C OMPOSITIO M ATHEMATICA
B. B RINDZA
Zeros of polynomials and exponential diophantine equations
Compositio Mathematica, tome 61, n
o2 (1987), p. 137-157
<http://www.numdam.org/item?id=CM_1987__61_2_137_0>
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Zeros of polynomials and exponential diophantine equations
B. BRINDZA
Mathematical Institute, Kossuth Lajos University 4010 Debrecen Pf.l2, Hungary Received 17 February 1986; accepted 27 March 1986
@ Martinus Nijhoff Publishers, Dordrecht - Printed in the Netherlands
Several
diophantine problems
can be reduced to theequation
where
f
is agiven polynomial
with rational(or algebraic) integer
coefficients.For the
study
of thisequation
the distribution of themultiplicities
of the zerosof the
polynomial f
isimportant.
From someinequalities
of Mason and Brownawell-Masserconcerning
J-unitequations
over function fields we de- rive some new bounds for the numbers of distinct andsimple
zeros ofpolynomials, respectively,
in§2.
In§3
we combine these results with known results onequation (1), proved by
the Gelfond-Baker method.Doing
so weobtain some new results on
equations
oftype (1)
where m, xand y
arevariables,
and on some relatedequations.
§2.
Lower bounds f or the number of distinct zerosLet k be an
algebraically
closed field of characteristic zero andk(z)
therational function field over k.
Further,
let K be a finite extension ofk(z)
ofgenus g and the
height
of a non-zero element a of K definedby
where v runs
through
the(additive)
valuations ofK/k
with value group Z.The
following
lemma is thepowerful inequality
of[Mason, 1984].
LEMMA 1 Let -1-1 denote a
finite
setof
valuations onK,
and suppose that yl, Y2 and 03B33 are non-zero elementsof
K such thatv(03B31)
=V(Y2)
=v(03B33)
= 0for
allThis paper has been written during a stay at the University of Leiden which was made possible by
a scholarship of the Hungarian Academy of Sciences.
v ~ J and yl + y2 + 03B33 = 0. Then either
YI/Y2
is elementof
k and soHK(03B31/03B32)
=
0,
orwhere 1 Y 1
denotes the numberof
elementsof
Y.The
following
result is aspecial
case of Lemma 1.COROLLARY A
(R.C. Mason)
LetPl, P2
andP3
becoprime polynomials
ink [ z ]
with
Pl
+P2
+P3
=0,
and such that theirproduct P1P2P3
has N distinct zeros ink. Then either
Pl, P2
andP3
are all constants, orFor elements UI’...’ un of
K,
not all zero, we define the(projective) height
as(cf. [Brownawell
andMasser,
toappear]).
It is clear that
We
adopt
theterminology
of Brownawell andMasser,
and say that elements ul, ... , un of Ksatisfying
give
rise to anon-degenerate
solution if there is nonon-empty
proper subsetof {u1,...,un}
whose elements arelinearly dependent
over k. Let J be afinite set of additive valuations of
K/k.
We recall that an element a of K is said to be J-unit ifv(03B1)
= 0 for all valuations v ofK/k
not contained in J.LEMMA 2
[Brownawell
andMasser,
toappear]. Suppose
ul, ... , un(n
>2) give
rise to a
non-degenerate
solutionof (2)
and that ul, ... , un are J-unitsfor
somefinite
set !/o ThenLEMMA 3
[Brownawell
andMasser,
toappear]. Suppose
u1,..., un(n
>2) give
rise to a
non-degenerate
solutionof (2)
and thatfor
1j
n the elementui
is anJj-unit for
somefinite
set9).
ThenBy using
Lemma 2 we shall prove thefollowing
result.THEOREM 1 Let
be a
polynomial
inXl, ... , Xn
with non-zerocomplex coefficients
and suppose thatwhere
f l, ... , fn E C[X ]
arepairwise relatively prime
non-constantpolynomials.
Then the
polynomial P(X)
=F(f1(X),
...f (X))
has at least( m(F) N(N-1)-1 L deg fj distinct zeros.
j=l
Proof.
