A Friendly Problem Book of Elementary Number Theory (With Hints)

XIONG Rui March 2, 2018 0 Preface

In 2017, I was honoured to become a teaching assistant of Elementary Number Theory (for freshmen major in mathematics), as a junior students (grade 3). The text teacher chosen is [30] which introduces a lot of domains of number theory, but in an elementary way. One of the shortcoming of this text is that such a good book covering so much domains is written for non-math-majors. So I was asked to present exercises weekly to supplement the course (Actually, the computer programme assignment takes the most exercise in our text). But almost every book, especially in China, titled “Elementary Number Theory”, has a lot of stupid exercises involving just tricks and tricks. I think too much tricks will separate us from the theory itself. So I want to select some I think “beautiful” exercises. And of course, some of them are required some tricks, as well. When I was a freshman to learn elementary number theory (used another book [18]). I was always frustrated by my unsolved exercises. And some of them can actually not be dealt with by method the book present. Then I loss all my interests in number theory until I learn more deep in Algebra. So, as a result, I give almost all subquestion a hint unless it is literally trivial, to protect the reader from frustrating by actually interesting number theory. Another feature of my selection is to give a lot of problems with deep background, and give suﬃcient reference for readers to broaden his or her horizons. Clearly, some of them is almost unreadable, but you can read a part, even a little∼little part, of their papers. The purpose of my attempt is to make freshmen familiar with the paper reading and critical thinking rather than some delicate tricky exercises. The text covers the discussion of Mersenne primes and perfect numbers. So I don’t view it as an excellent exercise but knowledge. If you are not a Silverman-user, you can read his book for some materials of this. And eventually, English is not my mother tongue, so beaucoup de grammar mistakes would exists, thank you.

Acknowledgement The picture in title is taken from Liangnian http:// www.zcool.com.cn/work/ZNTA3NDAwMA==.html. I want to thank professor Shoumin Liu, the teacher of the course, thank the freshmen major in Mathematics in class of Taishan college, and thank Guoxuan Li, another teaching assistant of our course. Without their support, this booklet could not be accomplished.

XIONG Rui Contents

0 Preface 2

1 Pythagorean Triples 1

2 Linear equation 4

3 Unique factorization 6

4 Congruence and Order 8

5 Chinese Remainder Theorem 12

6 Primitive Element 12

7 Quadratic Reciprocity 15

8 Euler’s ϕ-function 20

9 Arithmetic function 21

10 Sum of two squares 21

11 Diophantine Approximation 22

12 Gaussian Integers 25

13 Irrational number 30

1 Pythagorean Triples

Exercise 1 Prove that every primitive Pythagorean triple (a, b, c) with a odd and b even has the form

a = x2 − y2 b = 2xy c = x2 + y2 where a, b are coprime and consist exactly one odd number and one even number.

1 Exercise 2 (Hippasus) An integer a is a square of rational√ number, show that a is a square of integer number. As a consequence, a is either an irrational number or an integer.

Problem 3 Moreover, prove the following “rational root theorem”:

n p For any polynomial f(X) = anX + ... + a0 ∈ Z[X], if X = ± q is a root of f(X). Assume that p, q coprime, then q | |an|, p | |a0|. And derive (2) using it.

Exercise 4 For θ ∈ [0, π/2), show that

2ab a2 − b2 sin θ, cos θ ∈ ⇐⇒ ∃a, b ∈ , such that sin θ = , cos θ = Q Z a2 + b2 a2 + b2 Give a geometry interpretation.

Exercise 5 For n ≥ 3, their exists a Pythagorean triple (not necessarily primitive) (x, y, z) such that n ∈ {x, y, z}. ( 1 mod 4 n is odd, Exercise 6 Prove that for any n ∈ Z, n2 ≡ . 0 mod 4 n is even.

 1 mod 8 n is odd  Exercise 7 Prove that for any n ∈ Z, n2 ≡ 0 mod 8 n ≡ 0 mod 4.. 4 mod 8 n ≡ 2 mod 4

1Hippasus was a (tragical) Pythagorean philosopher. His discovery of irrational numbers is said to have been shocking to the Pythagoreans, and Hippasus is supposed to have drowned at sea, apparently as a punishment from the gods for divulging this.

1 Problem 8 (Fermat) The purpose of this problem is to show x4 + y4 = z2 has no nontrivial solution. (1)Show that it is without loss of generality that one assume x, y, z pairwise coprime. (Hint: Otherwise, p divides any two elements of {x, y, z}, then it divides the rest one. ) (2)Show that it suffices to deal with the case z is odd. (Hint: When z is even, then by mod 4, one can show that x and y both even. ) Then without loss of generality, one can assume x is odd and y is even, then x2 = u2 − v2 y2 = 2uv z = u2 + v2 With u, v coprime. (3)Show that u is odd and v is even. (Hint: Mod 4.) Then x = s2 − t2 v = 2st u = s2 + t2 with s, t coprime. Then y 2 = st(s2 + t2) 2 (4)Show that s2 + t2 is coprime to s and t respectively. Then s = S2 t = T 2 s2 + t2 = R2 one have S4 + T 4 = R2 √ √ √ But |R| = s2 + t2 = u < 2uv = |y|. Then, use Fermat’s favourite infinite descent. Or, equivalently, one can pick z as small as possible, then when you find another solution with smaller z, you will get a desired contradiction.

Problem 9 (Fermat) The purpose of this problem is to show x4 − y4 = z2 has no nontrivial solution2. (1)Show that it is still without loss of generality that one assume x, y, z pairwise coprime. (2)Show that it suﬃces to deal with the case z is odd. (Hint: When z is even, then z = 2pq y2 = p2 − q2 x2 = p2 + q2 then p4 − q4 = (xy)2 or q4 − p4 = (xy)2. Note that xy is odd. ) Then without loss of generality, one can assume x is odd and y is even, then x2 = u2 + v2 y2 = 2uv z = u2 − v2

2 The sketch of proof follows [14] P391, a little diﬀerent from our text book.

2 With u, v coprime. (3)Show that u is odd and v is even. (Hint: Mod 4.) Then x = s2 + t2 v = 2st u = s2 − t2 with s, t coprime. Then y 2 = st(s2 − t2) 2 (4)Show that s2 − t2 is coprime to s and t respectively. Then s = S2 t = T 2 s2 − t2 = R2 one have S4 − T 4 = R2 √ √ √ But |R| = s2 − t2 = u < 2uv = |y|. Then, use Fermat’s favourite inﬁnite descent. Problem 10 Show that there is no right triangle with rational sides and of 1 a2−b2 area 1. (Hint: It suﬃces to prove that 2 ab 2 fails to be square of rational number. Without loss of generality, assume

2 2 2 ab(a − b ) = c a, b, c ∈ Z with a, b coprime and consisting exactly one odd number and one even number. Then a, b, a+b, a−b are pairwise coprime, then a, b, a+b, a−b are all squares, then use the fact stated in (8) or (9). ) Remark. For which number can be the area of some right triangle with rational sides? Such number is called congruent number, cf [1]. The square-free congruent number less than 50 is given as follow (cf [2]) 5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31, 34, 37, 38, 39, 41, 45, 46, 47,... For example, 1 is not a congruence number, as we presented above. And, 5 is the area of 20/3, 3/2, 41/6-triangle, and 6 is the ares of 3, 4, 5-triangle. See [13] or [15], for a detailed introduction of the problem and its history in elementary number theory level. For another example, The congruent number 53, is the area of k2 − `2, 2k`, k2 + `2-triangle with 1873180325 1158313156 k = ` = 297855654284978790 297855654284978790 it implies the diﬃculty of direct computation to determine whether a number is congruent number. Actually, this problem is deeply connected with elliptic curves, more exactly, the elliptic curves y2 = x3 − n2x.

