8.

[5] H. FÜRSTENBERG: On the inﬁnitude of primes, Amer. Math. Monthly62 (1955), 353.

[6] I. SCHUR: Über die Existenz unendlich vieler Primzahlen in einigen speziellen arithmetischen Progressionen,Sitzungsberichte der Berliner Math.

Gesellschaft11(1912), 40-50.

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Bertrand’s postulate Chapter 2

Joseph Bertrand We have seen that the sequence of prime numbers2,3,5,7, . . .is inﬁnite.

To see that the size of its gaps is not bounded, letN := 2·3·5· · ·pdenote the product of all prime numbers that are smaller thank+ 2, and note that none of theknumbers

N+ 2, N+ 3, N+ 4, . . . , N+k, N+ (k+ 1)

is prime, since for2≤i≤k+ 1we know thatihas a prime factor that is smaller thank+ 2, and this factor also dividesN, and hence alsoN +i.

With this recipe, we ﬁnd, for example, fork = 10 that none of the ten numbers

2312,2313,2314, . . . ,2321 is prime.

But there are also upper bounds for the gaps in the sequence of prime numbers. A famous bound states that “the gap to the next prime cannot be larger than the number we start our search at.” This is known as Bertrand’s postulate, since it was conjectured and veriﬁed empirically forn <3 000 000 by Joseph Bertrand. It was ﬁrst proved for allnby Pafnuty Chebyshev in 1850. A much simpler proof was given by the Indian genius Ramanujan.

Our Book Proof is by Paul Erd˝os: it is taken from Erd˝os’ ﬁrst published paper, which appeared in 1932, when Erd˝os was 19.

Bertrand’s postulate

For everyn≥1, there is some prime numberpwithn < p≤2n.

Proof. We will estimate the size of the binomial coefﬁcient_2n

n

care- fully enough to see that if it didn’t have any prime factors in the range n < p≤2n, then it would be “too small.” Our argument is in ﬁve steps.

(1)We ﬁrst prove Bertrand’s postulate forn≤511. For this one does not need to check511cases: it sufﬁces (this is “Landau’s trick”) to check that

2,3,5,7,13,23,43,83,163,317,521

is a sequence of prime numbers, where each is smaller than twice the pre- vious one. Hence every interval{y:n < y≤2n}, withn≤511, contains one of these11primes.

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10 Bertrand’s postulate (2)Next we prove that

p≤x

p ≤ 4^x⁻¹ for all realx≥2, (1) where our notation — here and in the following — is meant to imply that the product is taken over all primenumbers p ≤ x. The proof that we present for this fact uses induction on the number of these primes. It is not from Erd˝os’ original paper, but it is also due to Erd˝os (see the margin), and it is a true Book Proof. First we note that ifqis the largest prime with q≤x, then

p≤x

p =

p≤q

p and 4^q⁻¹ ≤ 4^x⁻¹.

Thus it sufﬁces to check (1) for the case wherex=qis a prime number. For q= 2we get “2≤4,” so we proceed to consider odd primesq= 2m+ 1. (Here we may assume, by induction, that (1) is valid for all integersxin the set{2,3, . . . ,2m}.) Forq= 2m+ 1we split the product and compute

p≤2m+1

p =

p≤m+1

p·

m+1<p≤2m+1

p ≤ 4^m

2m+ 1 m

≤ 4^m2^2m = 4^2m. All the pieces of this “one-line computation” are easy to see. In fact,

p≤m+1

p ≤ 4^m holds by induction. The inequality

m+1<p≤2m+1

p ≤

2m+ 1 m

follows from the observation that_2m+1

m

= _m!(m+1)!^(2m+1)! is an integer, where the primes that we consider all are factors of the numerator(2m+ 1)!, but not of the denominatorm!(m+ 1)!. Finally

2m+ 1 m

≤ 2^2m holds since

2m+ 1 m

and

2m+ 1 m+ 1

Legendre’s theorem

The number n! contains the prime factorpexactly

k≥1

n p^k

times.

Proof. Exactly_n

p

of the factors ofn! = 1·2·3· · ·nare divisible by p, which accounts for_n

p

p-factors.

Next,_n

p²

of the factors of n!are even divisible byp², which accounts for the next _n

p²

prime factors p

ofn!, etc.

are two (equal!) summands that appear in

2m+1

k=0

2m+ 1 k

= 2^2m+1.

