86

E : y2 = x3 − Ax−B, A,B ∈ Z, ∆ = 16(4A3 − 27B2) 6= 0, D = 4A3 − 27B2

Map

θ : Z → Z�pZ = Fp = GF (p) (Zp before)

n 7→ [n ]p

Theorem (Nagell-Lutz) Points P = (x, y) of E(Q) of finite order, other than 0, i.e.the torsion points, have integer coordinates, i.e. are in E(Z), and either y = 0 or y | D.

We map E(Q) to E(Zp) through considering y2 ≡ x3 − Ax−B (mod p)

Theorem (Reduction Theorem) Let T ⊂ E(Q) be the subgroup of all points of finiteorder (the torsion subgroup). If p - 2D, p ∈ P then reduction mod p is an isomorphismof T onto a subgroup of E(Zp).

Theorem (Lagrange) If S ⊂ G and S is a subgroup of the finite group G, then#(S) |#(G)i.e. the order of S divides the order of G.

Corollary If P ∈ T and the order of P in E(Q) is m ∈ N then m |#E(Zp) ∀p - 2D.

These theorems can be used to determine the points of finite order of an elliptic curveE(Q).

Ex E : y2 = x3 + 3, D = −35 so let p > 5, p ∈ P. Then #E(Z5) = 6, #E(Z7) =13 ⇒ #T |6 and #T |13 ⇒ #T = 1 ⇒ T = {0} so T has no (finite) points of finiteorder. Note (1, 2) ∈ E(Q) since 22 = 13 + 3, so (1, 2) has infinite order and E(Q) hasan infinite number of points.

Ex E : y2 = x3−43x+166, D = 91215 ·13. Exploring small integers (x, y) ∈ Z2 we findP = (3, 8) ∈ E. Using the point doubling formula above the x-coordinates of 2P, 4P, . . .are x(P ) = 3, x(2P ) = −5, x(4P ) = 11, x(8P ) = 3 so x(P ) = x(8P ) ⇒ 8P = ±P soP is a point of finite order.

Since 3 - 2D, by the Reduction Theorem, T is isomorphic to a subgroup of E(Z3).#E(Z3) = 7 so #(T ) = 1 or 7. But 0 ∈ T and so does P = (3, 8) so #(T ) = 7. The onlyabelian group of order 7 is Z7, a cyclic group generated by P (which must be of order 7since its order divides 7). Computing {0, P, 2P, 3P, 4P, 5P, 6P} we get

T = {0, (3, ±8) . (−5, ±16) , (11, ±32)}.

Congruent Number Problem

87

Find a simple test to determine whether or not n ∈ N is the area of a right triangle, allof whose sides are of Q length.

Ex

6 = 12· 3 · 4 so 6 is congruent.

Ex Fermat n = 1 is not congruent. (X4 + Y 4 6= Z4 ∀X, Y, Z ∈ Z).

Ex Euler n = 7 is congruent.

{1, 2, 3, 4} are not congruent but {5, 6, 7, } are congruent.

Problem: Find a nice criteria to check n.

Theorem (Tunnell, 1983) Let n be an odd square-free natural number. Then if n iscongruent, the number of triples satisfying 2x2 + y2 + 8z2 = n is twice the number oftriples (x, y, z) satisfying 2x2 + y2 + 32z2 = n.

88

Let n be square-free and let X, Y, Z (X < Y < Z) be sides of a right triangle with arean. The number n ∈ N is fixed.

So n = 12XY, X2 + Y 2 = Z2.

Proposition There is a 1-1 correspondence between the right triangles given above andrational numbers x for which x, x + n, x − n are each the square of a rational number.The correspondence is

(X, Y, Z) 7→ x =

(Z

2

)2

x 7→ X =√x+ n−

√x− n

Y =√x+ n+

√x− n

Z = 2√x

In particular, n is congruent ⇔ ∃x ∈ Q+ such that x, x+n, x−n are squares of rationalnumbers.

