7. 6 Best Possible Approximations
1+
343
Hv's
= Hv's + 1)
1)
and this is a contradiction.
Theorem 7.18 The constant v's in the preceding theorem is best possible. In
other words Theorem 7.17 does not hold if v's is replaced by any larger value.
Proof It
5. 7 Elliptic Curves
277
We conclude this section with a description of some further properties
of the group E/0) of rational points on an elliptic curve. First we
introduce some notation taken from the theory of infinite abelian groups.
An abelian group
308
Farey Fractions and Irrational Numbers
Corollary 6.14 If a polynomial equation (6.6) with en = 1 has a nonzero
rational solution, that solution is an integer dividing c 0
Corollary 6.15 For any integers c and n > 0, the only rational solutions, if
an
288
group of points on the elliptic curve
Some Diophantine Equations
P) has order
5.9 CURVES OF GENUS GREATER THAN 1
Let f(x, y) be a polynomial of degree d whose coefficients may be
as a projective curve,
rational, real, or complex. We speak of the set
t
326
in the range 0
Simple Continued Fractions
i
j, then equations (7.1) become
1
(7.2)
=a;+,
If we take the first two of these equations, those for which i = 0 and
i = 1, and eliminate
we get
1
= ao + .1.
a+I
In this result we replace
with the replacem
274
Some Diophantine Equations
general equation
y2 =
x 3 + ax 2 + bx + c.
(5.50)
In order that this should define an elliptic curve, it is necessary and
sufficient that D =F 0 where
(5.51)
is the discriminant of the cubic polynomial in x, discussed in App
7.8
351
Pell's Equation
Solution Write this as 8 + x 1, so that x = (1, 16). Observing that
x = (1, 16, 1, 16) = (1, 16, x), we get the equation x = 1 + (16 + x 1 ) 1,
which is equivalent to the quadratic equation
x 2 + 16x 1

16
=
0.
Solving this f
6.2
303
Rational Approximations
have lg anfbnl ;. lg akjbkl for all n = 1, 2, 3, . But lg akjbkl
> 0 since g is irrational, and we can find an n sufficiently large that
This leads to a contradiction since we would now have
The condition that g be irrat
7.3
331
Infinite Continued Fractions
Again we use equations (7.6) to get h;k;_ 1  h;_ 1 k; = (a;h;_ 1 + h;_ 2)
ki1  h;l(a;ki1 + k;2) = (h;lki2 h;2ki1) = ( l)il. This
proves the first result stated in the theorem. We divide by k;_ 1 k; to get
6.2
301
Rational Approximations
4. Let ajb and a'jb' run through all pairs of adjacent fractions in the
Farey sequence of order n > 1. Prove that
a'
min ( h'
a)
 b
1
= n(n  1)
and
b'
b
n
5. Consider two rational numbers ajb and cjd such that ad be = 1,
5. 7 Elliptic Curves
261
13. Let the triple (Xn, Yn, Zn) of integers be defined as in the proof of
Theorem 5.16, and let Hn = max(IXnl, IYnl). Show that Hn+l
1
n 2
;. zH;. Deduce that Hn ;. 10 4  for n ;. 2.
14. Apply the tangent method to the curve x 3
Some Diophantine Equations
264
Solution
We write F(X, Y, Z) = 2X 3

2XZ 2

Y
3
+ Z 2 , and find
that
aF =6X 2

ax
2Z
aF
2
'

aY
= 3Y 2 +Z 2
aF
'

