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Ex Equations likey² =x³+ 7 are called “elliptic curves”. They arise in solving integrals for, say, the period of a body in a planetary orbit.

(Lebesgue, 1869) The equation y² =x³+ 7 is insoluble over Z.

Proof. If x is even, x = 2α ⇒ RHS = 8α³ + 7 = 8β + 7, where β = α³. But 0² ≡0, 1² ≡ 1, 2² ≡4, 3² ≡ 1, 4² ≡ 0, 5² ≡1, 6² ≡ 4 and 7² ≡ 1 (mod 8) so y² ≡ 7 (mod 8) has no solution. Hence x is odd. Write

y²+ 1 = x³+ 8

= (x+ 2)(x²−2x+ 4)

= (x+ 2)((x−1)²+ 3)

If x= 2n+ 1 (odd) then (x−1)²+ 3 = 4n² + 3 = 4m+ 3, m =n² so (see back) must have a prime factor of the form p = 4` + 3. But then y² + 1 ≡ qp ≡ 0 (mod p) But (lemma later) p≡3 (mod 4) ⇒ y² ≡ −1 (mod p) has no solution.

We frequently need to know the answer to the following: When does x² ≡ r (mod p) have a solution x? Or, more generally, x² ≡ α (mod m). The answer is given by the theory of quadratic reciprocity due to Gauss. This will be developed later.

8 Pell’s Equation

x²−N y² = 1 Trivial solution x= 1, y = 0, x, y>0.

N =−1 ⇒ (x, y) = (1, 0) or (0, 1) are trivial solutions only.

N 6−2 ⇒ (x, y) = (1, 0).

Let N > 0 and not a square: If N = M², M ≥ 1, x² −N y² = x² −(M y)² = (x− M y)(x+M y) = 1 ⇒ x−M y = 1 and x+M y = 1 so we can get all solutions. Indeed (x, y) = (1, 0) for x, y >0. So we always assume N ≥2.

Note: Solutions to Pell’s equation provide goodrational approximations for square roots, since x² =N y²+ 1

⇒

x

y 2

=N + 1 y²

⇒ ^x_y ≈√

N if y is large.

Note: This type of equation has a long and interesting history, and has lots of applica- tions, especially to fieldsF =Q(√

N).

Ex (Euler, 1770) A triangular number has the form ⁿ⁽ⁿ⁺¹⁾₂ . Which numbers are both triangular and square?

m² =n(n+ 1)/2

⇒ 8m²+ 1 = 4n²+ 4n+ 1 = (2n+ 1)²

⇒ x²−2y² = 1 where x= 2n+ 1, y = 2m.

So solutions to this Pellian equation produce (all) square triangular numbers.

Definition Afundamental solutiontox²−dy² = 1 is (r, s) where any other positive solution satisfies r < x and s < y.

Theorem 25 (Lagrange) Let (r, s) be the least positive (or fundamental) solution to x²−dy² = 1, where d is not a square. Then every solution to this equation is given by (x_n, y_n) where

xn+

√

dyn= (r+s

√ d)ⁿ for n= 1,2,3, . . .

Proof.

x²_n−dy²_n = (xn+yn

√

d)(xn−yn

√ d)

= (r+s√

d)ⁿ(r−s√ d)ⁿ

= (r²−s²d)ⁿ = 1ⁿ= 1

Hence (x_n, y_n) is a solution.

Let (a, b) be a solution. Suppose ∀n = 1,2,3, . . . , (a, b) 6= (x_n, y_n). Then there is a positive integer m with

(r+s√

d)^m < a+b√

d <(r+s√

d)^m+1 (21)

But (r+s√

d)^−m = (r−s√

d)^m so (21)⇒ 1<(a+b√

d)(r−s√

d)^m <(r+s√

d) (22)

Letu+v√

d= (a+b√

d)(r−s√ d)^m so u²−v²d = (u+v√

d)(u−v√ d)

= (a+b√

d)(r−s√

d)^m(a−b√

d)(r+s√ d)^m

= (a²−b²d)(r²−s²d)^m = 1·1^m = 1 Thus (u, v) is a solution.