(3)From Legendre’s theorem (see the box) we get that_2n

n

= ^(2n)!_n!n! contains the prime factorpexactly

k≥1

2n p^k

−2 n

p^k

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Bertrand’s postulate 11 times. Here each summand is at most1, since it satisﬁes

2n p^k

−2 n

p^k

< 2n p^k −2

n p^k −1

= 2,

and it is an integer. Furthermore the summands vanish wheneverp^k>2n.

Thus_2n

n

containspexactly

k≥1

2n p^k

−2 n

p^k

≤ max{r:p^r≤2n}

times. Hence the largest power ofpthat divides_2n

n

is not larger than2n. In particular, primesp >√

2nappear at most once in_2n

n

.

Furthermore — and this, according to Erd˝os, is the key fact for his proof

— primespthat satisfy ²₃n < p ≤ ndo not divide_2n

n

at all! Indeed,

Examples such as ₂₆

13 = 2³·5²·7·17·19·23 ₂₈

14 = 2³·3³·5²·17·19·23 ₃₀

15 = 2⁴·3²·5·17·19·23·29 illustrate that “very small” prime factors p < √

2ncan appear as higher powers in _2n

n , “small” primes with √ 2n <

p ≤ ²₃n appear at most once, while factors in the gap with ²₃n < p ≤ n don’t appear at all.

3p > 2nimplies (forn ≥3, and hencep≥3) thatpand2pare the only multiples ofpthat appear as factors in the numerator of^(2n)!_n!n!, while we get twop-factors in the denominator.

(4)Now we are ready to estimate_2n

n

, beneﬁtting from a suggestion by Raimund Seidel, which nicely improves Erd˝os’ original argument. For n≥3, using an estimate from page 14 for the lower bound, we get

4ⁿ 2n ≤

2n n

≤

p≤√ 2n

2n ·

√2n<p≤²₃n

p ·

n<p≤2n

p.

Now, there are no more than√

2nprimes in the ﬁrst factor; hence using (1) for the second factor and lettingP(n)denote the number of primes between nand2nwe get

4ⁿ 2n <

(2n)^√²ⁿ

· 4²³ⁿ

·(2n)^P(n), that is,

4ⁿ3 < (2n)^√^2n+1+P(n). (2) (5)Taking the logarithm to base2, the last inequality is turned into

P(n) > 2n

3 log₂(2n)−(√

2n+ 1). (3)

It remains to verify that the right-hand side of (3) is positive fornlarge enough. We show that this is the case forn= 2⁹ = 512(actually, it holds fromn= 468onward). By writing2n−1 = (√

2n−1)(√

2n+ 1)and cancelling the(√

2n+ 1)-factor it sufﬁces to show

√2n−1 > 3 log₂(2n) forn≥2⁹. (4) For n = 2⁹, (4) becomes 31 > 30, and comparing the derivatives (√

x−1) = ¹₂√¹x and(3 log₂x) = _{log 2}³ ¹_x we see that√

x−1 grows faster than3 log₂xforx >(_{log 2}⁶ )²≈75and thus certainly forx≥2¹⁰=

1024.

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12 Bertrand’s postulate One can extract even more from this type of estimates: Comparing the derivatives of both sides, one can sharpen (4) to

√2n−1 ≥ 21

4 log₂(2n) for n≥2¹¹, which with a little arithmetic and (3) implies

P(n) ≥ 2 7

n log₂(2n).

This is not that bad an estimate: the “true” number of primes in this range is roughlyn/logn. This follows from the “prime number theorem,” which says that the limit

lim

n→∞

#{p≤n:pis prime}

n/logn

exists, and equals1. This famous result was ﬁrst proved by Hadamard and de la Vallée-Poussin in 1896; Selberg and Erd˝os found an elementary proof (without complex analysis tools, but still long and involved) in 1948.

On the prime number theorem itself the ﬁnal word, it seems, is still not in:

for example a proof of the Riemann hypothesis (see page 64), one of the major unsolved open problems in mathematics, would also give a substan- tial improvement for the estimates of the prime number theorem. But also for Bertrand’s postulate, one could expect dramatic improvements. In fact, the following is a famous unsolved problem:

Is there always a prime betweenn²and(n+ 1)²? For additional information see [3, p. 19] and [4, pp. 248, 257].