Proof. (⇒) Let X, Y, Z ∈ Q+ be a triple with n = 12XY, X2 + Y 2 = Z2. Then

X2+Y 2 = Z2 and 2XY = 4n ⇒ (X±Y )2 = Z2±4n ⇒ (1)(

X±Y2

)2=(

Z2

)2±n = x±nif x =

(Z2

)2. So x, x± n are squares of rational numbers.

(⇐) Given x, x± n being squares, then

X =√x+ n−

√x− n

Y =√x+ n+

√x− n

Z = 2√x

satisfy X < Y < Z and X, Y, Z ∈ Q+. Finally

XY = (√x+ n−

√x− n)(

√x+ n+

√x− n) = (x+ n)− (x− n) = 2n

and

X2 + Y 2 = (x+ n) + (x− n)− 2√

(x+ n)(x− n)

+ (x+ n) + (x− n) + 2√

(x+ n)(x− n)

= 4x

= Z2.

89

�

Let n be a congruent number. By the above equation (1),(X ± Y

2

)2

=

(Z

2

)2

± n if n =1

2XY

Multiply these two equations together:(X2 − Y 2

4

)2

=

(Z

2

)4

− n2

so v2 = u4 − n2 has a rational solution v = X2−Y 2

4, u = Z

2. Now multiply by u2 :

u6−n2u2 = (uv)2. Let x = u2 =(

Z2

)2(as before) and y = uv = (X2−Y 2)Z/8 ⇒ a pair

(x, y) ∈ Q2 satisfying y2 = x3 − n2x—an elliptic equation y2 = x(x− n)(x+ n).

Hence if n is congruent, the curve y2 = x3 − n2x has a nontrivial rational point.Theconverse, that any point (x, y) ∈ Q2 must come from such a triangle is false in general.We need extra conditions equivalent to ∃Q ∈ En(Q) such that (x, y) = P = 2Q i.e. P isa (rational) point which is double a rational point.

Theorem 32B Let (x, y) ∈ Q2 be on y2 = x3 − n2x. Let x satisfy

(i) it is the square of a rational number,(ii) its denominator is even,(iii) its numerator is coprime with n.

Then there is a right triangle with rational sides and area n under the correspondence ofthe above Proposition.

Proof. Let u =√x ∈ Q+ (i) and let v = y

u∈ Q+. Since (x, y) is on En(Q) : v2 = y2

x=

x2 − n2 ⇒ v2 + n2 = x2 (1). Let t ∈ N be the denominator of u, i.e. the smallest N sotu ∈ Z. By (ii) t is even.Because n ∈ N, the denominators of v2 and x2 are the same by (1), namely t4.Hence (t2v)2 +(t2n)2 = (t2x)2 is a primitive Pythagorean triple with t2n even. (Primitivethrough (iii).) Hence ∃a, b ∈ Z such that t2n = 2ab, t2v = a2 − b2, t2x = a2 + b2. Thenthe right triangle with sides 2a

t, 2b

t, 2u has area 1

22at

sbt

= 2abt2

= n. Finally, the image of

this triangle with X = 2at, Y = 2b

t, Z = 2u is x =

(Z2

)2as required. �

90

Ex (i) and (ii) alone are not sufficient : n = 5, x = 254, y = 75

8⇒ X =

√x+ n −√

x− n =√

5 6∈ Q.

Back to Tunnell’s theorem n odd and square-free,

n congruent ⇒ #{(x, y, z) ∈ Z3 : 2x2 + y2 + 8z2 = n}= 2#{(x, y, z) ∈ Z3 : 2x2 + y2 + 32z2 = n} (B)

(A) ⇒ (B)

Then, subject to an unproved conjecture, (B) ⇒ (A). We can confidently use not(B) ⇒not(A)

Ex #{(x, y, z) : 2x2 + y2 + 8z2 = n} = #{(x, y, z) : 2x2 + y2 + 32z2 = n} if n < 8. Sonone of {1, 62, 3, 64, 5, 66, 7} can be congruent unless the size of each set is O(2 · 0 = 0).But x2, y2 ≡ 0, 1 or 4 (mod 8) ⇒ 2x2 + y2 + 8z2 6≡ 5, 7 (mod 8). So e.g. if n = 5 or 7both of the sets of triples are ∅.

