53 Ex Equations like y 2 = x 3 +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 y 2 = x 3 + 7 is insoluble over Z. Proof. If x is even, x =2α ⇒ RHS = 8α 3 +7=8β + 7, where β = α 3 . But 0 2 ≡ 0, 1 2 ≡ 1, 2 2 ≡ 4, 3 2 ≡ 1, 4 2 ≡ 0, 5 2 ≡ 1, 6 2 ≡ 4 and 7 2 ≡ 1 (mod 8) so y 2 ≡ 7 (mod 8) has no solution. Hence x is odd. Write y 2 +1 = x 3 +8 = (x + 2)(x 2 - 2x + 4) = (x + 2)((x - 1) 2 + 3) If x =2n + 1 (odd) then (x - 1) 2 +3=4n 2 +3=4m +3,m = n 2 so (see back) must have a prime factor of the form p =4‘ + 3. But then y 2 +1 ≡ qp ≡ 0 (mod p) But (lemma later) p ≡ 3 (mod 4) ⇒ y 2 ≡-1 (mod p) has no solution. We frequently need to know the answer to the following: When does x 2 ≡ r (mod p) have a solution x? Or, more generally, x 2 ≡ α (mod m). The answer is given by the theory of quadratic reciprocity due to Gauss. This will be developed later.
Transcript
53
Ex Equations like y2 = x3 +7 are called “elliptic curves”. They arise in solving integralsfor, 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. But02 ≡ 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) musthave 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 thetheory of quadratic reciprocity due to Gauss. This will be developed later.
54
8 Pell’s Equation
x2 −Ny2 = 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 − Ny2 = x2 − (My)2 = (x −My)(x+My) = 1 ⇒ x−My = 1 and x+My = 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 good rational approximations for square roots,since x2 = Ny2 + 1
⇒(x
y
)2
= N +1
y2
⇒ xy≈√N if y is large.
Note: This type of equation has a long and interesting history, and has lots of applica-tions, especially to fields F = Q(
√N).
Ex (Euler, 1770) A triangular number has the form n(n+1)2
. Which numbers are bothtriangular 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 A fundamental solution to x2−dy2 = 1 is (r, s) where any other positivesolution satisfies r < x and s < y.
Theorem 25 (Lagrange) Let (r, s) be the least positive (or fundamental) solution tox2 − dy2 = 1, where d is not a square. Then every solution to this equation is given by(xn, yn) where
xn +√dyn = (r + s
√d)n
for n = 1, 2, 3, . . .
Proof.
x2n − dy2
n = (xn + yn√d)(xn − yn
√d)
= (r + s√d)n(r − s
√d)n
= (r2 − s2d)n = 1n = 1
55
Hence (xn, yn) is a solution.
Let (a, b) be a solution. Suppose ∀n = 1, 2, 3, . . . , (a, b) 6= (xn, yn). Then there is apositive 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)
Let u+ 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 · 1m = 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 so u > 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) = (xn, yn) for some n. �
Finding the least positive solution is not easy however and requires the theory of continuedfractions of J. L. Lagrange. Frenicle’s table for non-square d up to 50 is given below.
56
Pell's equation"f,,'
~l:lPuler, after a cursory reading of Wallis's Opera Mathematica, mistakenlyr~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 hist~orks. He divised a cattle problem that would involve immense calculationj~?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 theplains of the Sicilian isle Trinacria and that were divided according to colorinto four herds, one milk white, one black, one yellow and one dappled,
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 'goforth 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, wasdiscovered by Carl Amthov in 1880. His solution implies that the numberof white bulls has over 2 X 105 digits. The problem becomes much moredifficult when the eighth and ninth conditions are added and the firstcomplete 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 offeredthe integral solution x = 2m2 + 1 and y = 2m. Pellian equations are found
in Hindu mathematics. In the fourth century, the Indian mathematican
--
57
58
9 Continued Fractions
Ex
1 +1
2 + 13+ 1
4
= 1 +1
2 + 113/4
= 1 +1
2 + 413
= 1 +1
30/13
= 1 +13
30
=43
30
looks silly until we consider some interesting continued fraction expansions
Definition By a simple continued fraction (or C.F.) we mean an expression
a0 +1
a1 + 1a2+···
= [a0, a1, a2, . . .]
where a0 ∈ Z and ai ∈ N for i > 1.