At first we show that the elementsaifk1l 1
...f,,k,,, (1
in)
arelinearly
independent
over C.Supposing
thecontrary
we havefor some
03BBl ~ C (1 i N)
with at least oneÀ
different from zero. More- over, we may assume that thecardinality
ofJ={1 | 03BBl ~ 0}
is minimal. Sincem(F)>0,
there are pqr such that03BBp03BBq03BBr ~0.
From Lemma 2 weobtain
where the valuation set J consists of the infinite valuations of
C( z )/C
and allthe finite valuation v such that
v(fi)
~ 0 for somei~ (1,.., n}
whenceFrom the definition of the
height
function we obtainComparing
theseinequalities
with(3),
we inferwhich contradicts
m ( F )
>N(N - 1).
Now,
we canapply
Lemma 2 to theequation
and we have
where s denotes the number of distinct zeros of the
polynomial P(X). Finally, (4) implies
the theorem.In the
special
case whenF( Xl, ... , Xn )
=a1Xk1
+ ··· +aNXkNN
we shallprove the
following stronger
result.THEOREM 2 Let
f1(X),..., fN(X)
be non-constantpairwise relatively prime polynomials
withcomplex coefficients
and ai, ... , aN non-zerocomplex
numbers.Suppose
thatThen the
polynomial Q(X)
=a1fk11(X)
+ ...+ aNfkNN(X)
has at least03BC N-1
distinct zeros.
Proof. Following
theargument
of theproof
of Theorem 1 we can see that there is no proper subsetf
of{1,..., NI
such that03A303BBiaifk1i
isidentically
zeroiE/
and
J-1 A =1=
0. Let t denote the number of distinct zeros of thepolynomial Q.
Then from Lemma 3 we have
Taking
the sum of theseinequalities
we obtainwhich
implies
our theorem.Now we shall
give
a lower bound for the number ofsimple
zeros of thepolynomial P(X)=aFn(X)+bGm(X)
wherea, b
are non-zerocomplex numbers;
n, m arepositive integers
andF, G~C[X]
are non-constantrelatively prime polynomials.
THEOREM 3
Suppose
thatpolynomial P(X)
has at leastSince n
deg
F is a trivial upper bound for the number ofsimple
zeros ofP,
our theorem shows that ’almost’ all zeros of P are
simple
anddeg
P 2 +Proof.
We assume thatP(X)
has therepresentation
with
By applying Corollary
AFrom
(5)
and(6)
we deduceand
Let 1 denote the number of
simple
zeros ofP(X).
Thenhence
This
implies
thatwhich proves the theorem.
COROLLARY 1
Suppose
that4 ( m - 2)(n - 2).
Then thepolynomial P(X)
hasat least two
simple
zeros.If
4 >( m - 2)(n - 2)
thenP(X)
may have less thantwo
simple
zeros.Proof.
In the case whenmin{m, n 1
> 3 ourcorollary immediately
followsfrom Theorem 3.
So,
we may assume that n = 3 and m 6.By Corollary
A wehave
It is easy to see that
deg F 2 3 ( k - 1)
anddeg G 1 3 (k - 1).
From(7)
weobtain
which proves the first
part
of thecorollary.
The
following examples
show thenecessity
of the condition4 ( m - 2)
(n - 2).
Case Take
Case Take
Case We may assume m = 2. Take
(cf.
F.Klein, Vorlesungen
über das Ikosaeder und dieAuflôsung
derGleichungen
vom fünftenGrade, Leipzig 1884.)
§3.
Power values ofpolynomials
Let
f
be apolynomial
with rationalinteger
coefficients and consider theequation
where b ~
0,
m >0, x
and zwith 1 z | >
1 are rationalintegers. [Tijdeman, 1976] proved m
is boundedby
acomputable
constantdepending only
onf
and b. He assumed that
f
has at least twosimple
rational zeros. Schinzel andTijdeman, 1976]
showed that thesupposition
of two distinct roots is sufficient.Later, [Sprindzuk, 1982]
and[Turk,
toappear]
gaveexplicit
upper bounds for theexponent
m. These results were extendedby [Shorey
andTijdeman,
inpress, Th.
