C OMPOSITIO M ATHEMATICA
R. T IJDEMAN
On integers with many small prime factors
Compositio Mathematica, tome 26, n
o3 (1973), p. 319-330
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319
ON INTEGERS WITH MANY SMALL PRIME FACTORS by
R.
Tijdeman
Noordhoff International Publishing Printed in the Netherlands
Let p
be aprime, p ~ 3,
and let n 1 = 1 n2 ... be the sequence of allpositive integers composed
ofprimes
p. It follows from a theorem of Stormer[22],
thatThis result was
improved
uponby
Thue[23 ].
He derived from his famous result on theapproximation
ofalgebraic
numbersby
rationals thatSee also
Pôlya [15]. Siegel’s
well knownimprovements [18, 19]
of Thue’sresult made it
possible
toimprove
upon(1).
Such animprovement
canbe obtained
immediately
from a result of Mahler[13, 1,
Satz2]. But,
as far as 1know,
it was not before 1965 that such a result was statedexplic- itly
in literature. In a survey paper[7,
p.218] Erdôs
gave thefollowing improvement
of(1).
Let 0 8- 1. ThenIn this paper we
give
severalimprovements
andgeneralizations
of in-equality (2).
In our Theorems 1 and 2 theproblem
offilling
the gap be- tween(2)
and theopposite
resultlimi~~ni+1/ni
= 1(cf. [15])
is solvedalmost
completely.
We shall prove that there exists aneffectively
com-putable
constant C =C(p)
such thatwhile we shall show that this
inequality
is false for Cn(p)- 1. (Here 03C0(x)
denotes the number ofprimes ~ x.)
It follows from
(1)
that for everyinteger k ~
1 there exists a numberAkp
such that ni + 1- ni > kfor ni
>Akp.
Because of its ineffective na-ture it is
impossible
to obtain estimates forAkp
from the methods of Thue andSiegel. Using
a resultof Gel’fond,
Cassels[5]
gave the first "effectiveproof" of (1).
In1964/65
D.H. Lehmer[11, 12]
gaveexplicit
upper boundsfor
A1p, A2p
andA4.
forall p.
In the Theorems 3 and 4 upper bounds aregiven
forAkp
for all k and all p and also forN~
in(2).
A consequence of this result is that if twointegers a
and b with 0 a b have the samegreatest prime factor,
then theinequality b - a ~ 10-6 log log a
holds.Professor Straus
suggested
to me ageneralization
of(2)
in anotherdirection, namely
forintegers ni
which have manyprime factors ~ p,
but are not
necessarily composed
ofprimes ~
p. Such a result isgiven by
Theorem 6.Finally
we solve aconjecture
of Wintner[7,
p.218] in
the afhrmative.We prove even much more in Theorem 7:
Let 0 ~ 1. There exists an
infinite
sequenceof primes
p, P2 ...such
that, if
ni n2 ... is the setof
allintegers composed of
theseprimes,
thenRecent results
proved by
the beautiful Gel’fond-Baker method form the basic tools fornearly
allproofs
in this paper. The results used for theproofs
of Theorems 1-5 have all beenpublished.
Since it is our main inter-est to indicate the
possibility
of theapplications,
we did not bother toomuch to get as
good
results aspossible,
but it should be noted that it ispossible
toimprove
upon our resultsby using
not yetpublished
resultsof Stark and
others,
orby improving
upon theauxiliary
results. Thisis
possible
because of additional information available here. For theproofs
of Theorems 6 and 7 a not yetpublished
result of Baker[3]
hasbeen used. This result is cited in section 7. I wish to express my thanks to Dr. A Baker for
communicating
the exact formulation of his result to me.1. As usual
03C0(x)
denotes the number ofprimes ~
x. The number ofdistinct
prime
factors and the greatestprime
factor of apositive integer
a are denoted
by 03C9(a)
andP(a), respectively.
For the sake ofbrevity
wewrite e.c. for
effectively computable.
In this paper all constants c, cl , C2,..and
C, C1, C2, ...
are e.c.2. We use a result of Fel’dman to prove
THEOREM 1. Let a and b be
positive integers,
3 a b. Put r =cv(ab),
p =
P(ab).
Thenwhere
Ci = cr41(log p)14r2 and Ci is
an e.c. absolute constant.PROOF. Without loss of
generality
we may assumeb ~
2a and r ~ 2.Let a =
03A0rj=1p03B1jj
and b =03A0rj=1p03B2jj
beprime
factorizations of a and b.Hence
According
to[9,
Theorem1 ]
thefollowing
estimate holds :where
and c is some e.c. absolute constant. Hence
This proves the theorem.
COROLLARY. Let ni n2 ... be the sequence
of integers composed of primes
not greater than p. Then there exists an e.c. constant C =C(p)
such that
PROOF. We have r ~
4pJ(3Iogp). (See
e. g.[17,
formula(3.6)] or [10])
3. The
following
theorem shows that the constantsC1
and C in thepreceding
section cannot bereplaced by
constants smaller than r -1 andn(p )-1, respectively.
