SIGNATURES
OF ITERATED TORUS KNOTS
R. A. Litherland
Department of Pure Mathematics
and Mathematical Statistics
16 Hill Lane
Cambridge CB2 ISB
By an iterated torus knot I mean a knot obtained by starting with a torus
knot, taking a cable about it, then a cable about the result,
reason why this class of knots is interesting
of all one-component
curves
links of isolated singularities
(i.e. all algebraic knots).
whether
A
is an independent
may be interpreted
Recently,
and so on.
is that it contains
of complex algebraic plane
Lee Rudolph has asked
set in the knot cobordism group.
in any sense you desire;
One
the class
in particular
[Ru]
Independence
Rudolph asks whether
n
[K]
implies
n = I,
K1 = K ;
linear independence
=
~
i=l
[Ki]
K,K i e A
this is of course weaker than the usual notion of
in a ~-module.
We shall give an affirmative
answer to this question.
that the ordinary torus knots are linearly independent.
result of Tristram
[Tr] that the
(2,k)
We shall also prove
This generalises
torus knots are independent.
proved that result using the signatures
which he introduced,
shall use the following notation.
L
for
L
and
~
a complex number of modulus
of the Hermitian matrix
the cases
p
~
Let
a p'th root of unity,
f
[KT].
due to Viro
p
V
Denote by
a Seifert matrix
~(L)
the signature
(In fact, Tristram considered
~p(L).)
of these signatures
[Vii and also ~for
It will also be important
Tristram
and for which I
a prime picking one such root for each
signature by
We will need a re-interpretation
Taylor
I.
(I - ~)V + (i - ~)V T.
and denoting the corresponding
covering spaces,
be a link,
the
~ = -i)
in terms of branched
to Kauffman and
to look at all the signatures
Research supported by a grant from the SRC.
available
72
so a word of caution is in order here.
cobordism invariant
then
for every
o (LI) = a (L2)
if we define
fL
L
~ (jump in
L1
and
~ (L)
L2
~ ~ S I.
is a
are cobordant,
In particular,
~ (L)
at
$ = e 2~ix)
is a cobordism invariant.
of the signatures
be a link, and suppose
N
F
and such that
= 0 c H2(N,~N ).
integer
m,
[F,~F]
The vector space
automorphism
~F = L,
•
N
of
N,
branched over
(given by the orientations
F,
of
N
H (N;¢)
splits as a direct sum of eigenspaces
of the
~, :
=
Consider the (sesquilinear)
denote its signature by
the Atiyah-Singer
[Vi;54.8]
with
are to be oriented and maps orientation-preserving).
H2(N;C)
combinations
N
HI(N ) = 0.
Then we can form, for any positive
covering transformation
all manifolds
~N = S 3,
is a properly embedded surface in
the m-fold branched cyclic cover
with canonical
F:
referred to above goes as follows.
is a 4-manifold with
Suppose further that
and
if
by
The re-interpretation
Let
It is not true that
however,
for all but finitely many
fL : R ÷ £
fL(x)
then
~;
Q
C=I
intersection
o (N, ~).
g-si~latures
thereof
Ker(T, - ~)
~) H ,
~m=l
form restricted
These signatures
of the
(see Rohlin
=
£m-manifold
[Ro;§4]).
say.
to each
H
,
we
are closely related to
~:
in fact they are linea~
Under the above conditions,
Viro
shows that
~ (L)
=
~ (~], ~) - o(N)
(See also Kauffman and Taylor
side is an invariant of
the G-signature
L
for
[KT; Theorem 3.1].)
m
= 1 .
Note that the right-hand
by Novitov additivity and (a very special case of)
theorem.
Now, for integers
is, for sufficiently
p, q, r > i,
small
over some surface spanning a
6 ,
the Pham-Brieskorn
manifold
an r-fold cyclic cover of the 4-ball branched
(p.q)
torus knot (or link, if
p,q
are not
73
coprime);
the covering transformation
factor by
e 2~i/r.