Ex The first congruent number n ≡ 1, 3 (mod 8) is n = 41:

If (B) ⇒ (A) is true, the above argument would imply all of the following (odd, square-free) numbers are congruent, through 2 · 0 = 0: {5, 7, 13, 15, 21, 23, 29, 31, 37, 39, 47}.

91

12 Numbers Rational and Irrational

If α ∈ R we say α ∈ Q if α = mn, m, n ∈ Z, n 6= 0.

Proposition√

2 6∈ Q.

Proof. Assume√

2 ∈ Q

⇒√

2 =a

b, (a, b) = 1 (???)

⇒ a =√

2b

⇒ a2 = 2b2 ⇒ 2 |a2 ⇒ 2 |a

So a = 2c and 4c2 = 2b2

⇒ 2c2 = b2 ⇒ 2 |b2 ⇒ 2 |b

Hence 2 |a and 2 |b so 2 |(a, b) so (a, b) 6= 1 (!!!). �

We say√

2 is irrational or√

2 ∈ I = R \Q.

We can generalise the above proposition to get a much wider family of irrational numbers:

Theorem 33 If x ∈ R satisfies the equation

xn + c1xn−1 + · · ·+ cn = 0

where ci ∈ Z, then x is either an integer or an irrational number.

Proof. Let x ∈ Q i.e. x = ab, b > 0, (a, b) = 1. Then

an = −b(c1an−1 + c2an−2b+ · · ·+ cnb

n−1)

If b > 1, then p | b ⇒ p | an ⇒ p | a but then p | (a, b) (!!!). Hence b has no primedivisors, so b = 1. �

Corollary If m ∈ N is not an nth power then m1/n = n√m ∈ I since α = m1/n satisfies

xn −m = 0.

Trigonometric function values and π

Lemma 1 Let g ∈ Z[x] (i.e. a polynomial with integral coefficients). Let h(x) = xng(x)n!

.If j 6= n, h(j)(0) is an integer divisible by (n + 1). If g(0) = 0, h(n)(0) is an integerdivisible by (n+ 1).

Proof. Let

xng(x) =1

n!(cnx

n + cn+1xn+1 + · · ·+ cjx

j + · · · )

where c0, · · · , cn−1 = 0 and the ci are integers.

92

Then the j’th derivative

h(j)(0) =cjj!

n!.

If j < n, cj = 0.

If j > n, n+ 1 |h(j)(0) since j!n!

= (n+ 1)(n+ 2) · · · (j).

If j = n ⇒ h(j)(0) = cj ⇒ h(n)(0) = cn but g(0) = 0 ⇒ xng(x) = xn[g1x+g2x2+· · · ] =

cn+1xn+1 + · · · ⇒ n+ 1 |h(j)(0). �

Lemma 2 If f(x) is a polynomial in (r − x)2, then, for any odd positive integer j,f (j)(r) = 0 i.e. f ′(r) = 0, f ′′′(r) = 0, · · · .Proof. Let j ≥ 0 be an integer and n ∈ N a positive integer.

f(x) = a0 =⇒ f (2j+1)(x) = 0 =⇒ f (2j+1)(r) = 0

f(x) = (r − x)2 =⇒ f ′(x) = −2(r − x) =⇒ f ′(r) = 0

f(x) = (r − x)2n, 2j + 1 > 2n =⇒ f (2j+1)(x) = 0 =⇒ f (2j+1)(r) = 0

f(x) = (r − x)2n, 2j + 1 < 2n =⇒ f (2j+1)(x) = (−1)2j+1 2n!

2j + 1!(r − x)2n−2j−1

=⇒ f (2j+1)(r) = 0.

Therefore if f(x) if a sum of even powers of (r − x) all of its odd derivatives vanish atx = r. �

Theorem 34 π is irrational, i.e. π ∈ I.Proof. Let f(x) = xn(1−x)n

n!where n ∈ N.