10.3.)
to the moregeneral
case, when theground ring
is aring
ofintegers
in anarbitrary algebraic
number fieldand,
moreover,where ’1TI’..., ’1Ts
(s 0)
aregiven
non-zeroalgebraic integers. Gyôry, Tijdeman
and the author
[Brindza
etal., 1985] proved
thefollowing
result which made itpossible
togive
effective bounds for the solutions of somediophantine equations
connected with the Fermatequation
THEOREM A Let
f(X)
EZ[X]
be apolynomial
with at least two distinct zeros.Let {P1,..., Ps}
be afinite
setof primes.
Put( if
s = 0put Y= (11
and P =1.) If
a,b,
m, w, x and y are rationalintegers
with ab =1=
0,
w e Yand |y| >
1 such thatthen
where A =
max{ 1 al, 1 b|, 4)
andCI, C2
arecomputable
constantsdepending only
on thedegree
andheight of f.
Here,
we extend Theorem A to the case of anyalgebraic
number field. LetL be an
algebraic
number field withring
ofintegers Z L
and letf ( X)
EZ L [X].
Further,
let a,b,
u1,..., Us(s 0)
begiven
non-zeroalgebraic integers
in Land consider the
equation
where x, z E
ZL
and al’...’ 03B1s, m arenon-negative integers.
THEOREM 4
Suppose f(X)
has at least two distinct roots and 0 =A z is not a unitin L. Then
(10) implies
thatwhere U =
max|NL/Q(ui)|, M=max{|NL/Q(a)|, INL/Q(b)|, 4}
andC3, C4
areeffectively computable
constantsdepending only
onf
and L.Further, if
zis a unit but not a root
of unity
and s = 0 thenwhere
Ml
=max{H(a), H(b), 2)
andC5
is acomputable
constantdepending only
onL and f .
*The
following
result is asimple
consequence of Theorem 4.THEOREM 5
* We denote by H( a) the height of a. By the height of an algebraic number a we mean the height
of the minimal defining polynomial of a with rational integer coefficients.
in x, y E
Z L
and rationalinteger
t with t > 1 such that 0 ~cd(B2(t)- 4cA(t))
and 0 =A x is not a unit in
ZL satisfy
where
C6 is
aneffectively computable
constantdepending only
onL, A, B,
c and d.In the
special
case when L =Q
andA(X), B(X)
are constantpolynomials
this
yields
a result of[Shorey
andStewart, 1983].
On the
equation
Let p
be a fixed oddprime. [Inkeri, 1946] proved
that there exist at most afinite number of
relatively prime positive integers
x, y, z whichsatisfy
theconditions
and for which at least one of the
differences 1 x - yi,
z - x, z - y is less thana
given positive
number -4Y.Later, [Everett, 1973]
gave a newproof
for thisresult
by using
Roth’s famous theorem onapproximation
ofalgebraic
num-bers.
[Stewart, 1977]
and[Inkeri
and van derPoorten, 1976] independently, proved that,
for anypositive
number.aet,
allpositive integer solutions x,
y, z and n > 2 of theequation
satisfy max(
n, x, y,z} C7
whereC7
is aneffectively computable
constantdepending only
on -4Y.[Inkeri, 1976]
studied the moregeneral equation
where h and g are
given
non-constantpolynomials
with rationalinteger
coefficients. He gave an upper bound for the solutions which
depends only
onp, h and g under some conditions. In
[Brindza, 1984a]
the authorproved
thatthese conditions are unnecessary. For some similar and more
general
resultswe refer to
[Brindza et al., 1985].
Let K be an
algebraic
number field andD1,...,Ds (s 0)
be distinctprime
ideals in K.
Further,
let Y denote the set of all valuations of Kcorresponding
to
D1,..., Ds sand OK,Y
thering
ofJ-integers
of K. We recall that an element aof K is said to be
J-integral
ifv(03B1)
0 for all v ~ J.Using Corollary
1 weshall prove the
following
result.THEOREM 6 Let
F,
G ~K[X] ]
non-constantpolynomials
which arerelatively prime
and m, ngiven positive integers
such that( m - 2)(n - 2)
4. Then all solutionsof
theequation
with z > 2 and 0 ~ y is not a root
of unity satisfy max{
where
Cg
is aneffectively computable
numberdepending only
on m, n,F, G, K,
sand
maxNK/Q(Dl).