Appendix: Some estimates

Estimating via integrals

There is a very simple-but-effective method of estimating sums by integrals (as already encountered on page 4). For estimating theharmonic numbers

H_n = n k=1

1 k we draw the ﬁgure in the margin and derive from it

1

1 2 n

H_n−1 = n k=2

1 k <

n 1

1

t dt = logn

by comparing the area below the graph off(t) = ¹_t (1 ≤t≤n)with the area of the dark shaded rectangles, and

H_n−1 n =

n−1

k=1

1 k >

n 1

1

t dt = logn

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Bertrand’s postulate 13 by comparing with the area of the large rectangles (including the lightly

shaded parts). Taken together, this yields logn+ 1

n < H_n < logn + 1.

In particular, lim

n→∞Hn → ∞, and the order of growth ofHn is given by

nlim→∞

H_n

logn = 1. But much better estimates are known (see [2]), such as

Here O ₁

n⁶ denotes a function f(n) such thatf(n) ≤ c_n¹₆ holds for some constantc.

H_n = logn+γ+ 1 2n − 1

12n² + 1 120n⁴ +O

1 n⁶

, whereγ≈0.5772is “Euler’s constant.”

Estimating factorials — Stirling’s formula

The same method applied to

log(n!) = log 2 + log 3 +· · ·+ logn = n k=2

logk yields

log((n−1)!) <

n 1

logt dt < log(n!), where the integral is easily computed:

n 1

logt dt =

tlogt−t n

1 = nlogn−n+ 1.

Thus we get a lower estimate onn!

n! > eⁿ^logⁿ⁻ⁿ⁺¹ = e n

e n

and at the same time an upper estimate

n! = n(n−1)! < neⁿ^logⁿ⁻ⁿ⁺¹ = en n

e n

.

Here a more careful analysis is needed to get the asymptotics ofn!, as given

byStirling’s formula _Here_f(n)_∼_g(n)_{means that}

n→∞lim f(n) g(n) = 1.

n! ∼ √ 2πn

n e

n

.

And again there are more precise versions available, such as n! = √

2πn n

e n

1 + 1

12n+ 1

288n² − 139 5140n³ +O

1 n⁴

.

Estimating binomial coefﬁcients

Just from the deﬁnition of the binomial coefﬁcients_n

k

as the number of k-subsets of an n-set, we know that the sequence _n

0

,_n

1

, . . . ,_n

n

of binomial coefﬁcients

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14 Bertrand’s postulate

• sums to n k=0

_n

k

= 2ⁿ

• is symmetric:_n

k

= _n

n−k

.

From the functional equation_n

k

= ⁿ⁻_k^k+1 _n

k−1

one easily ﬁnds that for everynthe binomial coefﬁcients _n

k

form a sequence that is symmetric andunimodal: it increases towards the middle, so that the middle binomial coefﬁcients are the largest ones in the sequence:

1 1 1

1 1 1 1

1 1 2 3 4 5 6 7

1510 6 3

1 1 4 10 20 35

155 1

1 6

7 1

1 21 35 21 Pascal’s triangle

1 =_n

0

<_n

1

<· · ·< _n

n/2

= _n

n/2

>· · ·> _n

n−1

>_n

n

= 1.

Here xresp.xdenotes the numberxrounded down resp. rounded up to the nearest integer.

From the asymptotic formulas for the factorials mentioned above one can obtain very precise estimates for the sizes of binomial coefﬁcients. How- ever, we will only need very weak and simple estimates in this book, such as the following:_n

k

≤2ⁿfor allk, while forn≥2we have n

n/2

≥ 2ⁿ n, with equality only forn= 2. In particular, forn≥1,

2n n

≥ 4ⁿ 2n. This holds since _n

n/2

, a middle binomial coefﬁcient, is the largest entry in the sequence_n

0

+_n

n

,_n

1

,_n

2

, . . . , _n

n−1

, whose sum is2ⁿ, and whose average is thus ²_nⁿ.

On the other hand, we note the upper bound for binomial coefﬁcients n

k

= n(n−1)· · ·(n−k+ 1)

k! ≤ n^k

k! ≤ n^k 2^k⁻¹,

which is a reasonably good estimate for the “small” binomial coefﬁcients at the tails of the sequence, whennis large (compared tok).

References

[1] P. ERD ˝OS:Beweis eines Satzes von Tschebyschef,Acta Sci. Math. (Szeged)5 (1930-32), 194-198.

[2] R. L. GRAHAM, D. E. KNUTH & O. PATASHNIK:Concrete Mathematics.

A Foundation for Computer Science,Addison-Wesley, Reading MA 1989.