But 1< u+v√

d ⇒ 0< u−v√

d <1 so 2u= (u+v√

d) + (u−v√

d)>1 + 0>0 And 2v√

d= (u+v√

d)−(u−v√

d)>1−1 = 0 sou >0, v >0 and u+v√

d < r+s√ d by (22), contradiction the assumption that (r, s) is the fundamental solution. Hence (a, b) = (x_n, y_n) for somen.

Finding the least positive solution isnot easy however and requires the theory ofcontinued fractions of J. L. Lagrange. Frenicle’s table for non-square d up to 50 is given below.

Pell's equation

"f,,'

~l:lPuler, after a cursory reading of Wallis's Opera Mathematica, mistakenly

r~buted the first serious study of nontrivial solutions to equations of the '~J;f°!In x2

-

dy2 = 1, where x

~

1 and y

~

0, to Cromwell's mathematician

~a,.John Fell. However, there is no evidence that Fell, who taught at the '~;~niversity of Amsterdam, had ever considered solving such equations.

~t[;rhey :would be more aptly called Fermat's equations, since Fermat first

~~(tlvestigated properties of nontrivial solutions of each equations. Neverthe- ,~\(tess, Pellian equations have a long history and can be traced back to the .;ff,.Greeks. Theon of Smyrna used x/y to approximate ~, where x and y

~'\gY(ere integral solutions to x2 - 2y2 = 1. In general, if x2 = dy2 + 1, then

;~~2/y =d+ 1/y2. Hence, for y large, x/y is a good approximation of

~'Yd, a fact well known to Archimedes.

JI(Archimedes's problema bovinum took two thousand years to solve.

itccording to a manuscript discovered in the Wolfenbiittel library in 1773 ,tRY Gotthold Ephraim Lessing, the German critic and dramatist, Archi-

~~edes became upset with Apollonius of Perga for criticizing one of his t~orks. He divised a cattle problem that would involve immense calculation j~?i solve and sent it off to Apollonius. In the accompanying correspon- r!~ence, A.rchimedes asked Apollonius to compute, if he thought he was

,ii' .

~ - -~

smart enough, the number of the oxen of the sun that grazed once upon the plains of the Sicilian isle Trinacria and that were divided according to color into four herds, one milk white, one black, one yellow and one dappled, with the following constraints:

white bull~ ~ yellow bulls + (~+ ~) black bulls,

(

^{1 1}

)

black bulls = yellow bulls + 4 + ^:5 dappled bulls,

i

(

^{1 1}

)

f^\

dappled bulls = yellow bulls + ^"6 + '7 white bulls,

white cows = (~+~) black herd,

black cows = (~+~) dappled herd,

dappled cows = (~+~) yellow herd, and yellow cows = (~+ ~) white herd.

Archimedes added, if you find this number, you are pretty good at numbers, but do not pat yourself on the back too quickly for there are two more conditions, namely:

white bulls plus black bulls is square and dappled bulls plus yellow bulls is triangular.

Archimedes concluded, if you solve the whole problem then you may 'go forth as conqueror and rest assured that thou art proved most skillful in the science of numbers'.

The smallest herd satisfying the first seven conditions in eight unknowns, after some simplifications, lead to the Pellian equation x2- 4729494 y2 = 1. The least positive solution, for which y has 41 digits, was discovered by Carl Amthov in 1880. His solution implies that the number of white bulls has over 2 X 105 digits. The problem becomes much more difficult when the eighth and ninth conditions are added and the first complete solution was given in 1965 by H.C. Williams, R.A. German, and C.R. Zarnke of the University of Waterloo.

In Arithmetica, Diophantus asks for rational solutions to equations of the type x2 - ^dy2 = 1. In the case where d = ^m2 + 1, Diophantus offered the integral solution x = 2m2 + 1 and y = 2m. Pellian equations are found in Hindu mathematics. In the fourth century, the Indian mathematican

--

9 Continued Fractions

Ex

1 + 1

2 + ₃₊¹1 4

= 1 + 1

2 + _13/4¹

= 1 + 1 2 + ₁₃⁴

= 1 + 1 30/13

= 1 +13 30

= 43 30

looks silly until we consider some interesting continued fraction expansions π : [3, 7,15, 1,292,1,1,1, . . .] i.e.