For clearly
{z~ + z~ + z 3 = ~} n D 6
z~ + z~ + z 3 = 0
V~
is given by multiplication
with the covering transformation
has no singular points,
this manifold
stated;
as required.
o
,
Brieskorn
[Br] has calculated
(p,q)
the signature of
number of triples
0 < j < q,
(i,j,k) of integers,
0 < k < r
such that
number of triples such that
Inspecting Brieskorn's proof
than sesquilinear,
eigenvectors;
V~
0 < i < p,
0 < _i + _J + _k < 1 rood 2
p
q
r
-i < ! + i + [ < 0
p
q
r
mod 2 .
i.
If
K
is
< i,
then
o
(K)
~+
=
number
0 < j
~-
=
rather
form) one sees that this formula ariscs from a basis of
those of eigenvalue
22~is/r
correspond
to triples
( c f . Lemma 2 of Goldsmith
number
a
(p,q)
= ~ +
of pairs
< q,
- ~
_ ,
(i,j)
such
of pairs
torus
that
such
of
x-i
that
knot
and
~ =
(i,j,s).
[Go]).
2~ix
Proposition
e
,
x
rational,
where
integers,
< ~
P
+ ~
q
x < ~
p
0 < i
< x
+ ~
q
As it stands this formula is rather unmanageable;
shape we look at the associated jump function
f
P,q
i
to be
(and allowing for his use of the ~-bilinear,
This gives the following result.
o
It is not
torus knot,
where
(~+
0 < x
z3
since
is a 4-ball.
hard to check that the branch set meets the boundary in a
~+
of the
is an r-fold cyclic branched cover of
fK '
mod
< x+l
< p,
2
mod
2 . Q
to get it into a useful
which we shall denote by
Prop.l tells you to count, with the indicated signs, the lattice points
j)
in the interiors of the regions of the unit square shown in Figure i.
0
X
Figure 1
74
From this picture
it is transparent
fp,q(X)
number
that
of lattice points
on lower diagonal
(number of lattice points
If
fp,q(X)
lines,
~ 0,
i
point
j
( wz~-'~-)
line).
lies on one of these
we have
i + j" = x or l+x ; it follows that pqx c £ but
P
q
(Note that this is as it should be;
o
can be discontinuous
px,qx ~ £ .
only at roots
these
these
so that some lattice
line
on upper diagonal
x.)
lines;
il = i2'
of the Alexander
Horeover
for given two we have
Jl = J2 "
does exist.
Finally,
We can write
0 < a < q.
polynomial,
if
(il - i2)/P
pqx c Z,
pqx = ap + bq
correspond
to precisely
+ (Jl - J2 )/q c Z,
px,qx ~ Z
whence
such a lattice point
for some integers
a,b
with
There are two cases.
b
a
(1)
0 < b < p .
Then
(~-,~-)
(2)
-p < b < 0 .
Then
( b~_
P
Noticing
which
there can be at most one lattice point on the union of
that
we see that
sf ,q(X) = +i
f
or
is a lattice point
a)
H
is a lattice point
-I
according
can be described
on the lower line.
on the upper
as we are in case
as follows.
line.
(i) or (2),
For any integer
n = ap + bq,
P,q
let
h p , q (n)
where
[..]
denotes
period
pq.
Then
(-1) [a/q] + [b/p] + [n/pq]
=
integer part;
ip~,q(x)
this is clearly
= ~lhp'q(pqx)
if
pqx c £ ,
well-defined
and of
px,qx ~ g
(*)
10
Lemma
i.
The f u n c t i o n s
otherwise
f
:R
÷
are l i n e a r l y
~
independent.
P,q
Proof.
Suppose
given some dependence
maximum of the product
enumerate
the distinct
pq
relation
over those
factorisations
f
amongst
the
appearing
f
's.
Let K
P,q
in the relation,
P,q
of K into two coprime
numbers
as
be the
and
75
{ p l , q l } . . . . , {pn,qn } ,
1 < Pl
1
fp,q ( ~ )
By (*),
+ ql
<
"'"
Pi+qi
= fp,q ( ~ )
with
< Pn
+ qn
'
= 0
for
pq < K ;
it follows that the (n+l
tuples
Pl+ql
1
(~)
(%i,q i (K)'
fPi,qi
( ~ )
. . . . .