[3] G. H. HARDY& E. M. WRIGHT:An Introduction to the Theory of Numbers, Fifth edition, Oxford University Press 1979.

[4] P. RIBENBOIM:The New Book of Prime Number Records,Springer-Verlag, New York 1989.

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Binomial coefﬁcients

are (almost) never powers

Chapter 3

There is an epilogue to Bertrand’s postulate which leads to a beautiful result on binomial coefﬁcients. In 1892 Sylvester strengthened Bertrand’s postulate in the following way:

Ifn≥2k, then at least one of the numbersn, n−1, . . . , n−k+ 1 has a prime divisorpgreater thank.

Note that forn = 2kwe obtain precisely Bertrand’s postulate. In 1934, Erd˝os gave a short and elementary Book Proof of Sylvester’s result, running along the lines of his proof of Bertrand’s postulate. There is an equivalent way of stating Sylvester’s theorem:

The binomial coefﬁcient n

k

= n(n−1)· · ·(n−k+ 1)

k! (n≥2k)

always has a prime factorp > k.

With this observation in mind, we turn to another one of Erd˝os’ jewels:

When is_n

k

equal to a powerm?

The casek = = 2leads to a classical topic. Multiplying_n

2

= m² by8 and rearranging terms gives(2n−1)²−2(2m)² = 1, which is a special case ofPell’s equation,x²−2y²= 1. One learns in number theory that this equation has inﬁnitely many positive solutions(xk, yk), which are given byxk+yk

√2 = (3 + 2√

2)^kfork≥1. The smallest examples are (x₁, y₁) = (3,2),(x₂, y₂) = (17,12), and(x₃, y₃) = (99,70), yielding ₂

2

= 1²,₉

2

= 6², and₅₀

2

= 35².

Fork = 2and > 2 there are no further solutions, and fork = 3it is known that_n

3

=mhas the unique solutionn= 50,m= 140,= 2, see Gy˝ory [3]. But now we are at the end of the line. Fork≥4and any≥2 no solutions exist, and this is what Erd˝os proved by an ingenious argument.

Theorem. The equation_n

k

= m has no integer solutions with ≥2 and 4≤k≤n−4.

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16 Binomial coefﬁcients are (almost) never powers Proof. Note ﬁrst that we may assumen≥2kbecause of_n

k

= _n

n−k

. Suppose the theorem is false, and that_n

k

= m. The proof, by contradiction, proceeds in the following four steps.

(1)By Sylvester’s theorem, there is a prime factorpof_n

k

greater thank, hencepdividesn(n−1)· · ·(n−k+ 1). Clearly, only one of the factors n−ican be a multiple of any suchp > k, and we concludep|n−i, and therefore

n ≥ p > k ≥ k².

(2) Consider any factor n−j of the numerator and write it in the form n−j =a_jm_j, wherea_jis not divisible by any nontrivial-th power. We note by(1)thata_jhas only prime divisors less than or equal tok. We want to show next thata_i =a_jfori=j. Assume to the contrary thata_i =a_j for somei < j. Thenm_i≥m_j+ 1and

k > (n−i)−(n−j) = a_j(m_i −m_j) ≥ a_j((m_j+ 1)−m_j)

> a_jm_j⁻¹ ≥ (a_jm_j)^1/2 ≥ (n−k+ 1)^1/2

≥ (ⁿ₂ + 1)^1/2 > n^1/2, which contradictsn > k²from above.

(3) Next we prove that thea_i’s are the integers1,2, . . . , kin some order.

(According to Erd˝os, this is the crux of the proof.) Since we already know that they are all distinct, it sufﬁces to prove that

a₀a₁· · ·a_k₋₁ divides k!.

Substitutingn−j=a_jm_jinto the equation_n

k

=m, we obtain

a₀a₁· · ·a_k₋₁(m₀m₁· · ·m_k₋₁) = k!m. Cancelling the common factors ofm₀· · ·m_k₋₁andmyields

a₀a₁· · ·a_k₋₁u = k!v

with gcd(u, v) = 1. It remains to show thatv = 1. If not, thenv contains a prime divisor p. Since gcd(u, v) = 1,pmust be a prime divisor of a₀a₁· · ·a_k₋₁ and hence is less than or equal tok. By the theorem of Legendre (see page 8) we know thatk!containspto the power

i≥1 k pⁱ. We now estimate the exponent ofpinn(n−1)· · ·(n−k+ 1). Letibe a positive integer, and letb₁ < b₂ <· · · < b_sbe the multiples ofpⁱamong n, n−1, . . . , n−k+ 1. Thenb_s=b₁+ (s−1)pⁱand hence

(s−1)pⁱ = b_s−b₁ ≤ n−(n−k+ 1) = k−1, which implies

s ≤ k−1 pⁱ

+ 1 ≤ k pⁱ

+ 1.