By Lemma 1 above, ∀j, f (j)(0) ∈ Z and f(x) = f(1 − x) ⇒ f (j)(1) ∈ Z. Since0 < x < 1 ⇒ 0 < xn < 1 and 0 < 1 − x < 1 ⇒ 0 < (1 − x)n < 1 we have0 < f(x) < 1

n!(1).

Let π2 = ab, a > 1, b > 1, a, b ∈ N (???). Let

F (x) = bn[π2nf (0)(x)− π2n−2f (2)(x) + π2n−4f (4)(x)− · · ·+ (−1)nf (2n)(x)].

93

So F (0) ∈ Z, F (1) ∈ Z. Now

d

dx

{F ′(x) sinπx− πF (x) cosπx

}=

{F (2)(x) + π2F (x)

}sinπx

= bnπ2n+2f(x) sinπx

= π2anf(x) sinπx

So

πan

∫ 1

0

f(x) sinπx dx =

[F ′(x) sinπx

π− F (x) cosπx

]1

0

= F (1) + F (0) ∈ Z.But by (1),

0 < πan

∫ 1

0

f(x) sinπx dx <πan

n!< 1 for n > n0

which is a contradiction. Hence π2 is irrational. �

Corollary π is irrational: If not π2 would be rational. ’

Note: With a similar, but more complex proof, we can show r ∈ Q \ {0} ⇒ cos r isirrational.

Corollary 1 to the note π is irrational, since if π ∈ Q, cosπ ∈ I but cosπ = −1.

Corollary 2 All trigonometric functions are irrational at non-zero rational values oftheir arguments.

Proof. r ∈ Q and sin r ∈ Q ⇒ cos2r = 1 − 2 sin2 r ∈ Q, which is false. Similarly,tan r ∈ Q ⇒ cos 2r = 1−tan2 r

1+tan2 r∈ Q. �

Corollary 3 Any non-zero value of an inverse trigonometric function is irrational atrational values of the argument.

Proof. Let r ∈ Q and arccos r = cos−1 r = s. Suppose s ∈ Q ⇒ cos s = r which is false.�

Exponential, hyperbolic and logarithmic functions

Note: e0 = 1 ∈ Q and sinh 0 = 0, cosh 0 = 1 but these are the only rational values atrational arguments. The proof is similar to Theorem34 based on cosh:

Corollary 4 er ∈ Q ⇒ e−r = 1er ∈ Q ⇒ er+e−r

2∈ Q but this is not possible if r ∈ Q.

Theorem 35 e is irrational, e ∈ I.Proof. Claim: ∀n ∈ N

0 < e−n∑

j=0

1

j!<

1

n · n!(1)

94

Represent e by an infinite series e = 1 + 11!

+ 12!

+ · · ·+ 1j!

+ · · · so

e−n∑

j=0

1

j!=

∞∑j=n+1

1

j!> 0

Also

e−n∑

j=0

1

j!=

1

(n+ 1)!+

1

(n+ 2)!+ · · ·

=1

n!

[1

n+ 1+

1

(n+ 1)(n+ 2)+

1

(n+ 1)(n+ 2)(n+ 3)+ · · ·

]<

1

n!

[1

n+ 1+

1

(n+ 1)2+

1

(n+ 1)3+ · · ·

]=

1

n!

[1/(n+ 1)

1− 1/(n+ 1)

](sum of a geometric series r = 1

n+1)

=1

n!

1

nwhich proves the claim.

Now let e = mn, m, n ∈ N, (m, n) = 1 (???), and assume n 6= 1. Let

η = n!

(e−

n∑j=0

1

j!

)By (1)

0 < η < n!1

n · n!=

1

nBut

η = n!

(m

n− 1− 1

1!− 1

2!− · · · − 1

n!

)∈ Z (!!!)

Hence e is irrational. �

Corollary√e is irrational, since otherwise e = (

√e)2 would be in Q.

Question: e seems to be ‘more’ irrational than√

2. We will explore families of irrationalnumbers below.

Let S ⊂ R be a subset. We say S has measure zero if it is possible to cover S with afinite or countable set of intervals of arbitrarily small total length. Write µ(S) = 0.