3 + 1

7 + ₁₅₊¹1 293+···

e: [2, 1, 2, 1, 1, 4, 1, 1, 6,1, 1, . . .]

√2 : [1, 2, 2, 2, 2, . . .]

√3 : [1, 1, 2, 1, 2,1, 2, 1,2, . . .]

√5 : [2, 4, 4, 4, . . .]

√n² + 1 : [n, 2n, 2n, . . .] (Euler)

Definition By a simple continued fraction (or C.F.) we mean an expression

a₀+ 1

a1+ _a ¹

2+···

= [a₀, a₁, a₂, . . .]

where a₀ ∈Z and a_i ∈N for i>1.

Note: [a0] = ^a₁⁰, [a0, a1] = ^a0a_a¹⁺¹

1 , [a0, a1, a2] = ^a2a1_a^a0+a2+a0

2a1+1

Generally, [a₀, . . . , a_n] = ^p_qⁿ

n wherep_nandq_nare polynomials in thea_i, linear in any given a_j, anda₀ doesnot occur in the denominatorq_n. (p_n, q_n) are called then^th convergents.

Note: [a0, . . . , an] = [a0, . . . , an−1+_a¹

n]

Proposition If [a₀, . . . , a_m] = [b₀, . . . , b_n], a_i, b_i ∈ N, a_m, b_n > 1 then m = n and a_i =b_i ∀i.

Proof. This follows by induction from

[a₀, . . . , a_m] =a₀+ 1

[a₁, . . . , a_m] =b₀+ 1 [b₁, . . . , b_n]

if we can show [a₁, . . . , a_m]>1 when a₁, . . . , a_m >1. But this is so since [a₁, . . . , a_m] = a₁+ _a ¹

2+···.

Let a_i > 0 and ∀n let τ_n = [a₀, . . . , a_n] then τ_n can be computed using the recursive formulas, for n≥2:

p₀ =a₀ p₁ =a₀a₁+ 1 p_n=a_npn−1+pn−2

q₀ = 1 q₁ =a₁ q_n=a_nqn−1+qn−2

so τ₀ = ^p_q⁰

0, τ₁ = ^p_q¹

1 and τ_n= ^p_qⁿ

n

Proof.

τ_n = [a₀, . . . , a_n] = [a₀, . . . , a_n−1+ 1 an

] = p⁰_n−1 q_n−1⁰ where these belong to a₀, . . . , an−2, an−1 +_a¹

n i.e. (induction) p⁰_n−1

q⁰_n−1 =

an−1 +_a¹

n

pn−2+pn−3

an−1 +_a¹

n

qn−2+qn−3

= a_n(a_n−1p_n−2+p_n−3) +p_n−2 a_n(an−1qn−2+qn−3) +qn−2

= anpn−1+pn−2

a_nqn−1+qn−2

(induction again!) Hence p_n=a_npn−1+pn−2 and q_n =a_nqn−1+qn−2

(p_n, q_n) are called the n^th convergents of the C.F.

Let θ ∈ R\Z, θ > 1. a₀ = bθc so θ = a₀ + _θ¹

1, θ₁ > 1 defines θ₁. Continue with θ₁ = a₁+ _θ¹

2 so a₁ =bθ₁c, θ₂ > 1 if θ₁ 6∈ Z etc θ_n = a_n+ _θ¹

n+1, a_n =bθ_nc, θ_n=1 >1 if θ_n 6∈Z. We get

θ =a₀+ 1

a1+ _a ¹

2+ ¹

...+ 1 an+ 1

θn+1

so θ= [a0, a1, . . . , an+ _θ¹

n+1]

Proposition The expansion stops if θ_n =a_n is in N and then θ∈Q⁺ i.e. is a positive rational number. Conversely, if θ∈Q⁺, the C.F. expansion is finite.