3 Problem 11 Show that for given n the followings are equivalent (1)There is a right triangle with rational sides of area n. (2)There is a isosceles triangle with rational sides of area 2n. (Hint: To a2+b2−c2 see (2)⇒(1), use the formula S = ab sin C, cos C = 2ab .) (3)Diophantus equation ab(a2 − b2) = nc2 has nontrivial solution. (4)The elliptic curve y2 = x3 − n2x has nontrivial rational point. (Hint: To see (3)⇒(4), show that

n2c2 na3 na = − n2 b2 b b And for any nontrivial rational point (x, y),

x2 − n2 2xn x2 + n2 , , y y y is a desired triangle. )

2 Linear equation

Exercise 12 (n-decimal system) For ﬁxed number r > 1, any positive number n can be written as

m n = amr + ... + a0 uniquely, with 0 ≤ ai < r. It is called the base r expansion of n. (Hint: Induction. )

Exercise 13 Show that the equation

ax + by = g with g = gcd(a, b), has a solution (x0, y0) such that 0 < |x0| < b, 0 < |y0| < a. Exercise 14 Generally, show that the equation

ax + by = g has solution iﬀ gcd(a, b)|g. And if (x0, y0) is one of the solution, any solution is of the form ( b x = x0 + gcd(a,b) t a t ∈ Z y = y0 − gcd(a,b) t

4 Exercise 15 If gcd(a, b) = 1, show that ax + by = n have nonnegative solution, i.e. x, y are both nonnegative, when n > ab−a−b. (Hint: draw a ﬁgure. ) Problem 16 (Euclidean proof of Dirichlet Theorem) It is well-known that Euclid proved that there are inﬁnite primes 3 approximately 2 thousand years ago. The sketch of his proof is

Otherwise, let p1, . . . , pn be all primes. Then P = p1 . . . pn + 1 is coprime to any primes. Thus P is a prime diﬀerent from p1, . . . , pn, by the unique factorization, a contradiction. A famous theorem proved using analytic method by Dirichlet claims

There are infinite many primes in {an + b : n ∈ Z} with a, b coprime. That is, there are infinite many primes p ≡ b mod a. The purpose of this problem is to generalize the ideal of Euclid to give an elementary attempt to Dirichlet’s theorem. (1)Prove that there are infinite many primes p ≡ 3 mod 4. (Hint: Other- wise, let p1, . . . , pn be all primes in the form of 4n+3. Then P = 4p1 . . . pn −1 is coprime to any pi, but P ≡ 3 mod 4, then there exists at least one prime, say p, in the factorization such that p ≡ 1 mod 4.) (2)Prove that there are infinite many primes p ≡ 5 mod 6. (Hint: Oth- erwise, let p1, . . . , pn be all primes in the form of 4n + 1. Then consider P = 6p1 . . . pn − 1.) (3)We will soon see that for any odd prime p, the equation x2 ≡ −1 mod p has a solution iff p ≡ 1 mod 4. Using this fact show that there are infinite 2 primes ≡ 1 mod 4. (Hint: p|4(p1 . . . pn) + 1 ⇒ p ≡ 1 mod 4.) (4)We will see that that for any odd prime p, the the dth cyclotomic polynomial Φd(X) such that

p|Φd(n)for some n ⇒ p6 | n, (p|d or p ≡ 1 mod d)

Show that there are inﬁnite primes ≡ 1 mod d. (Where Φd(n) > n − 1.) (Hint: For every k ≥ 2, p|Φkn(kn) implies p ≡ 1 mod kn, then p is a prime ≡ 1 mod n larger then kn.)

3What he literally said is that ‘prime numbers are more than any assigned multitude of prime numbers.’

5 Remark. If a nonconstant polynomial f(X) ∈ Z[X] satisfy for almost every (that is, except ﬁnite many cases) prime p,

p|f(n) for some n ⇒ p ≡ 1 or b mod a and inﬁnitely many primes of the latter type occur, then f(X) is called an Euclidean polynomial for b mod a. For instance, X + 1 is an Euclidean polynomial for 0 mod 1; 4X − 1 is an Euclidean polynomial for 3 mod 4; 6X − 1 is an Euclidean polynomial for 5 mod 6; 4X2 + 1 is an Euclidean polynomial for 1 mod 4. You can refer to [16] and [27] if you are interested in.

3 Unique factorization

Exercise 17 Show that there exists a prime between n and n!. (Hint: Oth- erwise, consider the product of all primes less than n and add 1.)

Exercise 18 Prove that there always exists a consequence of consecutive numbers of composite for arbitrary length. (Hint: For example, (n+1)!+2,..., (n+ 1)! + (n + 1) serves. )

Problem 19 (p-adic valuation) For a ﬁxed prime p, we deﬁne ordp n, the so-called p-adic valuation, for all n as follow

r r+1 ordp n = r ⇐⇒ p |n, p 6 | n that is, ordp n is the index of p in the decomposition of n. (0)Show that ordp(xy) = ordp(x) + ordp(y) (Hint: Do not expect extravagant praise when you ﬁnd a proof. ) (1)Show that

ordp(x + y) ≥ min(ordp(x), ordp(y)) x + y 6= 0

And when ord(x) 6= ord(y), the equality holds. If we accept the convention ordp 0 = ∞, then we can remove the assumption x + y 6= 0. (2)Show that

ordp(x1, . . . , xn) ≥ min (ordp(xi)) i=1,...,n

6 And when exists i such that j 6= i ⇒ ordp(xj) > ordp(xi), the equality holds. (3)One can extend ordp to Q by a ord = ord a − ord b p b p p show that it is well-deﬁned, and satisfy the condition in (1) and (2).

Problem 20 Show that for n > 1, 1 1 + ... + n fails to be an integer. (Hint: Use the fact stated in (19) where p = 2. Note that there is precisely one term achieving the minimum. )

P∞ Problem 21 (Legendre) In this problem, we will calculate ordp(n!) = i=1 ordp i. (1)Show that ∞ X n ord (n!) = p pi i=1 Pn Pn (Hint: Algebraically, use the identity, i=1 iki = i=1(ki+...+kn), a special case of Abel’s partial summation formula. ) (2)Assume m n = amp + ... + a0 is the base p expansion of n, see (12). Show that

n − (a + ... + a ) ord (n!) = m 0 p p − 1

(3)Without using the meanings of the binomial coeﬃcients, prove that

n! ∀m ≤ n ∈ , ∈ Z≥1 m!(n − m)! Z

n (4)When the binomial coefficient m is odd? Using your skilled computer technique give an diagram of the distribution in Pascal triangle, then you will surprisingly (may be not) find a Sierpińskitriangle. Explain why.

Problem 22 The purpose of this problem is to show that an − bn n|an − bn ⇒ n| a − b

7 (Other proofs are welcomed) (1)Show that when n is a prime p, the assertion holds. (Hint: A well- known identity xn − yn = (x − y)(xn−1 + xn−2y + ... + xyn−2 + yn−1) helps. an−bn n−1 It suﬃces to show when p|a − b, i.e. a ≡ b mod p, then a−b = a + . . . bn−1 ≡ pan ≡ 0 mod p) (2)Show that when n is a power of prime, saying pk, the assertion holds. (Hint: By induction, it still suﬃces to show when p|a − b, i.e. a ≡ b mod p. pi pi Note that a −b ≡ 0 mod p) api−1 −bpi−1 k k n n p m p m pk pk (3)Prove the assertion. (Hint: Let n = pkm, then a −b = (a ) −(b ) a −b a−b apk −bpk a−b reduce to the case of power of prime. )

Exercise 23 Show that, for any polynomial f with integral coefficients and of degree > 1, the set {|f(n)| : n ∈ Z≥0} has infinite composites. (Hint: Firstly, when the constant coefficient of f is not ±1, the exercise is done. Otherwise, note that the constant coefficient of f(x + k) cannot be ±1 invariably. )

4 Congruence and Order

Exercise 24 Show that p p| ∀1 ≤ i ≤ p − 1 i As a consequence, (x + y)p ≡ xp + yp mod p Derive Fermat’s little theorem by this fact. And show that

x 7→ xp

p p p p p is bijection on Fp. (Hint: If x = y , then 0 = x − y = (x − y) , then x − y = 0.)