(Pn+qn
fpi,qi " K
),
i = l,...,n
are linearly dependent.
But by (*) fp,q (x) ~ 0 for 0 ! x < P+q
Pq
p+q
f,q (~)
= +l . Hence the n × (n+l) matrix with rows (~) is
-ii
-i-II
-I-I-Ii
which has rank
Since
T h e o r e m i.
group
[]
is an additive cobordism invariant of
The torus k n o t s are l i n e a r l y
(in the usual
sense).
those of its constituent
independent
parts.
We suppose given an unknotted
solid torus
winding number
and any knot
we construct a satellite knot
Theorem
2.
If
~
with
f
q
(core of
=
of iterated torus knots.
V c S3
in the interior of
K*
V) = K ,
is a root o f unity,
o (K*)
in the knot c o b o r d i s m
Putting this together with the preceding
with (algebraic)
f :V + S 3
this proves:
of a satellite knot are determined by
calculation will give a formula for the signatures
embedding
K,
0
We will now show how the signatures
K
while
?
n.
fK
,
~ q(K) + ~ (k)
.
and a knot
V.
k
contained
From this "pattern"
by taking a faithful
and setting
K* = f(k).
76
Remarks
(I)
Shinohara
considering
(2)
If
k
[Sh] has proved the case
a Seifert surface of
(V i, k i)
of winding number
of
K.z ,'
Lk(fi(Pi), K) = 0 .
while if
cq = 1
2.
If
=
signature)
0
K
and an embedding
Pi
Then the corresponding
a knot,
a 2-disc
fi
in
is
N
a 4-manifold
with
If
is
of
V.z
V.z
onto a
we require
cq # i,
The proof is similar.
N
$D = K
K
using a pattern
result holds provided
to within ±2(n-l).
there
D
K.
i
is a narallel_ curve of
it only holds
is
and
q
if
by
K*.
link, we can replace each component
neighbourhood
HI(N)
(the ordinary
is a link, the same result holds with the same proof.
an n-component
Lemma
¢ = -i
and
with
~N = S 3,
[D,~D]
=
0 ~ H2(N,ZN).
Proof.
We can obtain
unknotted
curves
undercrossings
K
by starting with an unknot
., Jm
Jl''"
in
SB - K
to overcrossings);
K
and performing
with framings
we c a n a s s u m e
E.1 = +i
-
surgery on
(to change some
Lk(J i, K) = 0 = Lk(Ji,J j)
(i # j ) .
Let
N = D 4 + h 2 +...+ h 2
where the 2-handle
h ? is attached along
m '
1
~i ' and let D be a 2-disc in D 4 with boundary
K.
o
J.
1
with framing
Proof
of
Theorem
Take
D x B2
N
of
and
D
embedded in
2.
D
in
as provided by the lemma,
N
D × B2
with
with
~D x B 2 = f(V).
~G = f(k).
can form the m-fold cyclic cover
m
¢ = 1
~¢(N, ~)
Also,
since
embedding
Hence if
f :V ÷ ~N
of
X
D x B 2 = ~-I(D x B2),
X = cl(N - ( D x B2)).
corresponding
and
and so extends
o'¢(D × B2, T)
Now let
of
G
be a surface properly
[G,~G] = 0 c H2(N,~N),
N
branched over
~¢(K*) + ~(N).
is faithful
V ÷ a(D x B2),
Then
(N, ~)
and take a neighbourhood
Let
where
=
Then
o¢(k)
so we
and for
(1)
[D,~D] = O,
f
is faithful
to a h o m e o m o r p h i s m
~: N + N
as an
S 3 ÷ ~(D × B2).
is the projection,
.
~-I(X)
to the h o n o m o r p h i s m
G,
then
(2)
is the unbranched
HI(X ) ÷ Z m
m-fold cover
given by linking number
77
with
G.