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Binomial coefﬁcients are (almost) never powers 17 So for eachi the number of multiples ofpⁱ among n, . . . , n−k+1, and

hence among thea_j’s, is bounded by _p^k_i+ 1. This implies that the exponent ofpina₀a₁· · ·a_k₋₁is at most

−1

i=1

k pⁱ

+ 1

with the reasoning that we used for Legendre’s theorem in Chapter 2. The only difference is that this time the sum stops ati =−1, since theaj’s contain no-th powers.

Taking both counts together, we ﬁnd that the exponent ofpinvis at most

−1

i=1

k pⁱ

+ 1 −

i≥1

k pⁱ

≤ −1,

and we have our desired contradiction, sincevis an-th power.

We see that our analysis so far agrees with₅₀

3 = 140², as 50 = 2·5² 49 = 1·7² 48 = 3·4² and5·7·4 = 140.

This sufﬁces already to settle the case= 2. Indeed, sincek≥ 4one of theai’s must be equal to 4, but theai’s contain no squares. So let us now assume that≥3.

(4)Sincek≥4, we must haveai₁ = 1,ai₂ = 2,ai₃ = 4for somei₁, i₂, i₃, that is,

n−i₁=m₁, n−i₂= 2m₂, n−i₃= 4m₃.

We claim that(n−i₂)² = (n−i₁)(n−i₃). If not, putb = n−i₂and n−i₁=b−x,n−i₃=b+y, where0<|x|,|y|< k. Hence

b²= (b−x)(b+y) or (y−x)b=xy, wherex=yis plainly impossible. Now we have by part(1)

|xy| = b|y−x| ≥ b > n−k > (k−1)² ≥ |xy|, which is absurd.

So we havem²₂ =m₁m₃, where we assumem²₂> m₁m₃(the other case being analogous), and proceed to our last chains of inequalities. We obtain 2(k−1)n > n²−(n−k+ 1)² > (n−i₂)²−(n−i₁)(n−i₃)

= 4[m²₂ −(m₁m₃)] ≥ 4[(m₁m₃+ 1)−(m₁m₃)]

≥ 4m₁⁻¹m₃⁻¹.

Since≥3andn > k≥k³>6k, this yields

2(k−1)nm₁m₃ > 4m₁m₃ = (n−i₁)(n−i₃)

> (n−k+ 1)² > 3(n−n

6)² > 2n².

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18 Binomial coefﬁcients are (almost) never powers Now sincem_i≤n^1/≤n^1/3we ﬁnally obtain

kn^2/3 ≥ km₁m₃ > (k−1)m₁m₃ > n,

ork³> n. With this contradiction, the proof is complete.

References

[1] P. ERD ˝OS:A theorem of Sylvester and Schur,J. London Math. Soc.9(1934), 282-288.

[2] P. ERD ˝OS: On a diophantine equation, J. London Math. Soc. 26 (1951), 176-178.

[3] K. GY ˝ORY:On the diophantine equation _n

k = x^l, Acta Arithmetica 80 (1997), 289-295.

[4] J. J. SYLVESTER:On arithmetical series,Messenger of Math.21(1892), 1-19, 87-120; Collected Mathematical Papers Vol. 4, 1912, 687-731.

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Representing numbers as sums of two squares

Chapter 4

Pierre de Fermat 1 = 1²+ 0² 2 = 1²+ 1² 3 = 4 = 2²+ 0² 5 = 2²+ 1² 6 = 7 = 8 = 2²+ 2² 9 = 3²+ 10 = 3²+ 11 =

... Which numbers can be written as sums of two squares?

This question is as old as number theory, and its solution is a classic in the ﬁeld. The “hard” part of the solution is to see that every prime number of the form4m+ 1is a sum of two squares. G. H. Hardy writes that this two square theoremof Fermat “is ranked, very justly, as one of the ﬁnest in arithmetic.” Nevertheless, one of our Book Proofs below is quite recent.