Proof. Let θ = ^u_v ∈Q⁺, u, v∈N. Use division

u = a₀v+r₁ 0< r₁ < v v = a₁r₁+r₂ 0< r₂ < r₁ r₁ = a₂r₂+r₃ 0< r₃ < r₂

...

rn−1 = anrn+ 0

as if we were doing the Euclidean algorithm. These equations give θ=θ₀ = u

v =a₀+ r1

v =a₀ + 1

v/r₁ =a₀+ 1 θ₁ θ₁ = a₁+ r₂

r₁ =a₁+ 1

r₁/r₂ =a₁+ 1 θ₂ ...

θ_n = rn−1

r_n ∈N so the C.F. expansion is finite.

Proposition ∀n>2

θ = θ_np_n−1+p_n−2 θ_nqn−1+qn−2

Proof. The definition ofθ_n is θ= [a₀, . . . , an−1, θ_n] so θ =τ_n= ^p_qⁿ

n = ^θ_θ^{npn−1+pn−2}

nqn−1+qn−2 using a_n and θ_n for this particular C.F.

Ex √

2 = [1,2, 2, . . .]

(√

2−1)(√

2 + 1) = 2−1 = 1 ⇒√

2−1 = ¹

1+√ 2 so√

2 = 1 + ¹

1+√ 2. We now copy the expression for √

2 in the RHS into the √

2 on the RHS successively (photocopy model for recursion).

√

2 = 1 + 1

1 + 1 +₁₊^1√₂

= 1 + 1

2 + ¹

1+√ 2

= 1 + 1

2 + ₂₊¹1 1+√

2

etc. leading to √

2 = [1, 2, 2, 2, 2, . . . , 2, 1 +√

2]. If we continue indefinitely we obtain

√2 = [1, 2,2, . . .] = [1, 2 ].

Every quadratic irrational has a periodic continued fraction—this characterises quadratic irrationals.

Ex √

2 = [1,2, . . . ,2,1 +√

2] so a0 = 1, a1 = 2, . . . p0 =a0 = 1

q₀ = 1

τ0 = p₀ q₀ = 1

1 = 1

p1 =a0a1+ 1 = 3 q₁ =a₁ = 2

τ₁ = p₁ q₁ = 3

2 = 1.5 p₂ =a₂p₁+p₀ = 7

q₂ =a₂q₁+q₀ = 5

τ₂ = p₂ q₂ = 7

5 = 1.4 and the approximationτ_n ≈√

2 gets better.

Theorem 26 Let a0 ∈Z, ai ∈ N, i > 1. Then (τn) converges to an irrational number θ. The a_i are uniquely determined by the C.F. expansion of θ. Conversely, if θ is an irrational number, and τ_n= [a₀, . . . , a_n] are obtained by expanding θ as a C.F. then

θ= lim

n→∞τ_n.

Proof. The sequences (p_n) and (q_n) are both strictly monotonically increasing sequences of natural numbers.

Claim:

p_nq_n−1−p_n−1q_n = (−1)ⁿ⁻¹ (23)

∀n > 1. If n = 1 this is p₁q₀−p₀q₁ = (a₀a₁+ 1)1−a₀a₁ = 1 = (−1)¹⁻¹ which is true.

Assume it is true for n =m. Then

p_m+1q_m−p_mq_m+1 = (a_m+1p_m+pm−1)q_m−p_m(a_m+1q_m+qm−1)

= pm−1q_m−p_mqm−1

= −(p_mq_m−1−p_m−1q_m)

= −(−1)^m−1

= (−1)^m Hence, by induction, the claim is true ∀n >1.

Divide (23) by qnqn−1 to obtain pn

q_n − pn−1

qn−1

= (−1)ⁿ⁻¹ q_nqn−1

or

τ_n−τn−1 = (−1)ⁿ⁻¹ q_nqn−1

(24) Apply this toθ = [a₀, . . . , an−1, θ_n] to get

θ−τn−1 = (−1)ⁿ⁻¹

q_n−1(θ_nq_n−1+q_n−2) (25)

But θ_i >0 and q_i → ∞

∴ lim

n→∞τ_n =θ

since RHS of (25) → 0. The proof of uniqueness is similar to that given above when θ ∈Q⁺.