Ex S = N:

1 ∈(

1− ε

2, 1 +

ε

2

)2 ∈

(2− ε

22, 2 +

ε

22

)j ∈

(j − ε

2j, j +

ε

2j

)= Ij

95

So

N ⊂∞⋃

j=1

Ij

and

`(Ij) = length of Ij

= j +ε

2j−(j − ε

2j

)= 2

ε

2j

Then

∞∑j=1

`(Ij) = 2ε∞∑

j=1

1

2j

= 2ε

which can be made arbitrarily small by choice of ε > 0. Hence µ(N) = 0. We can replaceN by any countable set A = {an : n ∈ N} ⊂ R. by defining Ij =

(aj − ε

2j , aj + ε2j

)since

`(Ij) = 2ε2j .

Definition A property of real numbers is said to hold “almost everywhere” or toalmost all numbers, if the set of numbers which do not have the property has measurezero.

Ex µ(Q) = 0 since Q is countable. Hence almost all numbers are irrational.

Note We can count the numbers in Q+ via listing them and then counting the diagonals,skipping any already counted.

r1 r3 → r4↓ ↗ ↙r2 ×↙

r5↓

96

11

12→ 1

3· · ·

↓ ↗ ↙21

22

23· · ·

↙31

32

33· · ·

↓

1/1 1/2 → 1/3 · · ·↓ ↗ ↙

2/1 2/2 2/3 · · ·↙

3/1 3/2 3/3 · · ·↓

⇒ Q+ = {rn : n ∈ N}.

Since `([0, 1]) = 1 > 0, [0, 1] and (hence) R are not of measure 0. Hence, since the unionof any two countable sets is countable, the irrational numbers I are not countable.

Proof. A = {an : n ∈ N}, B = {bn : n ∈ N} ⇒ A ∪ B = {cn : c2n = an, c2n−1 = bn, n =1, 2, 3, . . .} so A ∪B is countable. �

Now let S ⊂ R. We say S is dense in R if ∀α < β ∃x ∈ S with α < x < β.

Archimedian Axiom (AA) ∀ε > 0 ∃n ∈ N such that 0 < 1n< ε.

Proposition Q is dense in R.

Proof. Let α < β¿ By AA ∃n ∈ N such that 0 < 1n< β − α. Let m ∈ Z satisfy

m < nβ 6 m + 1. Then α < β − 1n6 m+1

n− 1

n= m

nand m

n< β. Hence α < m

n< β and

we can let x = mn

. �

Proposition I is dense in R.

Proof. Let α, β ∈ R have α < β. Let α < mn< β as above, and using AA choose k ∈ N

so

0 <1

k<β − m

n√2.

Then α < mn< m

n+√

2k< β and x = m

n+√

2k∈ I. �

Definition A number is algebraic if it satisfies an equation

xn + a1xn−1 + a2x

n−2 + · · ·+ an = 0

with ai ∈ Q.

97

Ex√

2 satisfies x2 − 2 = 0.

The unique polynomial with leading coefficient 1 (called monic) inQ[x] of minimal degreewhich has a given algebraic number α as a root is called the minimal polynomial of α,and the degree of this polynomial is called the degree of α.

The set of all algebraic numbers is called A ⊂ R.

Proof. If An is the set of algebraic numbers of degree n for n = 1, 2, 3, . . . then

A =∞⋃

n=1

An

There are a countable number of polynomials of degree n with Q coefficients since p(x) =xn + a1x

n−1 + · · ·+ an ↔ (a1, . . . , an) ∈ Qn and the latter is a countable set.

But each polynomial has at most n roots in R ⇒ An is countable. To complete theproof we need to assume that a countable union of countable sets is countable. To seethis, use the diagonal counting trick:

A1 = {a11, a12, → a13, . . .}↓ ↗ ↙

A2 = {a21, a22, a23, . . .}↙

A3 = {a31, a32, a33, . . .}↓...