Remark. The mapping x 7→ xp is called Frobenius automorphism which is important in theory of ﬁnite ﬁelds.

Exercise 25 (Lucas) 4 For any integers m, n, and prime p, show that ∞ m Y mi ≡ mod p n n i=0 i

4See [8]

8 Where ∞ ∞ X i X i m = mip n = nip mi, ni ∈ {0, 1, . . . , p − 1} i=0 i=0 are the p expansion of m and n respectively. see (12). And we accept the convention 0 m = 1 m < n ⇒ = 0 0 n m (Hint: We will expand (1 + X) in Fp as follow

(1+X)m = (1+X)m0+pm1+... = (1+X)m0 (1+X)m1p ... = (1+X)m0 (1+Xp)m1 ... by (24). And in (1 + X)m0 , the degree of each monomials other then 1 is < p, and in (1 + Xp)m1 , the degree of each monomials other then 1 is ≥ p and < p2, and so on. Then, a easy calculation shows that the coeﬃcient of Xn is Q∞ mi, the proof is complete. ) i=0 ni Exercise 26 Prove the following facts about order where ord a denote the order of a modx with (a, x) = 1. (1)ord a = 1 ⇐⇒ a ≡ 1 mod n. (Hint: a fortiori. ) (2)ord a = n, then am = 1 ⇒ n|m. k n k m (3)ord a = n, then ord a = (n,k) . (Hint: Assume (a ) ≡ 1 mod x, then n k n n|km, then (n,k) | (n,k) m, then (n,k) | m.) (4)ord a = st, then ord as = t. (5)ord a = m, ord b = n, then ord ab|mn and when (m, n) = 1, the equality holds. (Hint: If (ab)s = 1, then (ab)sm = 1, then bsm = 1, then n|sm, then n|s when (m, n) = 1.) (6)ord a = m, ord b = n, then there exists an integer c such that ord c = mn. (Hint: Firstly, prove that there exists p, q such that (p, q) = 1, pq = [m, n], p|m, q|n, then use (4) and (5). )

Problem 27 This problem is on Fermat’s primes. (1)If 2m + 1 is a prime, show that m is a power of 2. 2n (2)Let Fn = 2 + 1, it is called Fermat number, show that Fn|Fm − 2 under the assumption m > n. (3)Show that for m 6= n, (Fn,Fm) = 1, then derive another proof of the inﬁnity of primes. (Hint: By (2), (Fn,Fm) ∈ {1, 2}. The stale assertion of inﬁnity of primes follows by the stale fact that Fn+1 = Fn ...F0 + 2.) n+1 (4)If a prime p|Fn, show that p is in the form of t2 + 1. (Hint: That n is 22 ≡ −1 mod n, then 2n+1|p − 1 by Euler’s theorem. )

9 (5)Show that all Fn are square-free. (Hint: You can read [19] page 9 for some help. )

Remark. The primes in {Fn}n≥0 are called Fermat’s primes. Up to today, the only known Fermat primes are

F0 = 3 F1 = 5 F2 = 17 F3 = 257 F4 = 65537 Fermat conjectured (but admitted he could not prove) that all Fermat numbers are prime. Euler found F5 = 4294967297 = 641 × 6700417, and proved (4) in the above problem. Exercise 28 Determine the last two digits of 3400 in decimal base. (Hint: 3400 ≡ 30 = 01 mod 100, since ϕ(100) = 40.) Exercise 29 Show that for n ≥ 3, ϕ(n) is even. (Hint: Pair. ) Problem 30 (Pick’s Theorem) 5 The purpose of this problem is to prove the famous theorem in combinatorics Given a simple polygon S constructed on lattice of integral points, that is, with integer coordinates, then b A = i + − 1 2 where A is the area of S, i is the number of lattice points in the interior of S, b is the number of lattice points on the boundary of S. For example, 8 S = 7 + − 1 = 10 2 (1)Show the equality holds for any m, n coprime.

n−1 m−1 X jm k X j n k (m − 1)(n − 1) i = i = n m 2 i=0 j=0 Please do not cheating by draw a ﬁgure. (Hint: Through direct calculate

Pn−1 m 1 Pn−1 x i=0 n i = n i=0 [im − (im mod n)] ∵ x = an + b, 0 ≤ b < n ⇒ b = a 1 Pn−1 1 Pn−1 = n i=0 im − n i=0 i m−1 n(n−1) 1 n(n−1) = n 2 = n 2 5You can refer [9] for more information.

10 n The equality of second line holds since {im}i=1 go through the remainers mod n.) (2)Give a geometry explanation of (1). (Hint: The number of integral points in a given right triangle. ) (3)Prove Pick’s Theorem.

Problem 31 For any odd prime p, if a nonempty subset T ⊆ Fp, satisfy that a + b a, b ∈ T, ∈ T 2 show that T is of one element or the entire Fp. (Hint: Actually, for any a ∈ Fp, T −a satisﬁes the property as well. Then when |T | ≥ 2, we can assume 0 ∈ T . And what’s more, we can show that for any n, and 0 ≤ m ≤ 2n m a ∈ T, a ∈ T 2n Since, by (12),

a0 a +...+ 2 m a 2n−1 + ... + a a + n−1 ... = n−1 0 = n−1 2 2n 2n 2 But 2n ≡ 1 mod p for some n by Fermat’s little theorem. Then so 22n ≡ 23n ≡ ... ≡ 1 mod p, as a result, for any m ≥ 0, a ∈ T, ma ∈ T

If |T | ≥ 2, and 0 ∈ T , one can take a a 6= 0, then ma go through Fp.)

11 Remark. The problem implies that the convex set in Fp is no more than one-point set and the entire set.

5 Chinese Remainder Theorem

Exercise 32 Show that the following equations

x ≡ b1 mod m1 x ≡ b2 mod m2 have solutions iﬀ (m1, m2)|(b1 − b2).

Exercise 33 For a, b coprime, given a list of finite primes {pi}, show that there are infinity many number coprime to all of pi, of the form an + b. Warning. If you exploit Dirichlet theorem, please prove it first. (Hint:

Just consider the equations x ≡ b mod a, x ≡ b mod p1 . . . pn.) Exercise 34 Prove that there always exists a consequence of consecutive num- 2 bers without square-free for arbitrary length. (Hint: x ≡ 0 mod p1, x ≡ −1 2 mod p2 etc. ) Exercise 35 For coprime m, n, one have mϕ(n) + nϕ(m) ≡ 1 mod mn (Hint: It is equivalent to mod m and mod n by CRT)

6 Primitive Element

Exercise 36 (Wilson’s theorem) For any prime p, Show that (p − 1)! ≡ −1 mod p (Hint: There are beaucoup de proofs of this argument. The most elementary method is to pair the element with its inverse, then only 1 and −1 can not be paired with a distinct element. The easiest proof is use the primitive element’s argument, then

2 p−1 p(p−1) p−1 (p − 1)! ≡ gg . . . g = g 2 ≡ g 2 mod p

p−1 Since ord g = p − 1, thus g 2 ≡ −1 mod p. The last proof is from the polynomial identity xp−1 − 1 = (x − 1) ... (x − (p − 1)) in Fp, then the assertion follows by Vilta’s theorem. )

12 Exercise 37 For odd prime p, the n-th Newton’s summation of Fp, ( n n n −1 mod p p − 1|n Sn = 1 + 2 + ... + (p − 1) ≡ 0 mod p p − 16 | n