But this is just
depends on
~-l(x)
q
q
times
linking number with
but not otherwise on
: _~q '
and also write
assert that the following
o~ (Xq,
k
•q
and
G.
D,
To emphasize
for the restriction
of
T
so
~-I(x)
this we write
to
Xq
We
two formulae hold.
"rq)
=
o q(X1,
"Cl)
(3)
,--......___./
o~¢ [N, -c)
=
o g ( D x B 2 , • ) + o ¢ ( X~ q ,
(4)
-Cq) .
Now (i) - (4) give
oK(K* ) + or(N)
Taking
k
to be a core of
c~ (K)
+
V
o'g(k) + o q ( X ] ,
T1)
(s)
.
in (5) gives
a(N)
=
o (XI, "el)
and substituting
in (5) from this gives the desired result.
is ie~ediate
q
that
~I
is coprime to
corresponds
considering
For
if
suitable
m,
because
then
Notice that
Xq ~ Xl
(3)
in such a way
to ~q . The general case follows from this by
q
t-fold covers of X, where
t = m/hcf(m,q).
(4), consider the Mayer-Vietoris
sequence
H 2 (Xq n D × B 2) +H2(~q) Q H 2 [ D
× B 2] ÷H2([) ÷
(6)
÷ HI(X q n D x B 2)
Now
X n D × B 2 = D x ~B 2
union of solid tori;
Let
X ÷ S1
(Fig.2),
in particular
be a map inducing
D x ~B 2 ----~ X . >
t
so
X
q
n D x B 2 = D x ~B 2
is a disjoint
H2(X q n D x B 2) = 0 .
Lk(-, D)
S 1 q-~s 1
t
on
HI ;
we have the diagram
78
Figure 2
Pulling back the standard m-fold cover
I
Hence
mono and
m > S1
gives a diagram
!
D x ~B 2 ~
~
> ~I
D x SB 2 ~
X
> S~
H.(D × ~B 2) ÷ H.(~ )
> S1
q>
S1
is mono, so the homomor~hism
H2(~q) @ H 2 ( D x B 2) + H2(N )
the intersection
S1
form and commutes with
is an isomorphism.
T, ,
~
of (6) is
Since this preserves
(4) follows.
O
Now let
{PI' ql; "''; Pn' qn }
be a
knot.
(Pl,q])
(A
cable about a
(p,q) cable goes
about its companion.)
(P2,q2) cable about
p
times meridianally
We will only consider
generality we may assume
qi > I,
Pn > qn "
Then by induction
n
fK
=
Z
f
i=l Pi'qi;ri
... about a
and
q
pi,qi > 0 ;
Set
(pn,qn)
torus
times longitudinally
without
r i = qlq2
loss of
"'" qi-I
79
where
fp,q;r(X)
=
fp,q(rX)
.
Note that the signatures will certainly not suffice to give linear independence
of
A
in the usual sense, as a sum of
a sum of
in more than one way.
fK's
linearly independent.
In fact
f2,3;5
(Hence, for instance,
{6,5} ;
but
f
's can often be reassembled into
p,q;r
Also, the functions fp,q;r are not
=
f6,5 - f2,3 - f2,s - f3,5 "
{2,5;3,2} # {3,2} # {5,3}
{2,5;3,2} ~ A .)
has the same signatures as
However, the class
A
is sufficiently
restricted that the following rather weak result is enough.
Lemma
3.
coprime
If
pairs,
pq = p'q'r
then
with
k
p,q,p',q',r
coprime
to
> 1
K
and
(p,q),
(p',q')
with
k
f
P,q
unless
= K
there exists
{p,q} = {6,5},
(K)
¢
{p',q'}
fP ',q '; r (K)
= {3,2}
.
The proof of this is rather long and tedious, and is relegated to an
appendix.
Let
B
be the class of those
Pi > Pi+lqi+l
(i = 1,...,n-l).
{pl,ql;p2,q2;
... :pn,q n }
with
L~ [LDT;~I]
that
It follows from
A~B=,
so the following result includes the answer to Rudolph's question.