1
In the
special
case when K =Q,
s =0,
m = n and z( = n)
is fixed we havethe above mentioned result of the author.
On the
equation
Let K be an
algebraic
number field withring
ofintegers Z K
and letf1(X),...,fN(X)~ZK[X], (N > 1)
be non-constantpolynomials
which arepairwise relatively prime. Further,
let UI’...’ Us(s 0)
begiven
non-zeroalgebraic integers
in K. Write U =max |NK/Q(ul)| 1 again. Combining
Theo-rems 2 and 4 we
immediately
have thefollowing
result.THEOREM 7
Suppose min kl
>N( N - 1).
Then all solutionsof
theequation
in x, y E
ZK
andnon-negative integers
03B11,..., as, z with 0 =A y is not a unit(in ZK) satisfy
where
C9
andCl.
arecomputable
numbersdepending only
on thepolynomial
fk11(X)
+...+fkNN(X)
and K.Auxiliary
resultsLet K be an
algebraic
number field ofdegree
d and letR K, h K
and rK denote theregulator,
class number and unit rank ofK, respectively. By 1 a 1
we shalldenote the maximum of the absolute values of the
conjugates
of analgebraic
number a and we denote
by N(a)
the norm of a. It is known that if a EZK
with 0 :A a then
LEMMA 4
Suppose rK
1. There existmultiplicatively independent
units~1,...,
~rk in
K such thatfor
everyQ-isomorphism
cpof
K whereCil
is acomputable
constantdepending only
ond; further,
there areQ-isomorphisms
CPI’...’CPrK
such thatwhere
C12
is an absolute constant.Proof.
See[Stark, 1973]
and[Zimmert, 1981].
LEMMA 5 Let a be a non-zero element
of
K. There exists a unit E in themultiplicative
groupgenerated by
’TJI’...’’TJrK
such thatfor
everyQ-isomorphism cp of
K whereC13
is acomputable
numberdepending only
on d.Proof.
See e.g.[Gyôry, 1980].
The
following
lemma is a consequence of Baker’s estimateconcerning
linearforms.
LEMMA 6 Let al’...’ aN
(N 2)
be non-zero elementsof
K and letA1,..., AN (each 3)
be upper boundsfor
theheights of
al’...’ aN,respectively.
PutThere exist
computable positive
absolute constantsC14
andC15
such thatfor
every B 2 the
inequalities
have no solution in rational
integers bl, ... , bN
with absolute values at most B.LEMMA 7
If
xl, X2 and x3 are non-zeroalgebraic integers
in Ksatisfying
xi + X2 + X3 = 0 and Xl’ X2, X3 are J-units then
for
some 03C3 ~Z K
and Pi EZ K
we havewhere
C16
is acomputable
numberdepending
on K and J.Further, if max 1 N(Xl) |
Mfor
somepositive M,
then;
where
C17
is acomputable
numberdepending
on K and J.LEMMA 7 is a
special
case of Lemma 6 of[Gyôry, 1979].
LEMMA 8 Let a be a non-zero
algebraic integer of degree
n which is not a rootof unity.
There exists aneffectively computable positive
numberC18 depending only
on n such that
Proof.
See e.g.[Schinzel
andZassenhaus, 1965].
The
following
lemma is an effective version of a well-known theorem ofLeVeque.
Letf(X) ~ K[X] and
assume thatf
has therepresentation
with
aN ~ 0,
n > 0 and03B1l ~ 03B1j
fori ~ j. Further,
let m > 1 be a natural number andput
LEMMA 9
[Brindza, 1984b] Suppose that {t1,..., tn}
is not apermutation of
then-tuples
Then all
J-integral
solutionsof
theequation
satisfy
where P =
max N.p ( if
s =0,
put P =1)
andC20
is acomputable
constanti
depending
onK, f
and m.Proof.
See[Brindza, 1984b] (cf. [Shorey
andTijdeman,
inpress]).