(Hint: Use the primitive element’s argument,

p−1 p−1 ( n X n X nk −1 g ≡ 1 i.e. p − 1|n S = i ≡ g ≡ np mod p n g −1 ≡ 0 otherwise i=1 k=1 gn−1 Where used the formula you must know in high school. )

m Problem 38 (Chevalley-Warning theorem) Let {fi}i=1 ⊆ Fp[X1,...,Xn] Pm be polynomials in n variables such that i=1 deg fi < n, deﬁne

n V = {(x1, . . . , xn) ∈ Fp :, fi(x1, . . . , xn) = 0 for any i = 1, . . . , m} Show that p divides the cardinality of V . (Hint: One can construct

m Y p−1 P (X1,...,Xn) = (1 − fi ) i=1 One can check that

fi(x1, . . . , xn) = 0 for any i = 1, . . . , m ⇐⇒ Pi(x1, . . . , xn) ≡ 1

fi(x1, . . . , xn) 6= 0 for some i = 1, . . . , m ⇐⇒ Pi(x1, . . . , xn) ≡ 0 Pm But deg P = i=1 deg fi < n, then X #V ≡ P (x1, . . . , xn) n (x1,...,xn)∈Fp

You will show the right hand side equals to 0 using (37). )

Remark. A remarkable trick in number theory when the problem involves counting V ⊆ A, or showing V 6= ∅ is that one can construct a ( 1 x ∈ V “characteristic function” χ(x) = (In the above problem, χ(x) 0 x ∈ A \ V takes to be P ). Then X #V = χ(x) x∈A

13 Sometimes, χ(x) has some nice formation. For example, one can utile the identity ( Z 1 1 a = b e(ax)e(bx)dx = δab = a, b ∈ Z≥0 0 0 a 6= b or 1 X i i e b e b = δ n n n ab i∈Z/nZ where e(x) = e2πix. Since e(x) has a nice analytic property, one can estimate the order of it. because of this typical example, the trick is called “exponential sums”.

Exercise 39 (Wolstenholme) Show that when prime p > 3, 1 1 ord 1 + + ... + ≥ 2 p 2 p − 1

Pp−1 1 1 Pp−1 1 (Hint: Pair them. More exactly, 2S = i=1 i + p−i = p i=1 i(p−i) . It Pp−1 1 suﬃces to show ordp i=1 i(p−i) ≥ 1. But

p−1 p−1 p−1 p−1 X 1 X 1 X 1 X ≡ ≡ − ≡ − i2 ≡ 0 mod p i(p − i) i(−i) i2 i=1 i=1 i=1 i=1 the proof is complete. Where using (37) or the formula of summation of squares. )

Remark. The readers may complain why one can mod p for a rational a a number. Actually, for b with p - b, then one can write b ≡ ac mod p, where c is the unique number mod p with bc ≡ 1 mod p. More algebracally, we can module p over more big ring than Z, na o = : p b = {x ∈ : ord x ≥ 0} Z(p) b - Q p Then npa o p = : p b = {x ∈ : ord x ≥ 1} Z(p) b - Q p we say x ≡ y mod p ⇐⇒ x − y ∈ pZ(p) Then it is not diﬃcult to built up the theory of congruence theory of it.

14 Problem 40 (Why 142857) A well-known observation (maybe in primary school) is 142857 × 1 = 142857 142857 × 2 = 285714 142857 × 3 = 428571 142857 × 4 = 571428 142857 × 5 = 714285 142857 × 6 = 857142 (1)Prove that 142857 × 7 = 999999 and 1 = 0.14285˙ 7˙ 7 (2)Show that a . . . a 0.a˙ a ... a˙ = 1 n 1 2 n 10n − 1 (Hint: Up to now, you must get to know why two deﬁnitions of rational number coincide. ) (3)What’s the length of the minimal period of 1/n in base m expansion? 1 ` (Hint: That is, the minimal number k, such that n = mk−1 , for some ` ∈ Z. That is the minimal number k, such that mk ≡ 1 mod n, i.e. ord m in mod n) (4)Explain why 142857. (Hint: Because, simply, 10 is a primitive element for 7.)

7 Quadratic Reciprocity

Exercise 41 Show that in the sense mod a prime p, any integer is a sum of two squares. That is, ∀n ∈ Z,

2 2 ∃x, y ∈ Z, n ≡ x + y mod p

2 p−1 p+1 (Hint: One know that |{x mod p}| = 2 + 1 = 2 , then by pigeonhole principal, {n − x2 mod p} ∩ {x2}= 6 ∅.) Problem 42 Show that there are inﬁnite many primes of the form 2 (1)8k + 3. (Hint: Consider m = (p1 . . . pn) + 2 ≡ 3 mod 8, for any p 6≡ ±1 mod 8 but p|m, one must have p ≡ 3 mod 8.)

15 2 (2)8k + 5. (Hint: Consider m = (p1 . . . pn) + 4 ≡ 5 mod 8, for any p 6≡ ±1 mod 8 but p|m, one must have p ≡ 3 mod 8.) 2 (3)8k + 7. (Hint: Consider m = (p1 . . . pn) − 2 ≡ 7 mod 8, for any p|m, p ≡ ±1 mod 8, but p can not always equiv to 1.) 4 (4)8k + 1. (Hint: Consider m = 16(p1 . . . pn) + 1 ≡ 1 mod 8. If p|m, then X4 ≡ −1 mod p has solutions, i.e. there exists integer of order 8, then 8|p − 1, then p ≡ 1 mod 8.)

Exercise 43 For any prime p > 2, show that for any a, b ∈ Z (X2 − a)(X2 − b)(X2 − ab) ≡ 0 mod p always has solution.

4 Exercise 44 Show that for any prime p, X + 1 is reducible over Fp. More precisely, X4 + 1 = (X2 + aX + b)(X2 + cX + d)

−1 2 −2 for some a, b, c, d ∈ Fp. (Hint: Since p p = p , one of Legendre symbols equals to 1, thus one of the following decomposition is over Fp  √ √ √ X4 + 1 = (X2)2 − ( −1)2 = (X2 + −1)(X2 − −1)  √ √ √ X4 + 1 = (X2 + 1)2 − ( 2X)2 = (X2 + 2X + 1)(X2 − 2X + 1) √ √ √ X4 + 1 = (X2 − 1)2 − ( −2X)2 = (X2 + −2X − 1)(X2 − −2X + 1)

)

Remark. When deciding a polynomial f(X) over Z to be irreducible or not, a very useful observation is that if f(X) is reducible, then f(X) is clearly reducible over any Fp. In other word, if f(X) is irreducible over some of Fp, then f(X) is reducible. It is called “the principal of locality”. The example of above exercise is the most famous example that “the prin- 4 cipal of locality” fails.√ √ Firstly, X + 1 is clear an irreducible over Z, since 4 Q 2 2 X + 1 = (X ± 2 ± 2 ). But, secondly, it always reducible over Fp. Problem 45 We will show that f(X) = (X2 − 2)(X2 − 7)(X2 − 14) always has zeros over any Z/nZ. (1)Show that for prime p, and p - a, if there exists solutions for x2 ≡ a mod p then, show that, so does x2 ≡ a mod pk

16 2 for any k ∈ Z≥1. (Hint: Using induction, for example, solve (x + py) ≡ a mod p2, a equation of y.) (2)Show that for any prime p, positive integer k, f(X) has zeros over k 2 7 14 Z/p Z. (Hint: Since p p = p , one of Legendre symbols equals to 1. And for p = 2, X2 − 7 has zeros, for p = 7, X2 − 2 has zeros. ) (3)Show that for any nature number6 f(X) has zeros over Z/nZ. (Hint: Chinese remainders theorem. )

Remark. This problem indicates that “the principal of locality” of zeros fails.

Problem 46 Fibonacci sequence cannot be a more famous sequence in Math- ematics. Let {Fn}, F1 = F2 = 1,Fn+2 = Fn + Fn−1,

Fn : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,... Show that

• When p ≡ ±1 mod 5, p|Fp−1;

• When p ≡ ±2 mod 5, p|Fp+1. (Hint: It is widely known that " √ !n √ !n# 1 1 + 5 1 − 5 Fn = √ − 5 2 2

thus √ √ 1+ 5 p 1− 5 p 1 2 2 F ≡ √ √ − √ p−1 5 1+ 5 1− 5 2 2 √ p √ p ≡ 1 √ 1+ √5 − 1− √5 2p−1 5 1+ 5 1− 5 p−1 1−5 2 ≡ 2p−2(1−X2) mod p where used (24) For the case of p ≡ ±2 mod 5, the argument is similar,

p−1 1 + 5 2 F ≡ mod p p+1 2p p−1 when p ≡ ±1 mod 5, 5 is a square in Fp, thus, by Euler’s citizen, 5 ≡ 1 mod p, then Fp−1 ≡ 0. Otherwise, p ≡ ±2 mod 5, 5 is not a square in Fp, p−1 thus 5 ≡ −1 mod p, then Fp+1 ≡ 0.)