Aside Numbers of the form α+β√

d, d ∈ N, d 6= m² are a field, F = Q(√

d), the

“extension” of Q by√ d:

1 α+β√

d = α−β√ d α² −β²d =

α

α²−β²d

−

β

α²−β²d

√

d∈ {α₁+β₁√ d}

Diophantine Approximation

Equation (25) implies

θ− pn

q_n

= 1

q_n(θ_n+1q_n+qn−1)

< 1

q_nq_n+1 (26)

The numbers q0, q1, . . . are strictly increasing in N. The continued fraction process pro- vides us with aninfinite sequence ofrational approximations to an irrational number,θ, namely the convergents ^p_qⁿ

n ∈Q. How rapidly do they approach θ?

By (26), if ^x_y is a convergent,

θ− x y

< 1 y²

It is possible to prove that (Hurwitz, 1891] any irrational numberθhas an infinite number of rational approximations which satisfy

θ− x y

< 1

√5y² (27)

This is the best possible: If we choose β > √

5 then there are numbers η ∈ R\Q for which there are only a finite number of rationals ^x_y with

η− ^x_y

< _βy¹2. e.g. the golden ratio

g = 1 + 1 1 + ¹

1+_1+...¹

= 1 + 1 g so g²−g−1 = 0 ⇒ g = ¹⁺

√ 5 2 .

Inequalities of the form (27) will be very important later when we study rational, alge- braic, irrational and transcendental numbers such as ⁴⁰¹₄₀₃, ¹⁺

√5

2 and e orπ.

Quadratic Irrationals

• solutions to quadratic equations with Z coefficients e.g. x²−2 = 0 ⇒ x=√ 2.

• simplest type of irrational e.g. (√

4 + 7^1/3)^1/5 is ‘more’ irrational as isπ (see later)

Ex θ= ²⁴⁻

√15

17 : 3<√

15<4 ⇒ bθc= 1 and θ= 1 + 1

θ₁

θ₁ = 1

θ−1 = 17 7−√

15 = 7 +√ 15 2 bθ₁c = 5

⇒ θ1 = 5 + 1 θ₂

θ₂ = 1

θ1−5 = 2

√15−3 =

√15 + 3 3 bθ₂c = 2

⇒ θ2 = 2 + 1 θ₃

θ₃ = 1

θ2−2 = 3

√15−3 =

√15 + 3 2 bθ₃c = 3

⇒ θ3 = 3 + 1 θ₄

θ₄ = 1

θ3−3 = 2

√15−3 =

√15 + 3

3 so θ₄ =θ₂

⇒ 25−√ 15

17 = 1 + 1

5 + ₂₊ ¹1 3+ 1

2+ 13

Ex

√

2 = [1, 2 ]

√

3 = [1, 1,2 ]

√

5 = [2, 4 ]

√6 = [2, 2,4 ]

H. Davenport, The Higher Arithmetic

Ex √

50 = [ 7,14 ]

Purely periodic fractions

Ex

√

2 + 1 = 2 + 1 2 + ₂₊¹1

√ ···

6 + 2 = [ 4,2 ]

These numbers are easier to deal with than those with a ‘preperiod’.

Ex

α = 4 + 1

1 + ₃₊ ¹1 4+ 1

1+ 1 3+···

= [4,1,3, α]

using the recursive equations we get convergents ₄

1,⁵₁,¹⁹₄,⁵₁, . . . . α= 19α+ 5

4α+ 1 ⇐ α= αpn−1+pn−2

αqn−1+qn−2

Hence 4α²−18α−5 = 0 and α is a quadratic irrational.

Now consider the number β which has the period of α reversed: β = [ 3,1,4 ] ⇒ β = 19β+ 4

5β+ 1

⇒ 5β²−18β−4 = 0

The equations are the same if −_β¹ =α ⇒ −¹_β is the second root of the equation for α called the (algebraic) conjugateof α orα.