Let 03B31,..., 1 Yn ’1TI’...’ 03C0s
( n
>1,
s0)
bealgebraic integers
in K with Yl =1=Y J
for i
=1= j
and suppose that 0 =A 03C0l is not a unit in K(1
is).
PutConsider the
equation
where z, y
~ J*,
r1,..., rn,m E N,
e is a unit and0 ~ y
EZK
is not a unit.Let T be a
positive
number.LEMMA 10
([Shorey
andTijdeman,
inpress]
Th.10.3)
Iffor
allprime
ideals D then all solutions x, z, E, y, y, mof equation (14)
underthe above mentioned conditions
satisfy
mC21
whereC21
is aneffectively computable
constantdepending only
onK, J*,
T and thebinary form
Proof of
Theorem 4We shall follow the
proof
of Theorem A(see [Brindza, 1985]).
We assumethroughout
thatf(X)
has therepresentation
with
ak ~
0 and03B3l ~ 03B3j
fori ~ j.
Let K =L(-yl,..., 03B3n)
and d =[K : Q].
Thenwe have
where 03B2i
=ak03B3i ~ Z K
and the number xcorresponds
to akx in(10).
In thisproof
cl, C2, ... will denotepositive computable
numbers whichdepend only
on
f
and L.Let m, ai 1... 1 a,, x, z be an
arbitrary
but fixed solution of(15). Using
square brackets to indicate
principal
ideals inK,
weget
At first we suppose that z is a unit but not a root of
unity
and s = 0. ThenNow we can
apply
Lemma 7 to theequation
We
obtain x - f31
= op, and03B21
-f32
= o?2 wherehence
rxl c2 Ml . Since |~(b-1)| 1
>(2M1)-1
for everyQ-isomorphism
(p of Kand |~(z)| >1+c3
for some (p, we obtain from(15)
thathence m c6
log MI.
This proves the secondpart
of Theorem 4.In the
sequel
we assume z is not a unit inZ K. Then |N(z)| 2
and from(15)
we deduceIf
H(x) c9M,
then our theoremimmediately
follows. We may thereforeassume without loss of
generality
thatH(x)
>clo M
for asufficiently large
clo which will be determined later. ThenLet
D1,...,Dt
t be the distinctprime
ideals of K which divide[ak][u1]... IU,Ilâl
whereThen we have
NDl c12Uc13
and tc14(S
+1).
We write the ideals[a] and [b] ]
in the
following
formwhere vl, w are
non-negative integers
andA,
9 areintegral
ideals which arerelatively prime to D1,..., Pt.
From(16)
we obtainwhere
A* 1 d, B*|B
arerelatively prime ideals,
9-n is anintegral
idealwhich is
relatively prime to D1,...,Dt and d l, ... , d
arenon-negative integers.
The
g.c.d.
of any two factors[x - 03B2l], [x - f on
the left hand side of(18)
is adivisor of the ideal
[ à ] and
eachprime
ideal divisor of A*B* divides at most one of the ideals[ x - 03B2l].
Hence we can writewhere
Al, B1, mi
i areintegral
ideals withAl|A, Bi|B, M1|M
fori=1,..., n
and thedji
arenon-negative integers.
We take the norm andlogarithm
of both sides ofequation (19);
if clo islarge enough
then we haveFor
symmetry
let usset r 1 = r2
andr2 =
rl. From(19)
we obtainPutting
By applying
Lemma 5 we may assume thatfor every
Q-isomorphism
ip of K and Thenand
By (21)
we havewhere h
= hKr1r2
and ~1, E 2 are units in K.By
Lemmas 4 and 5 fi i can be written in the formwhere ~1,..., 11r are units in K with
max H(~j)
c24, wi1,.., wir are rationalintegers
andmaxH(03BCi)
C25. From(23)
and(24)
we deducefor every
Q-isomorphism
T of K. Let us consider the absolute values of theexpressions
on theright-hand
side of(25).
It is clear thatand
(where
theproduct
is taken over allQ-isomorphisms
(J =1= (p ofK);
on the otherhand
Comparing (27)
with(28)
we haveHence, by (22)
and(23),
we obtainNow,
we consider theequation (25)
for r = rKappropriate Q-isomorphisms
ofK.