6positive integer

17 Remark. You may ask why the hint can calculate a number in Fp such arbitrarily. Actually, You can set a rational fraction 1 1 + X n 1 − X n f (X) = − ∈ (X) n X 2 2 Z then by putting them into Fp[X], one can get the desired formula such as 1 + Xp−1 f (X) = p+1 2p 2 then one can take√ X = 5 in. More precisely, we were actually consider the number in Fp[ 5], where √ ( Fp p ≡ ±1 mod 5 Fp[ 5] = 2 Fp[X]/(X − 5) p ≡ ±2 mod 5 Problem 47 Now, we will decide the cardinality of quadratic curve over the plane Fp, where p is an odd prime. (1)Show that the number of the solution(s) of X2 ≡ a mod p equals to a 0 p + 1. (Where we accept the convention p = 0.)

• Now we will consider the curve

2 2 2 γ = {(X,Y ) ∈ Fp : aX + bY + cXY + DX + EY + F = 0} A consequence of analytic geometry claims that any quadratic equation in two variables are in not more than the following three forms up to a invertible transform (thus, do not change the cardinality) 1. parabola X2 + bY = c. 2. ellipse or hyperbola X2 + bY 2 = c 3. line X + bY = c.

(2)Show that the line X + bY = c is of cardinality p. (3)Show that the parabola X2 + bY = c is of cardinality

p−1 X c − bY + 1 = p p Y =0 • For ellipse or hyperbola case, we ﬁrstly deal with easy case X2 − Y 2 = a 6= 0 and X2 − Y 2 = 0.

18 (4)Show that U 1 1 X = V 1 −1 Y is invertible. (5)Show that the hyperbola X2 − Y 2 = a 6= 0 is of cardinality

p−1 X a − Y 2 + 1 = p − 1 p Y =0 and the double line X2 − Y 2 = 0 is of cardinality 2p − 1. (It is amazing that the former result does not depend on a.) • In general case, it suﬃces to calculate p−1 X f(i) p i=0 for some quadratic function f(n) = an2 + bn + c where a 6= 0. Let d = b2 − 4ac be its discriminant. (6)Show that  p−1 a X f(i) − p d 6= 0, i.e. p6 | d, = p a i=0 (p − 1) p d = 0, i.e. p|d. (7)Conclude that the hyperbola X2 + bY 2 = c is of cardinality

 −b p + p bc 6= 0, −b p + (p − 1) p bc = 0. (8)In particular the circle X2 + Y 2 = a is of cardinality p + 1 a 6= 0, p ≡ 1 mod 4,  p − 1 a 6= 0, p ≡ 3 mod 4, 2p − 1 a = 0, p ≡ 1 mod 4,  1 a = 0, p ≡ 3 mod 4.

−1 √ Remark. When p = 1, that is, p ≡ 1 mod 4, then −1 ∈ Fp, which −1 means the geometry is similar to C. And when = −1, that is, p ≡ 3 √ p mod 4, then −1 ∈/ Fp, which means the geometry is similar to R. You may understand the results above in some way by this principal.

19 8 Euler’s ϕ-function

Exercise 48 Show that ϕ(n) lim = 1 n→∞ n Exercise 49 Show that

∀n ∈ N, #{k ∈ N : ϕ(k) = n} < ∞ (Hint: Note that ϕ(k) = k Q(1 − 1/p) ≥ Q p(1 − 1/p) = Q(p − 1), then the possible primes divides k is of ﬁnite number, then ϕ(k) = k Q(1 − 1/p) ≥ k Q(1 − 1/P ) with the largest possible prime P , then k is ﬁnite. )

Exercise 50 Show that ϕ(k) n ∀n, m ∈ , # (k, `): = = ∞ N ϕ(`) m

(Hint: Consider (nama+1, na+1ma).)

Exercise 51 Show that

∀k ∈ N, ∃n ∈ N, such that ϕ(n) = ϕ(n + k) (Hint: Let p be the lest prime satisfying p 6n, then n = (p − 1)k holds. )

Remark. Why I don’t ask for a proof of #{n ∈ N : ϕ(n) = ϕ(n + k)} = ∞? It is simply because it is still an unsolved problem, cf [19], B36 Page 90. And there are series of unsolved problem on ϕ. Such as

Are there any k ∈ N such that ϕ(n) = k has only one solution? And Are there any composite n such that ϕ(n)|n − 1? You may interest in [12] and [22].

Exercise 52 Find all n ∈ N, such that ϕ(n) is a power of prime, at least the all that human nowadays knows. (Hint: Recall that ϕ(n) is even when n ≥ 3. That is, to solve ϕ(n) = Q(p − 1)pk−1 = 2m, then p − 1 = 2` for some ` and with related k = 1 or p = 2. Thus p = 2` + 1 is a Fermat’s prime. See remark after (27) Therefore, all n such that ϕ(n) is a power of prime, is of the form k 2 d1 . . . d`, where every di is a Fermat’s prime. )

20 9 Arithmetic function

Exercise 53 Show that X µ2(d) n X µ(d) ϕ(n) = = ϕ(d) ϕ(n) d n d|n d|n

(Hint: Check it when n is a power of prime. )

Exercise 54 Show that for any arithmetic function f X X f(d) = µ(d)Sd Sd = f(d) + f(2d) + ... + f(n) d∈[1,n],(d,n)=1 d|n

10 Sum of two squares

Exercise 55 Show that for prime p the solution to the equation x2 + y2 = p is unique up to sign and transposition. (Hint: Let (x, y) and (X,Y ) be two positive solutions. Let c be the solution of c2 + 1 = 0 mod p in [0, p/2], then x ≡ cy mod p or x ≡ −cy mod p, then

p2 (a2 + b2)(A2 + B2) aA − bB 2 aB + bA2 aB − bA2 aA + bB 2 1 = = = + = + p2 p2 p p p p both brackets in one of last or last but the last are integers. )

Exercise 56 Show that the circle x2 +y2 = p consists no rational point when p ≡ 3 mod 4 and consists inﬁnite rational points when p ≡ 1 mod 4. (Hint: For the assertion of inﬁnity, use the fact for x2 + y2 = 1.)

Problem 57 Let h(n) be the number of solutions of x2 + y2 = n. Show that X √ h(n) = πx + r(x) |r(x)| < 12 x n

P 2 2 (Hint: n

Remark. General result of the problem of lattice point inside a curve is considered by Jarnlk, see [20] page 124.

21 11 Diophantine Approximation

Problem 58 The purpose of this problem is to show the following statement which implies the index 2 is best in Dirichlet’s theorem.

For any > 0, almost all of x ∈ R do not satisfy the property p that there exists inﬁnitely many fractions q with relation prime integers p, q such that

p 1 x − ≤ q q2+ More precisely, we will show for any > 0 ( )! ∃ inﬁnite p, q ∈ Z, (p, q) = 1 µ x ∈ : = 0 R such that x − p ≤ 1 q q2+

where µ is the standard Lebesgue measure over R. 7 The process used follows [31] page 42 exercise 16 and page 46 problem 1. ∞ (1)Suppose {En}n=1 is a countable family of subsets of R, show that ∞ ∞ \ [ lim sup En := Ek = {x ∈ R :#{n : x ∈ En} = ∞} n=1 k=n (Hint: A set-theory fortiori shows

E = {x ∈ R : ∀n > 0, ∃k > n, such that x ∈ Ek} Another set-theory fortiori gives

∞ ∞ ∞ \ \ [ E = {x ∈ R : ∃k > n, such that x ∈ Ek} = {x ∈ R : x ∈ Ek} n=1 n=1 k=n this proves the assertion. )

Assume, in addition to (1), En = (an1, bn2) ∪ ... ∪ (anmn , bnmn ) such that

∞ ∞ m X X Xn µ(En) = |bni − ani| < ∞ n=1 n=1 i=1

7You can now just think of the Lebesgue measure as a map assigning ‘measurable’ subset of R to its ‘length’. For example, µ((0, 1)∪[2, 3]) = 2. In this problem, you are not assumed to be equipped with any knowledge of this.