Theorem
is i n d e p e n d e n t
3.
in
the c o b o r d i s m
group,
in the s e n s e
that
n
[K]
=
Z
i=l
implies
that
n = i,
Observe that
[Ki]
K,~ i ~
K1 = K .
Pi > Pi+lqi+l
is equivalent to
piqiri > Pi+lqi+iri+l ;
it is then not difficult to deduce Theorem 3 from the lemma.
A small amount
of care is needed to ensure that the exception to the lemma causes no trouble.
80
REFERENCES
[Br]
E. B r i e s k o r n ,
B e i s p i e l e zur D i f f e r e n t i a l t o p o l o g i e
I n v e n t . Math.2 (1966) 1-14.
[Go]
Deborah L. G o l d s m i t h , Symmetric f i b r e d l i n k s , i n Knots, groups and
3 - m a n i f o l d s , Ann. o f Math. S t u d i e s 84, ed. L.P. Neuwirth
( P r i n c e t o n 1975).
[Hi]
F. H i r z e b r u c h ,
(1966/7).
[KT]
Louis H. Kauffman and Laurence R. T a y l o r ,
Amer.Math. Soc. 216 (1976) 351-365.
[LDT]
L~ Dang Tr~ng,
281-321.
[Ro]
V.A.
[Ru]
Lee Rudolph,
[Sh]
Y. Shinohara,
On the signature
156 (1971) 273-285
[Tr]
A.G. Tristram,
66 (1969)
[Vii
O.Ja.
APPENDIX
Proof
Lemma
by
A1
Sur l e s noeuds a l g ~ b r i q u e s , Comp.Math. 25 (1972)
Notices
submanifolds
39-48.
Amer. Math. Soc.
Some cobordism
251-264.
of four-dimensional
23 (1976)
of knots
invariants
manifolds,
410.
and links,
for links,
Trans.Amer.Math. Soc.
Proc. Camb. Phil. Soc.
3.
all variables
< a,b > .
Let
a,b
b
=
2
denote
integers.
We denote
the h.c.f,
of
a
in
We start with two lemmas.
> i,
and suppose
[ b " a (i - ~1 )] .
the interva~l
;
(ii)
a = 4 , b = 3 ;
(iii)
a = 6 , b < 5 ;
(iv)
Trans.
:
b
(i)
Signature of links,
Viro,
Branched coverings of manifolds with boundary and link
invariants I, }~ath.USSR Izvestija 7 (1973) 1239-1255.
of Lemma
and
and e x o t i c s p h e r e s , Sem. Bourbaki 314
Rohlin,
Two-dimensional
Func. Anal.Appl. 5 (1971)
In this appendix,
a
Singularities
yon S i n g u l a r i t a t e n ,
a = I0, b = 3.
Then
there
is n o i n t e g e r
one of the following
coprime
holds:
to
81
We c o n s i d e r
Proof.
three
cases.
a
(1)
a =
n > O.
2n+l,
n < ~a .
assumption
(2)
a
=
2(2n+1),
But
1
n
is
If
We h a v e
~
n
> 0,
to
4
Thus
b
case
n ~ 2 ;
< 2 + - 2n-1
if
a
= 10
and
a
=
n
If
n
a,
so
<2n-l,
b
= 1
b < 2 + n_i '
Thus
O.
coprime
< n,a >
= 0,
a>
a
1
so
= 2
we h a v e
=
and
,
so b y
b = 2.
and
a
n < ~
a
so
~
1
~ 1 £ a(1
-~
).
contradiction.
and
a
< ~
2n-1
Hence either
b = 2
"
n = 1
then
a = 6
and
or
b
,
so
2n-1
< 6,
2n-1
a
<
B
< 4.
In
the
while
if
n
latter
= 2,
~ 3 .
a
(3)
4n,
< ~a
2n-1
In
the
Y2 < Y1
'
and
Suppose
further
Lemma
A2
> O.
.
We h a v e
Thus
latter
< 2 + 2 n2 - ~
b
case
Suppose
that
<2n-l,a
a = 4
that
a
there
is
>
Hence
"
and
= 1
b
2n-1
either
b
are
coprime
< 7
= 2
'
or
so
n = 1
< 4.