Let’s start with some “warm-ups.” First, we need to distinguish between the primep = 2, the primes of the formp = 4m+ 1, and the primes of the formp= 4m+ 3. Every prime number belongs to exactly one of these three classes. At this point we may note (using a method “à la Euclid”) that there are inﬁnitely many primes of the form4m+ 3. In fact, if there were only ﬁnitely many, then we could takepk to be the largest prime of this form. Setting

Nk := 2²·3·5· · ·pk−1

(wherep₁= 2,p₂= 3,p₃= 5, . . . denotes the sequence of all primes), we ﬁnd thatNkis congruent to3 (mod4), so it must have a prime factor of the form4m+ 3, and this prime factor is larger thanpk— contradiction.

Our first lemma characterizes the primes for which−1 is a square in the fieldZp (which is reviewed in the box on the next page). It will also give us a quick way to derive that there are infinitely many primes of the form 4m+ 1.

Lemma 1. For primesp= 4m+ 1the equations²≡ −1 (modp)has two solutionss∈ {1,2, . . ., p−1}, forp= 2there is one such solution, while for primes of the formp= 4m+ 3there is no solution.

Proof. Forp= 2takes = 1. For oddp, we construct the equivalence relation on{1,2, . . . , p−1}that is generated by identifying every element with its additive inverse and with its multiplicative inverse inZp. Thus the

“general” equivalence classes will contain four elements {x,−x, x,−x}

since such a4-element set contains both inverses for all its elements. How- ever, there are smaller equivalence classes if some of the four numbers are not distinct:

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20 Representing numbers as sums of two squares

• x≡ −xis impossible for oddp.

• x≡xis equivalent tox²≡1. This has two solutions, namelyx= 1 andx=p−1, leading to the equivalence class{1, p−1}of size2.

• x≡ −xis equivalent tox²≡ −1. This equation may have no solution or two distinct solutionsx₀, p−x₀: in this case the equivalence class is{x₀, p−x₀}.

Forp= 11the partition is {1,10},{2,9,6,5},{3,8,4,7};

forp= 13it is

{1,12},{2,11,7,6},{3,10,9,4}, {5,8}: the pair{5,8}yields the two solutions ofs² ≡ −1 mod 13.

The set{1,2, . . . , p−1}hasp−1elements, and we have partitioned it into quadruples (equivalence classes of size 4), plus one or two pairs (equivalence classes of size2). Forp−1 = 4m+ 2we ﬁnd that there is only the one pair{1, p−1}, the rest is quadruples, and thuss²≡ −1 (modp)has no solution. Forp−1 = 4mthere has to be the second pair, and this contains the two solutions ofs²≡ −1that we were looking for.

Lemma 1 says that every odd prime dividing a numberM²+ 1must be of the form4m+ 1. This implies that there are inﬁnitely many primes of this form: Otherwise, look at(2·3·5· · ·q_k)²+ 1, whereq_kis the largest such prime. The same reasoning as above yields a contradiction.

+ 0 1 2 3 4

0 0 1 2 3 4 1 1 2 3 4 0 2 2 3 4 0 1 3 3 4 0 1 2 4 4 0 1 2 3

· 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 3 1 4 2 4 0 4 3 2 1

Addition and multiplication inZ5

Prime ﬁelds

Ifpis a prime, then the setZp={0,1, . . . , p−1}with addition and multiplication defined “modulop” forms a finite field. We will need the following simple properties:

• Forx ∈ Zp,x = 0, the additive inverse (for which we usually write−x) is given byp−x∈ {1,2, . . . , p−1}. Ifp >2, thenx and−xare different elements ofZp.

• Eachx∈Zp\{0}has a unique multiplicative inversex∈Zp\{0}, withxx≡1 (modp).

The deﬁnition of primes implies that the mapZp →Zp,z →xz is injective forx = 0. Thus on the ﬁnite setZp\{0} it must be surjective as well, and hence for eachxthere is a uniquex = 0 withxx≡1 (modp).

• The squares0²,1²,2², . . . , h²deﬁne different elements ofZp, for h= ^p₂.

This is sincex²≡y², or(x+y)(x−y)≡0, implies thatx≡y or thatx ≡ −y. The1 + ^p₂elements0²,1², . . . , h²are called thesquaresinZp.

At this point, let us note “on the ﬂy” that forallprimes there are solutions for x²+y² ≡ −1 (modp). In fact, there are ^p₂+ 1distinct squares x²inZp, and there are ^p₂+ 1distinct numbers of the form−(1 +y²).

These two sets of numbers are too large to be disjoint, sinceZphas onlyp elements, and thus there must existxandywithx²≡ −(1 +y²) (modp).