22 then ∞ ∞ ! ∞ ! ∞ \ [ [ X µ (lim sup En) = µ Ek = lim µ Ek ≤ lim µ (Ek) = 0 n→∞ n→∞ n=1 k=n k=n k=n

This shows that almost none of x appear inﬁnite times in En. This is known as Borel-Cantelli lemma. (2)Prove the statement. (Hint: Fix x ∈ [0, 1], and take

q [ i 1 i 1 E = − , + q q q2+ q q2+ i=0

∞ X 1 Then use Borel-Cantelli lemma. A fact you must know is < ∞.) q1+ q=1

Exercise 59 Hurwitz theorem claims8 that for any irrational number x, there are inﬁnite fractions p/q with (p, q) = 1 such that

p 1 − x < √ q 5q2 √ √ 5−1 We will show that 5 is best by show x = 2 such that p δ 1 x = + |δ| < √ q q2 5 has only ﬁnite solutions. (1)Show that δ2 √ − 5δ = pq − q2 + p2 q2

2 √ (2)prove the assertion. (Hint: when q 0, δ − 5δ < 1.) q2

Problem 60 When a sequence {xn} dense? (1)show that there always exists a subsequence {ak} of

{nγ − bnγc : n ∈ Z} such that ak → x. Or, equivalently, {nγ − bnγc : n ∈ Nγ} = [0, 1]. (Hint: Use Dirichlet’s argument. )

8see, for example [20] or [7]

23 (2)For any γ irrational, for any x ∈ [−1, 1] prove that there exists a subsequence {bk} of {sin 2πγn : n ∈ N} such that bk → x. (Hint: Let ak = nkγ − bnkγc be a sequence in (1) such that 2πnk → arcsin x.) 9 (3) Show that, for any x ∈ [0, 1] there always exists a subsequence {cn} of 1 1 1 1 1 + + ... + − 1 + + ... + : n ∈ 2 n 2 n N 1 1 such that ck → x. (Hint: Denote Sn = 1 + 2 + ... + n . Note that Sn goes to 1 1 inﬁnity slowly. A fact that n → 0 means for any , for n 0, n < /2. But

1 1 |Sn+k − Sn| = + ... + → ∞ k → ∞ n + 1 n + k then for k 0, 1 < Sn+k − Sn. Assume x + h ∈ (Sn,Sn+k) for some h ∈ Z. Then there exists n+1 ≤ N ≤ n+k −1 (since, for example, take the minimal number N such that x + h ∈ (Sn,Sn+N+1)), such that

SN−1 ≤ x + h ≤ SN+1

Thus 1 1 x + h − 1 + + ... + ≤ |SN+1 − SN−1| ≤ 2 N Then take 0 < 1, then we can assume that

h ≤ SN−1 ≤ x + h ≤ SN+1 < h + 1

Then 1 1 1 1 x − 1 + + ... + + 1 + + ... + ≤ 2 N 2 N This is long desired result. ) (4)Show that there are some number x ∈ [0, 1] such that there does not exists a subsequence {dn} of ( √ !n $ √ !n% ) 5 + 1 5 + 1 − : n ∈ 2 2 N

9This subquestion essentially has nothing to do with number theory (in narrow sense), so you can ﬁgure out it in the margin of your homework paper (If you write in a very tiny font and do not complain with the small margin like Fermat).

24 such that dn → x. (Hint: When n 0, show that $ √ !n% √ !n √ !n 5 + 1 1 + 5 1 − 5 = + (−1) 2 2 2

√ n √ n √ n 5+1 j 5+1 k 1− 5 Then, 2 − 2 = − 2 (+1) → 0, 1.)

Remark What is the distribution of a sequence {xn}? We say {xn} ⊆ [0, 1] is equidistributive when for any 0 ≤ a ≤ b ≤ 1, #{1 ≤ r ≤ n : a ≤ x ≤ b} lim r = b − a n→∞ n

That is, the probability of {xn} appear in [a, b] is b − a. Weyl shows that #{1 ≤ r ≤ n : a ≤ rγ − brγc ≤ b} lim = b − a n→∞ n which is surprisingly proved by means of Fourier Analysis10. But up to now, whether {γn − bγnc} is equidistributive is still an open problem.

12 Gaussian Integers

Exercise 61 Get the very ﬁrst statement on Pythagorean triples from the ring of Gaussian integers. (1)Show that, in the ring Z[i], the relation αβ = γn for some α, β rela- tively prime numbers and a unit, implies α = 0ξn and β = 00ηn, with ; , 00 units. (2)Show that the integer solutions of the equation

x2 + y2 = z2 such that x, y, z > 0 and (x, y, z) = 1 (“Pythagorean triples”) 11 are all given, up to possible permutation of x and y, by the formulæ

x = u2 − v2 y = 2uv z = u2 + v2 where u, v ∈ Z, u > v > 0, (u, v) = 1, u, v not both odd. 10see, for example, [32] page 105. 11this exercise is taken from [28] page 5, so the notation in this exercise may not suitable for ours.

25 Problem 62 (Congruence in Gaussian integer ring) We will now study the congruence in Z[i], for a Gaussian integer π, say α ≡ β mod π ⇐⇒ π|α − β It is a routine to check that ≡ is an equivalent relation (if you know). Then we can divide Z[i] into [ Z[i] = α + (π) α∈Z[i] where α + (π) := {α + πβ : β ∈ Z[i]} It deﬁnes a quotient ring

Z[i]/(π) = {α + (π): α ∈ Z[i]} (1)The first problem is for a fixed π = a+bi, how many different equivalent class? That is #{α + (π): α ∈ Z[i]} (Hint: See the figure as follow The number equals to N(π) = a2 + b2. You

b a

should prove that when a, b are diﬀerent points in the square π|a − b implies a = b.) (2)Show that, when π = p with p a prime number ≡ 3 mod 4, show that

Z[i]/(π) = {a + bi : a, b ∈ Z/pZ} (3)Show that, when π = a + bi with a2 + b2 is a prime ≡ 1 mod 4, show that

Z[i]/(π) = {0 + (π), 1 + (π), 2 + (π), 3 + (π), 4 + (π), . . . , p − 1 + (π)}

26 Exercise 63 Show that, for any square-free integer D, Let √ √ Z[ D] := {m + n D : m, n ∈ Z} Show that any element can be written as a product of irreducible elements, where irreducible elements stands for any element can NOT be written as a product of two nonunit elements. (Hint: For any element r, if r is irreducible, then r = r is desired. Otherwise, r = r1r2, if r1 and r2 is irreducible, then r = r1r2 is desired, . . . . Since |N(r)| = |N(r1)| · |N(r2)| such that |N(r)| decrease, thus, the process is terminal. ) √ Exercise√ 64 Show that Z[ 2] is a Euclidean ring. (Hint: Use the “norm” N(a + b 2) = |a2 − 2b2|.)

Remark. It is a diﬃculty to determine whether a ring is Euclidean. The only trick of our exercise is to take the norm to be the Euclidean value. There exists, of course, a ring is Pid but Euclidean, see, for example, [11]. And, you may be content of a throughout review on Euclidean and the history of the research on it, cf [23].