Note: [a0] = a0
1, [a0, a1] = a0a1+1
a1, [a0, a1, a2] = a2a1a0+a2+a0
a2a1+1
Generally, [a0, . . . , an] = pnqn
where pn and qn are polynomials in the ai, linear in any given
aj, and a0 does not occur in the denominator qn. (pn, qn) are called the nth convergents.
Note: [a0, . . . , an] = [a0, . . . , an−1 + 1an
]
59
Proposition If [a0, . . . , am] = [b0, . . . , bn], ai, bi ∈ N, am, bn > 1 then m = n andai = bi ∀i.Proof. This follows by induction from
[a0, . . . , am] = a0 +1
[a1, . . . , am]= b0 +
1
[b1, . . . , bn]
if we can show [a1, . . . , am] > 1 when a1, . . . , am > 1. But this is so since [a1, . . . , am] =a1 + 1
a2+··· . �
Let ai > 0 and ∀n let τn = [a0, . . . , an] then τn can be computed using the recursiveformulas, for n ≥ 2:
p0 = a0 p1 = a0a1 + 1 pn = anpn−1 + pn−2
q0 = 1 q1 = a1 qn = anqn−1 + qn−2
so τ0 = p0q0, τ1 = p1
q1and τn = pn
qn
Proof.
τn = [a0, . . . , an] = [a0, . . . , an−1 +1
an] =
p′n−1
q′n−1
where these belong to a0, . . . , an−2, an−1 + 1an
i.e. (induction)
p′n−1
q′n−1
=
(an−1 + 1
an
)pn−2 + pn−3(
an−1 + 1an
)qn−2 + qn−3
=an(an−1pn−2 + pn−3) + pn−2
an(an−1qn−2 + qn−3) + qn−2
=anpn−1 + pn−2
anqn−1 + qn−2
(induction again!)
Hence pn = anpn−1 + pn−2 and qn = anqn−1 + qn−2 �
(pn, qn) are called the nth convergents of the C.F.
Let θ ∈ R \ Z, θ > 1. a0 = bθc so θ = a0 + 1θ1, θ1 > 1 defines θ1. Continue with
θ1 = a1 + 1θ2
so a1 = bθ1c , θ2 > 1 if θ1 6∈ Z etc θn = an + 1θn+1
, an = bθnc , θn=1 > 1 ifθn 6∈ Z. We get
θ = a0 +1
a1 + 1a2+ 1
...+ 1
an+ 1θn+1
so θ = [a0, a1, . . . , an + 1θn+1
]
60
Proposition The expansion stops if θn = an is in N and then θ ∈ Q+ i.e. is a positiverational number. Conversely, if θ ∈ Q+, the C.F. expansion is finite.
as if we were doing the Euclidean algorithm. These equations give
θ = θ0 =u
v= a0 +
r1v
= a0 +1
v/r1= a0 +
1
θ1
θ1 = a1 +r2r1
= a1 +1
r1/r2= a1 +
1
θ2
...
θn =rn−1
rn∈ N
so the C.F. expansion is finite. �
Proposition ∀n > 2
θ =θnpn−1 + pn−2
θnqn−1 + qn−2
Proof. The definition of θn is θ = [a0, . . . , an−1, θn] so θ = τn = pnqn
= θnpn−1+pn−2
θnqn−1+qn−2using an
and θn for this particular C.F. �
Ex√
2 = [1, 2, 2, . . .]
(√
2− 1)(√
2 + 1) = 2− 1 = 1 ⇒√
2− 1 = 11+√
2so√
2 = 1 + 11+√
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 + 11+√
2
= 1 +1
2 + 11+√
2
= 1 +1
2 + 12+ 1
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 ].
61
Every quadratic irrational has a periodic continued fraction—this characterises quadraticirrationals.
Ex√
2 = [1, 2, . . . , 2, 1 +√
2] so a0 = 1, a1 = 2, . . .
p0 = a0 = 1q0 = 1
}τ0 =
p0
q0=
1
1= 1
p1 = a0a1 + 1 = 3q1 = a1 = 2
}τ1 =
p1
q1=
3
2= 1.5
p2 = a2p1 + p0 = 7q2 = a2q1 + q0 = 5
}τ2 =
p2
q2=
7
5= 1.4
and the approximation τn ≈√
2 gets better.