In general letα= [a₀, . . . , a_n, α] be purely periodic, then α = pnα+pn−1

q_nα+qn−1

Letβ = [a_n, . . . , a₀] = [a_n, . . . , a₀, β] then (Ex) β = p_nβ+q_n

pn−1β+qn−1

As before −_β¹ is the conjugate of the root α.

Note: If β >1 then−1<−_β¹ <0.

Theorem 27 Any purely periodic continued fraction represents a quadratic irrational number α > 1 with a conjugate α satisfying −1 < α < 0. This conjugate is α = −_β¹ where β is defined by the C.F. of α with the period reversed.

Remark (Galois, 1828) This property characterises numbers with purely periodic con- tinued fractions.

Definition A quadratic irrational α is reduced if α >1 and−1< α <0.

Theorem 28 If α is reduced, its C.F. expansion is purely periodic.

Proof. There are integersa, b, c such that a α² +b α+c= 0. Solving for α:

α= −b±√

b²−4ac

2a = P ±√

D Q

where P, Q∈Z, D ∈N, D 6=m². Assume the sign is positive, else multiply by ⁻¹₋₁ so α= P +√

D Q so α, the other root, is

α= P −√ D

Q .

Note that

P² −D

Q = b²−(b²−4ac)

2a = 2c ⇒ Q|P²−D But 1< αand −1< α <0 so

(i) α−α >0 ⇒

√D

Q >0 ⇒ Q >0 (ii) α+α >0 ⇒ ^P_Q >0 ⇒ P >0 (iii) α <0 ⇒ P < √

D (iv) 1< α ⇒ Q < P +√

D <2√ D

⇒ P, Q∈N, P <√

D, Q <2√

D and Q|P² −D. (28)

Now expand α as a C.F.

α = a₀ = 1

α₁, a₀ =bαc, α₁ >1

⇒ α = a₀+ 1 α₁

⇒ α₁ = − 1

a₀−α ⇒ −1< α₁ <0

Hence α₁ is reduced also. Similarly α₂, α₃, . . . are reduced.

Now

1 α1

=α−a₀ = P +√ D

Q −a₀ = P −Qa₀+√ D Q

so let P₁ =−P +Qa₀ so

α₁ = Q

−P₁+√

D = P₁+√ D

Q₁ (29)

where Q₁Q=D−P₁² and Q₁ ∈Z since Q|D−P² and P₁ ≡ −P (mod Q).

Then

α₁ = P₁+√ D Q₁

and since α₁ is reduced, P₁ > 0, Q₁ > 0 and get the conditions (28) above using (29).

We carry on with the C.F. process, using α₁ instead of α, . . .. Each complete quotient

pn

qn has the form

α_n= P_n+√ D Q_n

whereP_n, Q_n satisfy (28)There are only afinite set of possibilities for the pairs (P_n, Q_n) so eventually we come to a pair (Pm, Qm) = (Pn, Qn), m > n so αm = αn and so the C.F. isperiodic from this point on.

Claim: The C.F. is purely periodic.

Subclaim: αn−1 =αm−1. If this were so we would be able to work back to get, eventually, α₀ =α_m−nproving pure periodicity. Proof of the subclaim: α_n=a_n+_α¹

n−1 ⇒ α_n =a_n+

1

αn+1. Letβ_n =−_α¹

n then−1< α_n<0 ⇒ 1< β_nand−_β¹

n =a_n−β_n+1orβ_n+1 =a_n+_β¹

n

so a_n = bα_nc = bβ_n+1c. Now let n < m and α_n = α_m so α_n = α_m ⇒ β_n = β_m and a_n−1 =bβ_nc=bβ_mc=a_m−1. Butα_n−1 =a_n−1+_α¹

n, α_m−1 =a_m−1+_α¹

m ⇒ α_n−1 =α_m−1. Applying this again successively α=α₀ =αm−n=α_r say, and

α = [a₀, a₁, . . . , ar−1, α_r]

= [a₀, a₁, . . . , ar−1, α]

= [a₀, a₁, . . . , ar−1] pure periodic with period length r.