12 Problem 65 Explain the following ‘proof√ without words’ of Euclidean di- −1+ −3 13 vision in Z[i]. And, assuming ρ = 2 , show that

a r

12If you ignore the letters a, b, r.

27 Z[ρ] = {a + bρ : a, b ∈ Z} is an Euclidean domain. They are called Eisenstein integers, see [3]. (Hint: )

Problem 66 (The false proof of Fermat’s Last Theorem) Let’s go back to 1847, 7 years later after his giving his correct proof of the Fermat’s Last Theorem of the case n = 7, Lam´eclaimed that he proved the Fermat’s assertion, cf [21]. The proof is as following14. 2πi th For a ﬁxed odd prime n. Let r, for example, e n be the n be one of the root of unit. (1)Denote riA = A(i), riB = B(i), show that

An + Bn = (A + B)(A0 + B(p−1)) ... (A(p−1) + B0)

And denote the `th factor by M (`). ` n−` (2)Denote z` = r + r , and show that

(i) (j) ( i+j ) M + M = z i−j M 2 ( 2 )

Where i, j is assumed to be the element in Z/nZ. (3)Show that if a ∈ Z[r] such that a|Mi, a|Mj, then a divides all M`. (Hint: That is, a subset T of Z/nZ such that a, b ∈ T implies 2a−b = b+2(a−b) ∈ T . Then, b ∈ T b+(a−b) ∈ T b+2(a−b) ∈ T b+3(a−b) = 2(b+2(a−b))−a ∈ T...

13if you like, without words. 14where the notation is as classical (stupid) as he use more than one century ago.

28 thus, b + x(a − b) ∈ T , then T = Z/nZ.) His ﬁrst gap is assuming that Z[r] is a unique factorization domain. Let us see what happen if the assumption holds. By (3) We can assume the common divisor of M (`) is k. And assume M (`) = km(`), with m` pairwise “coprime”. th (`) (`)n And then by the unique factorization, m` is an n power, say m = µ , then µ` is pairwise “coprime”. Then (2) gives

n (i)n (j)n ( i+j ) µ + µ = z i−j µ 2 ( 2 ) And, he said that

0n 000n 00n (11) µ + µ = z1µ

However, to make this equation (11) possible, it is necessary that the sum of the nth powers of the complex numbers µ0 and µ000 jn+1 divided by z1 , which is impossible. We also show that z1 = r + rn−1 can not be the nth power of a complex number. 15

Which is the second gap. I believe that his logic breaks because he used more language rather than the mathematical symbols 16 Anyway, maybe the most insightful step is (3). However, one can prove that if x, y is coprime, then x + ζiy is coprime to x + ζjy, which makes his idea go broke.

Remark. However, Kummer use the idea to prove the following fact correctly.

For regular prime p, Xp + Y p = Zp has no nontrivial solutions. Where a regular prime is a prime satisfying some condition of algebra. It is expected that 61% of primes are regular. You can refer to some algebraic number theory texts for help, for example, [26] Page 101, if interested in.

15The paper was written in Fran¸cais,and the original sentences are “Or, pour que cette équation(11) fûtpossible, il faudrait nécessairment que la somme des niéme puissances des 0 000 jn+1 nombres complexes µ et µ fútdivisible par z1 , ce qui est impossible. On démontre n−1 iéme d’ailleurs que z1 = r + r ne peut êtrela n puissance d’un nombre complexes. ” 16And, doubtlessly, the two tricks (using more language rather than the mathematical symbols and using stupid notations) are both good ways to protect your paper from the referees who want to check whether you are right. See [10] for more tricks.

29 13 Irrational number

Exercise 67 Show that log2 3 is irrational.

Problem 68 Show that x is infinite non-repeating decimal iff x is irrational17. (Hint: It suffices to show that x is rational iff x is finite decimal or infinite repeating decimal. Clearly, x is finite decimal, then x is rational. And when x is ... −m infinite repeating, we have see in (40) that x = finite decimal + 10n−1 × 10 a for some m, n. And conversely, for a rational number x = b , times it suffi- ciently many 10s, so that the denominator is coprime to 10. Then it suffice a ... to shoe any b with b coprime to 10 can be written as 10n−1 for some n. It is not impossible, for example, n = ϕ(b), then 10ϕ(b) ≡ 1 mod b, by Euler’s theorem. )

Exercise 69 Show that the Champernowne constant

C = 0.123456789101112131415161718 ... is irrational. Where C is obtained by concatenating the positive integers and interpreting them as decimal digits to the right of a decimal point. (Hint: Show that there are arbitrary many 0s between some of two 1s. )

Remark. Someone believes that the digits of π contains any ﬁnite sequences of numbers. But the assertion has not been proved up to 2017. But the digits of Champernowne constant contains any ﬁnite sequences of numbers, clearly. And a website to search, for example, your birthday in π is https://www.angio.net/pi/digits.html.

Problem 70 The purpose of this problem is to show that e is irrational using the following equality 1 1 1 e = + + + ... 0! 1! 2!

(1)Show that ∞ 1 X 1 1 < < N! n! (N − 1)! n=N

1 1 1 1 1 1 1 1 (Hint: n! = n(n−1) (n−2)! ≤ n(n−1) (N−2)! = (N−2)! n−1 − n .)

17that is, can not written as a rate of two integers

30 (2)Show that for any N, N!e fails to be integer. (Hint: One have

∞ X 1 N!e = (some integer(s)) + N! n! n=N then use (1)) (3)Then, as a result, e is irrational.

Exercise 71 Show that sin 1 is irrational. (Hint: It is similar to above problem. )

Exercise 72 Show that sin 1◦ is irrational. (Hint: Otherwise, cos 2◦ is ra- √ 3 ◦ tional, then 2 = cos 30 is rational. For general result, see the following problem. )

Problem 73 The purpose of this problem is to show for rational number r ∈ , Q π π cos 2πr ∈ ⇐⇒ θ ∈ , Q 2 Z 3 Z (1)For any θ ∈ R, note that 2 cos 2θ = (2 cos θ)2 − 2 2 cos 3θ = (2 cos θ)2 − 3(2 cos θ) show that there exists monic polynomial fn(X) with integer coeﬃcients such that 2 cos nθ = fn(2 cos θ) p (2)Prove the assertion. (Hint: Assume r = q , then 2 cos q(2πr) = 0, that is, 2 cos 2πr is a root of fq(x), but (3) make sure that 2 cos 2πr ∈ Z. The only values can be taken is 0, ±1, ±2. Then, it is easy to check. )

Remark. The proof above essentially uses Chebyshev polynomials. The proof using the theory of primitive polynomials can be found in [24] P268. and a proof involving cyclotomic ﬁelds can be found in [25] P353 proposition 9.8.5, which is very brief but of course using ﬁeld theory that is a little abstract.

Problem 74 The purpose of this problem is to show that the set √ { n : n is square-free} √ √ √ is linearly independent over Q. Then, as a result, for example, 2 + 3 + 6 is irrational.

31 For a linear equation

k X √ λi ni = 0 λi ∈ Q (∗) i=1

We say it is a equation involving n primes if P = {prime p: for some i, p|ni} is of cardinality n. We will use induction on number of primes involving. The case |P| = 1 is trivial. Now let us assume |P| ≥ 2. (1)Assume more λi 6= 0. Pick any p 6= q ∈ P, show that (∗) is equivalent to ` ! ` ! √ X p √ X p p = αr hr + q βr hr r=1 r=1 involving no more primes than P, and

∀r, p6 | hr, q6 | hr, αr, βr ∈ Q (Hint: You can show that (∗) ﬁrstly is equivalent to X √ √ X √ 0 = µj mj + p νj mj with mj run over all square-free number over P\{p}. Then is equivalent to P √ √ µj mk p = − P √ νj mj By “numerator rationalization”, it is eventually equivalent to

` √ X √ p = µj mj j=1 Then, by same way, pick another q ∈ P, then you will get the desired form. ) (2)Prove the assertion. (Hint: Let the equation in (1) to be √ √ p = A + qB

Square it, one has √ 0 = (A2 + qB2 − p) + AB q it is a linear equation involving less prime. One have AB = 0 and A2+qB2−p. If B = 0, then A2 = p becomes a contradiction. If A = 0, then B2 = p/q is still a contradiction. This means, there is no linear equation like (∗) with λi 6= 0.)