> 2 ,
Y1
no
coprime
to
Then
either
or
6
and
y2
and
a
to
with
a
T2 < ~ < Tl
with
a
Y1
that
- T2 > 2
"
(i)
a =
3
and
Yl
~ 1
mod
3 ;
(ii)
a =
6
and
Y1
~ 5
mod
6 .
Proof.
We m a y
assume
0 < Y2
< a .
There
so
2 > ~- .
are
two cases.
a
(I) Y2 = a-i
.
Then
Y1 = a+l,
(2) Y2 ~ a-i .
Then
Y1
Hence
a = 3
and
Y1 = 1
rood 3.
a
< a .
We m u s t
have
Y2 < ~ < ¥I,
from which
a
follows
is
no
Lemma A1
From now
p,q,p',q'
further
with
b
>~ 2, < p , q >
assume
is to u s e
P,q
cases,
coprime
that
does
but
the
( k..)
{p,q}
that
This
we
the
Hence
a
=
i,
and
of Lemma
r => 2
( Kk )
or
I4
'
hence
3.
and
That
5, 2 < ~ .
Thus
3a
~1"
Applying
there
Y1
[]
= 5 rood 6 .
is,
pq = p'q'r
whenever
<k,K
= K .
>
=
I.
We
Our
and
{ p ' , q ' } = {2,3} . The b a s i c i d e a
l
has period
r ' a n d try to p r o v e t h a t
obscured
of Lemma
interval
a = 6,
= ~f , ,q';r
fp',q';r
it is the b a s i s
~ < a-2y 2 ,
the
situation
= {6,5}
becomes
in
find
<p',q'>
fp , q
that
fact
not.
=
"
to
= 4
on we c o n s i d e r
a i m is to p r o v e
f
Y1 = a - Y 2
that
integer
it
a
in the c o n s i d e r a t i o n
A 3 below.
of several
special
82
Let
s ,
Lemma
If
p = stu ,
and
<u,s>
A3.
< ~ ,q > =
Set
B1q
-
1
s = <p,p'q'
,
every prime
0 < B l < #2 < P • < ~i,P > =
I~l g
factor of
t
divides
f
and
~I
B2
~
mod s .
~2 q ]
~--I
~ ~ [~"
then
k i = (q-~)p + Biq ;
then
<ki,K
> =
I.
Suppose
#2q
<
~
<
-
P
-
P
2pq,
0 < k I < pq < k 2
Hence
fp,q (-~-)
But
>
I.
Suppose
Proof.
-
where
=
g I 5 g2
fp~,q,;r
mod s
(~)
implies
= fp~,q,;r
=
(-1)
_-
(_1) i - 1
that
(~
k I ~ k2
mod p'q' ,
a contradiction.
],
and hence
Since
a #
that
l
P
,
we are done.
we now distinguish
In cases
From
(1)
t > 1
(2)
t
=
(5)
t
= i,
(4)
t = I,
(i) and
(*)
z~
~,
possibilities
u = 3,
s = 4
-- 1,
that
/ (3,4)
c~ #
~ #
,q(l -~-)
s > I)
p', q'
exclude all combinations
~hen
t > l ~
(p,q) = (3,4),
for
(u,s)
u ~ 2 .
<.,q>
(I~ note that
then gives
u ~ 3,
(2) we shall prove
(*) it follows
(in case
four cases.
;
and
r
since
and since
(5,6) or (3,10).
except
I~, q]
in these cases
< p,q > =
I,
p > 2
Lemma A1
There are only finitely many
in each case,
(p,q) = (5,6),
and direct
{p',q'}
calculation
= (2,3}
.
will
83
(l)
t > 1
(2)
Lemma
.
t = i,
with
81 = I,
for s o m e
coprime
8 =
then
u
and
(u + Y 2 ) s
;
Lemma
there
u =
and
(s-3,2s-3)
=
Yl
Yl
(1,2s+l)
t = i,
B 1 = l,
Lemma
A1 no
arguments.