Using
Cramer’srule,
Lemma4, (26), (29)
and(30)
we haveLet us write
Wi’
in theform Wi’ = mw(1)ij+w(2)ij
withw(1)ij, w(2)ij~Z
and0 w(2)ij
m.Similarly,
f or 1 -1, 2 and j = 1,...,
tput dij = md(1)ij
+d(2)ij
withnon-negative integers d(1)ij, d(2)ij
such thatd(2)ij
m. Forbrevity
let us writeBy applying
Lemma 4 andinequalities (20), (22), (30)
and(31)
we haveThis
implies
thatFurther, by
the construction of the elements81, i
andby (22),
Setting wj
=w15) - w(2)2j
andDj
=d(2)1j - d(2)2j
we rewrite(23)
in the formIf
( x - 03B21)h
=( x - 03B22) h
then x =(~03B22 - 03B21)(~ - 1)-1
where 1 ~ e is a root ofunity
in K and soH(x)
C38. We have seen that clo may be chosenarbitrarily large. So,
we may assume that( x - 03B21)h~(x- 03B22)h.
For aQ-isomorphism
(p such thatx = 1 cp(x) we
obtainby taking
CIOlarge enough
Now,
weapply
Lemma 6 togive
a lower bound forFinally, comparing (33)
with(34)
we have Theorem 4.Proof of
Theorem 5From
(11)
we havefor
some 03BE ~ ZL.
It is clear thatwhere c44 and c4, are
computable
constantsdepending only
onL, B,
A and c.From Theorem 4 with s = 0 we infer that t is bounded. We may therefore
assume that t is fixed.
Hence,
Lemma 9implies
Theorem 5.Proof of
Theorem 6We assume the
polynomial P(X)
=Fm(X)
+Gn(x)
has therepresentation
with a ~ 0 and
03B1i~03B1j
for i=1= j.
FromCorollary
1 we haveP(X)
has at leasttwo
simple
zeros. Let K be thesplitting
field ofP(X)
withring
ofintegers ZK,
and write xl = ax
and Oi = aai,
i =1,...,
k. Then the elements03B21, ... , f3k
arealgebraic integers
in K andequation (12)
can be written in the formwhere d =
deg
P.Using
square brackets to indicateprincipal
ideals in K weobtain
where al,
bl, ... , as, bs
arenon-negative integers
and¥,
Wè areintegral
idealssuch that
It is known that there is an
integral
ideal q with bounded norm such thatDa11...DassD
=[ w ] for
somealgebraic integer
w. Thenand x2:=
x103C9~ Z K . Further,
it is clear that thegreatest prime
factor of thenorm of the
g.c.d.
of[x2] and [ w is
bounded. Let q 1, ... , abe
theprime
idealdivisors of
[ a d - lwd ] and
let h denote the class number of K. Writea h
=[03C0l].
i =
1,...,
t.By
Lemma 5 we may assume that maxfifl
is bounded. From(35)
we have
where e is a unit and
81, ... , 8t
arenon-negative integers. If y
is not a unitthen from Lemma 10 we have z is bounded.
If y
is a unit then x2 -f31l.AJ,
X2 -,82w
and(03B22 - 03B21)
areJ1-units
whereJ1
consists of all the valuations v of K such thatv(03C903C01
...03C0t(03B22-03B21)) ~
0.By applying
Lemma 7 towe infer that
is bounded. It follows that
H(x1), H(x)
andH(yz)
are also bounded. Then there is a bounded non-zero rationalinteger
A such thatAyz ~ Z K
thereforeDblzl| [A],
i =1,...,
s. If at leastone bl
ispositive
then we have z isbounded;
ifbl
= ...= bs
= 0then y E ZK and y
is not a root ofunity,
so,by
Lemma 8we obtain
which also
implies
z is bounded.Finally,
from Lemma 9 we have the theorem.Acknowledgements
1 would like to thank Prof. Dr. R.
Tijdeman
and Dr. F. Beukers for their valuablesuggestions.
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