32 Problem 75 The purpose of this problem is to show that π is irrational. b Assume π = a . Let xn(a − bx)n f(x) = n!

(1)Show that, f (i)(π) and f (i)(0) are always integers. (2)Show that if u, v is continued diﬀerential n times, show the following formula Z Z uv(n)dx = [uv(n−1) − u0v(n−2) + ... + (−1)n−1u(n−1)v] + (−1)n u(n)vdx

(3)Show that Z π f(x) sin xdx 0 is a positive integer. (4)Show that when 0 < x < π,

πnan bn 0 < f(x) sin x ≤ f(x) < = n! n!

bn (5)Show that π is irrational. (Hint: When n → ∞, one have n! → 0.) Remark. The proof is from [29]. It essentially uses the Legrendre polynomials, see [17].

Exercise 76 18 Are there any irrational number a and b with ab rational? A famous proof is as follow √ √ √ 2 √ 2 √ If 2 is rational, the proof is complete. Otherwise, since ( 2 ) 2 = √ √ 2 √ 2 √ 2 = 2, a = 2 , b = 2 is desired. Anyone ﬁrstly reading it may make tremendous complaint on it or make high comment on it. One may ask, √ √ 2 Is 2 rational? To be, or not to be? In 1900, Hilbert announced a list of twenty-three outstanding unsolved problems. The seventh problem is

18If you have no interests in the stories, you can skip the materials up to the end of this exercise.

33 Is ab always transcendental, for algebraic a 6∈ {0, 1} and irrational algebraic b? The problem is settled by Gelfand in 1934, and reﬁned by Schneider in 1935, cf [4]. In 1934, Gelfand proved the problem is positive, that is The number ab is always transcendental, for algebraic a 6∈ {0, 1} and irrational algebraic b. √ √ 2 Use this theorem to show 2 is irrational (and actually transcendental).

Problem 77 A theorem in transcendental number theory, Hermite Theorem, [5] and [6], claims that

if α1, . . . , αn are algebraic numbers linearly independent over Q, then eα1 ,..., eαn are algebraically independent over Q.

Where “algebracally independent” stands for (eα1 ,..., eαn ) is not a zero of any nonzero polynomial in n variables. Clearly, x, 1 are algebraically independent iﬀ x is transcendental. Use this fact show that e is transcendental, and use the identity e2iπ = 1, show that π is transcendental.

P∞ 1 Exercise 78 Show that n=1 10nn is transcendental. (Hint: Argue like text book )

34 References

[1] Congruent number. https://en.wikipedia.org/wiki/Congruent_ number. [2] Congruent number. https://oeis.org/A003273.

[3] Eisenstein integer. https://en.wikipedia.org/wiki/Eisenstein_ integer. [4] Gelfond-schneider theorem. https://en.wikipedia.org/wiki/ Gelfond-Schneider_theorem.

[5] Hermite-lindemann theorem. https://mathworld.wolfram.com/ Hermite-LindemannTheorem.html. [6] Hermite-weierstrass theorem. https://en.wikipedia.org/wiki/ Lindemann-Weierstrass_theorem. [7] Hurwitz theorem. https://en.wikipedia.org/wiki/Hurwitz’s_ theorem_(number_theory). [8] Lucas’s theorem. https://en.wikipedia.org/wiki/Lucas’s_theorem. [9] Pick’s theorem. https://en.wikipedia.org/wiki/Pick’s_theorem. [10] Tips for authors - j.s. milne. https://www.jmilne.org/math/tips. html. [11] Oscar A Campoli. A principal ideal domain that is not a euclidean domain. The American Mathematical Monthly, 95(9):868–871, 1988. [12] R. D Carmichael. Note on eulers ϕ-function. Bulletin of the American Mathematical Society, 28(3):109–110, 1922. [13] V Chandrasekar. The congruent number problem. Resonance, 3(8):33– 45, 1998. [14] Henri Cohen. Number theory: Volume I: Tools and diophantine equations, volume 239. Springer Science & Business Media, 2008.

[15] Keith Conrad. The congruent number problem. The Harvard College Mathematics Review, 2:58–74, 2008. [16] Keith Conrad. Euclidean proofs of dirichlet’s theorem. 2012.

35 [17] Beiye Feng. Polynomials and irrational numbers (in Chinese). Ha’erbing Unversity press, 2008. [18] Keqin Feng and Hongbing Yu. integers and polynomials (in Chinese). Higher Education Press, 1999. [19] Richard Guy. Unsolved problems in number theory, volume 1. Springer Science & Business Media, 2013. [20] Luogeng Hua. Introduction to number theory (in Chinese). 1979.

[21] G. Lam´e. Dmonstration gnrale du thorme de fermat sur limpossibilit en nombres entiers de lquation xn + yn = zn. C. R. Acad. Sci. Paris, 24:310C314, 1847. Available at https://gallica.bnf.fr/ark:/12148/ bpt6k29812/f310.image. [22] D. H Lehmer. On eulers totient function. Bulletin of the American Mathematical Society, 38(1):745–751, 1932. [23] Franz Lemmermeyer. The euclidean algorithm in algebraic number ﬁelds. Expositiones Mathematicae, 13:385–416, 1995. [24] Shangzhi Li. Linear Algebra (for Department of Mathematics)(in Chi- nese). Beijing, Higher Education Press, 2006.

[25] Wen-Wei Li. Methods of algebra: Volume 1, in Chinese. Beijing, Higher Education Press (unpublished), 2016. Available at http://www.wwli. url.tw. [26] James S. Milne. Algebraic number theory (v3.07), 2017. Available at https://www.jmilne.org/math/. [27] Ram Murty. Primes in certain arithmetic progressions. 01 2006. [28] J¨urgenNeukirch. Algebraic number theory, volume 322. Springer Science & Business Media, 2013.

[29] Ivan Niven. A simple proof that n is irrational. Biscuits of Number Theory, (34):111, 2009. [30] Joseph H Silverman. A friendly introduction to number theory, volume 10. 2006. [31] Elias M Stein and Rami Shakarchi. Real analysis: measure theory, inte- gration, and Hilbert spaces. Princeton University Press, 2009.

36 [32] Elias M Stein and Rami Shakarchi. Fourier analysis: an introduction, volume 1. Princeton University Press, 2011.

37 Index e is irrational, 30 Gaussian integers, 25 π is irrational, 33 Gelfand, 34 n-decimal system, 4 p-adic valuation, 6 Hermite Theorem, 34 Hilbert, 34 142857, 15 Hurwitz theorem, 23

Arithmetic function, 21 Jarnlk, 21

Borel, 23 Legendre, 7 Borel-Cantelli lemma, 23 Legrendre polynomials, 33 Lucas, 8 Cantelli, 23 Champernowne constant, 30 order, 6, 9 characteristic function, 13 Chebyshev polynomials, 31 Pick’s Theorem, 10 Chevalley, 13 Primitive Element, 12 Chevalley-Warning theorem, 13 Pythagorean Triples, 1 Chinese Remainder Theorem, 12 congruent number, 3 Quadratic Reciprocity, 15 convex set, 12 rational root theorem, 1 cyclotomic polynomial, 5 Schneider, 34 Diophantine Approximation, 22 Dirichlet Theorem, 5 the principal of locality, 16 Dirichlet’s theorem, 22 Unique factorization, 6 Eisenstein integers, 28 equidistributive, 25 Warning, 13 Euclid, 5 Wilson’s theorem, 12 Euclidean proof, 5 Wolstenholme, 14 Euler’s ϕ-function, 20 exponential sums, 14

Fermat, 1, 2 Fermat’s primes, 9 Fibonacci sequence, 17 Frobenius automorphism, 8