(3.1)
There
Hence
we c a n
8 < p •
(*)
a n d so
integer
Also
Since
8 ~ 1
is p r o v e d .
U
- ~2 > ~ '
~i
A3
so s e t t i n g
n
factor
q
(1)
mod
s,
If not,
By
Since
s = 4
(81,82)
=
is
(l,s+l)
(2)
that
if
<~,q
A3 w i t h
complete.
case
can o b t a i n
Lemma
(*).
is n o w
In this
Apply
6.
>
D = 12.
= 1
Apply
L e m m a A3
5q
~ ~ Iii22
I
then
a contradiction
by similar
two s u b c a s e s .
Take
~ i.
are b o t h
take
of
3.
s z 5 mod
to p r o v e
and
but we
~ = n
T
~
= l,
q
3n+l
n < - 12
either
is c o p r i m e
and again
obtain
this
exhausts
Let
q = ozv ,
divides
assume
Then
= n.
or
absurd.
Since
n ~ i.
we may
s ~ 1 mod
A3 to the h a i r s
Then
6.
u -< 2.
in
and
,
Then
we f i n d
applies
< q,12 >
every prime
p
so
Lemma
1 = yl s + 6u
greatest
- u ,
1 < 8 •
P8 => ~1 ,
Lemma
s = 4.
These
t = i,
be the
and
apply
so t h a t
(*).
(,).
in c a s e s
are
3n+l,
Since
(4)
p
and
(s+2,3s+2)
B2 = 5 "
q = 3n+2,
odd.
Y2
Y2 > Yl
If
3.
~ 5 mod
u = 3,
n > 5(3n+])
12
(3.2)
to
.
Apply
and
longer
q =
~ I mod
to p r o v e
6,
with
a ~ |" ~v , ~ -S |q
s ~ 7.
The proof
(3)
is c o p r i m e
s = i,
Y1
gives
are two possibilities.
we h a v e
u =
(BI,B2)
Let
Then
(u + Y2 - YI )s < ~
3,
(3),
(2.2)
YI.
SU
i.e.
case
If
choose
= 1 .
than
8 2 = 1 + s(t-l)u
(3,4).
s > I,
< Yl,u >
that
8 - 1 <
(2.1)
~
If
8
B l = I,
+ ~u = s u + i = p + i,
A3 shows
A2,
(u,s)
less
+ 6u ,
8 < (u + Y l ) S
Lemma
3,
82 = p-l.
~ ;
to
u ~
A3 w i t h
that
~,
and
• = l
case
to
n
must
a contradiction.
(3).
where
< ~ ,o >
and
p = 12,
o =
= i .
v -< 2.
Now
<q,p'q'
>
,
By s y m m e t r y
if
u = w = 1
be
84
then
so
r = i,
r = 2
p' < q',
contrary
and
p'q'
and notice
to hypothesis.
is odd.
that
Also,
Assume
(without
u
and
v
cannot both be 2,
loss of generality)
2p' + q' < p' + 2q' < p'q'.
Now,
p+q
characterised as the least positive
k such that
( k,K} =
k
,q ( ~ ) = +i ; similarly
2p' + q'
is the least positive
<k,K
) = 1
and
p+q = 2p' + q' ;
f',q';r
Now consider
< k ,K ) =
1.
= +i
k = p' + 2q'.
We have
and
k
with
~ p' + q',K ) ~ i ).
pq = K = (2p')q' ,
k
(K)
= (-1)
l+q'
k = ~
2p' + (2-p')q'
%,q
(since
can be
1
we have
Hence
{p,q} = {2p',q'}
0 < k < p'q'
< K
and
So
%',q';r
But also
(~)
since also
that
k
(K)
rhis contradiction
= f2p',q'
disposes
k
(K)
[ 1 / q ' ]+
,
=
+1 .
so
= (-1)
of case
[2/p' ] + [ k / p ' q ' ]
[(l+q')/2q']+[(2-D')/2p']+[k/K]
(4) and completes
-
= -
the p r o o f of Lemma 3.