Theorem 26 Let a0 ∈ Z, ai ∈ N, i > 1. Then (τn) converges to an irrational numberθ. The ai are uniquely determined by the C.F. expansion of θ. Conversely, if θ is anirrational number, and τn = [a0, . . . , an] are obtained by expanding θ as a C.F. then
θ = limn→∞
τn.
62
Proof. The sequences (pn) and (qn) are both strictly monotonically increasing sequencesof natural numbers.
Claim:pnqn−1 − pn−1qn = (−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
since RHS of (25) → 0. The proof of uniqueness is similar to that given above whenθ ∈ 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
63
Equation (25) implies ∣∣∣∣θ − pnqn
∣∣∣∣ =1
qn(θn+1qn + qn−1)
<1
qnqn+1
(26)
The numbers q0, q1, . . . are strictly increasing in N. The continued fraction process pro-vides us with an infinite sequence of rational approximations to an irrational number, θ,namely the convergents pn
qn∈ Q. How rapidly do they approach θ?
By (26), if xy
is a convergent, ∣∣∣∣θ − x
y
∣∣∣∣ < 1
y2
It is possible to prove that (Hurwitz, 1891] any irrational number θ has an infinite numberof 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 xy
with∣∣∣η − x
y
∣∣∣ < 1βy2
.
e.g. the golden ratio
g = 1 +1
1 + 11+ 1
1+...
= 1 +1
g
so g2 − g − 1 = 0 ⇒ g = 1+√
52
.
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
2and e or π.
Quadratic Irrationals
• solutions to quadratic equations with Z coefficients e.g. x2 − 2 = 0 ⇒ x =√
2.
• simplest type of irrational e.g. (√
4 + 71/3)1/5 is ‘more’ irrational as is π (see later)
These numbers are easier to deal with than those with a ‘preperiod’.
Ex
α = 4 +1
1 + 13+ 1
4+ 1
1+ 13+···
= [4, 1, 3, α]
using the recursive equations we get convergents[
41, 5
1, 19
4, 5
1, . . .
].
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 − 1β
= α ⇒ − 1β
is the second root of the equation for α
called the (algebraic) conjugate of α or α.
In general let α = [a0, . . . , an, α] be purely periodic, then
α =pnα + pn−1
qnα + qn−1
Let β = [an, . . . , a0] = [an, . . . , a0, β] then (Ex)
β =pnβ + qn
pn−1β + qn−1
66
As before − 1β
is the conjugate of the root α.
Note: If β > 1 then −1 < − 1β< 0.
Theorem 27 Any purely periodic continued fraction represents a quadratic irrationalnumber α > 1 with a conjugate α satisfying −1 < α < 0. This conjugate is α = − 1
β
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 integers a, b, c such that aα2 + 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 thatP 2 −DQ
=b2 − (b2 − 4ac)
2a= 2c ⇒ Q |P 2 −D
But 1 < α and −1 < α < 0 so
(i) α− α > 0 ⇒√DQ> 0 ⇒ Q > 0
(ii) α + α > 0 ⇒ PQ> 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 |P 2 −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
67
Hence α1 is reduced also. Similarly α2, α3, . . . are reduced.
Now1
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 − P 21 and Q1 ∈ Z since Q |D − P 2 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 quotientpnqn
has the form
αn =Pn +
√D
Qn
where Pn, Qn satisfy (28)There are only a finite set of possibilities for the pairs (Pn, Qn)so eventually we come to a pair (Pm, Qm) = (Pn, Qn) , m > n so αm = αn and so theC.F. is periodic 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 = αm−n proving pure periodicity. Proof of the subclaim: αn = an+ 1
αn−1⇒ αn = an+
1αn+1
. Let βn = − 1αn
then −1 < αn < 0 ⇒ 1 < βn and − 1βn
= an−βn+1 or βn+1 = an+ 1βn
so an = bαnc = bβn+1c. Now let n < m and αn = αm so αn = αm ⇒ βn = βm andan−1 = bβnc = bβmc = am−1. But αn−1 = an−1+ 1
αn, αm−1 = am−1+ 1
αm⇒ αn−1 = αm−1.
Applying this again successively α = α0 = αm−n = αr say, and