az
= 4XZ + 2YZ.
Suppose that X 0 : Y0 : Z 0 is a point of IP'zcfw_C) at which these three expressions
7.8
353
Pell's Equation
are real numbers, not necessarily integers. Then
E
M 
IP
u
=
M( E
+ M/P)
and hence
0
E
< M 
IP
IP < M( E + M/P)
1
Also 0 < EjM /P implies Ej(M/P) > 1, and therefore
By Theorem 7.14, EjM is a convergent in the continued fraction
300
Farey Fractions and Irrational Numbers
Corollary 6.6 The nth row consists of all reduced rational fractions ajb
such that 0 ajb 1 and 0 < b n. The fractions are listed in order of
their size.
Definition 6.1 The sequence of all reduced fractions with d
Simple Continued Fractions
340
Proof We have x = (a 0 , a 1, ) and 1jx = (0, a 0 , ap ). If hn/kn
and
are the convergents for x and 1jx, respectively, then
=
0,
k 0 = 1,
h'I = 1,
k 0 = 1,
ki = a 0 ,
ho = ao,
k1
= ap
= ap
h1
= a0 a 1 +
=
a0 a 1
+
+ kn3
=
7.3
333
Infinite Continued Fractions
Theorem 7.9 Two distinct infinite simple continued fractions converge to
different values.
Proof Let us suppose that (a 0 , ap a 2 ,
by Lemma 7.8, [8] = a 0 = b 0 and
1
) =
(b 0 , b 1, b 2 ,
) =
8. Then
1
Hence ( a 1
5.8
Factorization Using Elliptic Curves
infinity, namely 0: 1: 0, so that the group E/ZP)
2p + 1. One would expect that the right side
residue (mod p) for roughly half the residues
E/ZP) should be close to p. Indeed, it is known
285
has order between 1 an
330
Simple Continued Fractions
which is true by equations (7.6). If n
(ao,x)
=
=
1, the result is
xh 0
k
X 0
+ h_ 1
k
+ 1
which can be verified from (7.6) and the fact that ( a 0 , x) stands for
a 0 + 1jx. We establish the theorem in general by induction
Farey Fractions and Irrational Numbers
306
For integers h, k with k > 0, we then have
The expression on the left in (6.3) is not zero because both g and
irrational. The expression lh 2  hk k 2 1 is a nonnegative
integer. Therefore lh 2  hk el
1 and w
266
Some Diophantine Equations
To construct a point B such that A+ B = 0, let L 1 denote the tangent
line to 6j(IR) at 0, as in Figure 5.4(b). The further intersection of this line
with 6j(IR) is called 00. Let L 2 denote the line through A and 00, wh
292
Some Diophantine Equations
divisor, and thus to settle the problem completely it suffices to show that
for each prime p > 2, the equation xP + yP = zP has no solution in
positive integers. Euler settled the case p = 3 in 1770, and Dirichlet and
Legend
8.1
361
Elementary Prime Number Estimates
Definition 8.1 The von Mangoldt function A(n) is the arithmetic function
A(n) =log p if n = pk, A(n) = 0 otherwise. We let 1/J(x) = I: 1.;n.;xA(n),
it(x) = Lp.;x log p, and '7T(x) = Lp.;xl.
The motivation for cons
278
Some Diophantine Equations
points, for which Y5 does not divide D, but from a general theorem of
Siegel it follows that the number of such points is at most finite. A precise
description of groups that can occur as tors(E/0) is provided by:
Mazur's Th
341
7. 6 Best Possible Approximations
< y b. Prove that every good approximation is a fair
approximation. Prove that every fair approximation is either a convergent or a secondary convergent to f
(d) Prove that not every secondary convergent is a "fair ap
390
Primes and Multiplicative Number Theory
the asymptotic mean value
that fit this description.
Theorem 8.24
E7 f(d)jd.
We consider several applications
For x ;. 2,
L
rp(n)jn = 
n
6
7T'
2x
+ O(log x).
Proof Taking F(n) = rp(n)jn, by (4.1) we see that (8
362
Primes and Multiplicative Number Theory
log n, so that
N1
T(x) =log N
N1
E
+
log n
x
+
n=1
E j n+1 log udu
n=1
= logx + J.Nlogudu
logx
n
+ J.xlogudu,
1
1
and hence T(x) x log x x + 1 + log x. The stated estimate follows on
combining these two bound
267
5. 7 Elliptic Curves
P6 = O(AB)
Figure 5.5. The elliptic curve 2x(x 2  1) = y(y 2
A = ( 1,  1), B = (  1, 1), C = (1, 1).

1), with 0 = (0, 0),
cubic 6"h(IR) has the four distinct points P0 , P 1 , P 2 , P3 in common with L, it
6"h(IR). Not onl
332
Simple Continued Fractions
( a 0 , ap ,an) = hn/kn = rn is called the nth convergent to the infinite
continued fraction. We say that the infinite continued fraction converges
to the value limn . " rn. In the case of a finite simple continued fraction
7.2
327
Uniqueness
Such a finite continued fraction is said to be simple if all the
integers. The following obvious formulas are often useful:
X;
are
1
The symbol [x 0 , xi>, x) is often used to represent a continued fraction.
We use the notation (x 0 , x