�

Since µ(A) = 0, almost all numbers are not algebraic. We call these numbers transcen-dental and the set of all such numbers T = R \ A.

Ex21/3 +

√2√

3∈ A, π and e ∈ T

The former is not difficult, but π and e are both very difficult.

BothQ and I (and T) are dense in R. This implies each real number can be expressed as thelimit of rational numbers : Let α ∈ R then give n ∈ N ∃rn ∈ Q with α− 1

n< rn < α + 1

n

so |α− rn| < 1n⇒ α = limn→∞. But this universal fact gives little insight into the

difference between Q, A and T.

Definition A real number α is said to be approximable by rationals to order n ∈ N

98

if ∃ a constant C = C(α) > 0 such that the inequality∣∣∣∣α− h

k

∣∣∣∣ < C

kn

has infinitely many rational solutions hk

where k > 0, (h, k) = 1.

Note Approximable to order 3 ⇒ Approximable to order 2 and 1.

Theorem 36 If α ∈ I, ∃ infinitely many hk∈ Q with∣∣∣∣α− h

k

∣∣∣∣ < 1

k2

i.e. α is approximable to order 2.

Proof. See page 75. If α ∈ I its continued fraction expansion is infinite so the set ofconvergents pn

qnis infinite and ∣∣∣∣α− pn

qn

∣∣∣∣ < 1

qnqn+1

<1

q2n

so we can let hk

= pn

qn.

OR Let n ∈ N. Consider the n+ 1 real numbers

S = {0, α− bαc , 2α− b2αc , . . . , nα− bnαc}

and their distribution in the intervals jn6 x < j+1

n, j = 0, . . . , n − 1 which cover [0, 1),

so contain all of the numbers in S. Hence (by the Dirichlet pigeon-hole principle) twonumbers lie in the same interval, say 0 6 n1 < n2 6 n, n1α − bn1αc , n2α − bn2αc ∈[

jn, j+1

n

). The length of this interval is 1

nso

|(n2α− bn2αc)− (n1α− bn1αc)| <1

n

Let k = n2 − n1 and h = bn2αc − bn1αc , k ∈ N, h ∈ Z. so |kα− h| < 1n

andk 6 n (1) ⇒

∣∣α− hk

∣∣ < 1nk6 1

k2 . Suppose there were only a finite number of such pairs(h, k) : (h1, k1) , . . . , (hr, kr). Let

ε = min

{∣∣∣∣α− h1

k1

∣∣∣∣ , . . . , ∣∣∣∣α− hr

kr

∣∣∣∣} > 0.

Use AA to find n ∈ N with 0 < 1n< ε so ∃h, k by (1) so

∣∣α− hk

∣∣ < 1nk6 1

n< ε so h

k6= hi

ki

(!!!). �

Theorem 37 Any rational number is approximable to order 1, but not to any higherorder.

99

Proof. Let α = ab, (a, b) = 1, b > 1 be rational. Then there are infinitely many solutions

(x, y) to ax − by = 1 (x = x0 + bt, y = y0 + at, t ∈ Z if(x, y) is one solution) andinfinitely many with x > 0. Then

ax− by = 1 ⇒∣∣∣ab− y

x

∣∣∣ =1

bx<

2

x

Hence α is approximable to order 1.If y

x∈ Q and y

x6= a

bthen ∣∣∣a

b− y

x

∣∣∣ =

∣∣∣∣ax− bybx

∣∣∣∣ > 1

bx

there is no constant C such that 1bx< C

x2 for infinitely many x ∈ N. Henceab

= α is notapproximable to any order higher than 1. �

Ex ξ = 110

+ 1102 + 1

204 + · · ·+ 1102m + · · · ∈ I. Let rm = (m+ 1)th partial sum of ξ, rm ∈

Q. |ξ − rm| = 10−2m+1+ 10−2m+2

+ · · · < 2 · 10−2m+1= 2(10−2m

)2. rm = an

102m , an ∈ N. Sothis inequality shows we can approximate ξ to order 2 at least. Hence ξ 6∈ Q ⇒ ξ ∈ I.