The gap meantby
Erdôs at[7,p. 128 ]
is therefore filled up almostcompletely.
THEOREM 2. Let P
= {p1, ···, pr}
be agiven
setof primes,
r >1,
and put p = maxjpj. Then there existinfinitely
manypairs of integers
a, b such thatPROOF. Let T be a
positive integer
and consider all numbers of the formt1 log p1 + t2 log p2 + ··· +tr log pr
with0 ~ tj ~
Tfor j
=1, ···,
r.We have
(T+1)r non-negative
numbers of absolutevalue ~ rT log
p.It follows that there are two among
them,
allog p1
+ ··· + ocrlog pr
andfli log p1 + ··· + 03B2r log pr
say, such thatHence,
Without loss of
generality
we may assume thatcejflj
= 0for j
=1,
..., r.Putting
a =03B203B111
...prr
and b =poi ... p03B2rr,
we obtaina2 ~
ab ~prT .
Hence,
T ~ 2log al(r log p). Substituting
this estimate in(5)
we see thatWe know from
(5)
thatlog (bla)
-0,
if T - oo.Hence,
The
inequality (4)
follows from(6)
and(7). By letting
T - oo we obtaininfinitely
manypairs a, b subject
to(4).
4. One cannot deduce an
explicit
bound for theN~
in(2)
from theproof
ofSiegel.
Thefollowing proof
based on the Gel’fond-Bakertheory
shows that
N~
can be chosen such thatTHEOREM 3. Let 0 8 1. Let a and b be
positive integers
such thatPROOF. We have a > 1 and r ~ 2. Let a =
03A0rj=1p03B1jj
and b =03A0rj=1p03B2jj
be
prime
factorizations of a and brespectively.
Hence,We have
and therefore
where 03B4 =
~/3
and H = maxj(03B2j-03B1j) ~
3log
a. It follows from Baker’s well known result[1]
thatHence,
This proves the theorem. We recall that r ~
4p/(3 log p).
5. The next theorem
gives
an upper bound for a similar to the one in Theorem 3. Thepresent
estimate is better if b - a is rather small and ris
large.
Aysimilar
method ofproof
wasgiven by
Ramachandra[16,
Theorem
3].
Here we use Baker’s result on theinteger
solutions of thediophantine equation y2 = X3 + k.
THEOREM 4. Let a and b be
positive integers,
a b. Put r =ce(ab)
andp =
P(ab).
ThenPROOF. Let a =
pil
...prr
and b =p03B211 ··· 0"
beprime
factorizations of a and brespectively.
Let a =ve3 and
b =wf2,
where e,f,
v, w arepositive integers
such that v is cubefree and w issquarefree.
We haveThus
(x, y) = (vwe, vw2f)
is aninteger
solution of thediophantine
equa-tion y2
=x3+k, where k
=(b - a)v2w3.
It follows from Baker’s estimatefor such solutions
[2],
thatWe
have |v| ~ p2r, |w| ~ pr. Hence,
which
implies
our assertion.So we obtain
COROLLARY. Let k
and p
bepositive integers.
LetAkp
be the smallestinteger
a such thatif b - a
= k andP(ab)
= p then a ~Akp.
ThenAkp .
Heproved
forexample log A1p ~ c2p2eP!2
for some absolute con-stant c2 and
all p.
He also included tables of numerical resultsfor p ~
41.It appears that the upper bounds for
Akp
are rather crude. It would be of great interest toreplace
the factor epby
some constant powerof p.
This would be
possible,
if Hall’sconjecture [4]
on the solutions ofy2 = x3+k
wereproved. (M.
Hall Jr. hasconjectured
thatif x, y
areintegers with x3 ~ y2, then |x3-y2|
>|x|
for x >xo . )
6. The
following
theorem has direct consequences for the number theo- reticfunction f3(n)
introducedby
Erdôs andSelfridge [8 ].
In their no-tation it
implies f3(n)
>10 - 6 loglog
n, which is the first non-trivial lower bound forf3(n). They conjectured f3 (n)
>(log n)’3
for some absoluteconstant C3 and all n.
(This
would also be a consequence of Hall’s con-jecture.)
THEOREM 5. Let a and b be
positive integers
such that a b andP(a)
=P(b).
ThenPROOF. We know from the above
Corollary
thatHence, by P(a) ~ 2,
This
implies (i). Since P(a) ~ b-a,
we have7. In this section we
generalize (2)
in adirection
which wassuggested by
Professor E. G. Straus. As Professor Bakerpointed
out in adiscussion
with Professor Straus the
following
recent result of his makes itpossible
to
give
aninequality
like(2)
forintegers containing
many smallprime
factors.
If 03B11, ···,
an are non-zeroalgebraic
numbers withdegrees
at mostd,
and
if
theheighths of
03B11, ···, 03B1n-1 and an are at most A’ andA(~ 2),
respectively,
then there is aneffectively computable
numberC, depending only
on n, d and A’ suchthat, for any ô
with 0 ô2,
theinequalities
have no solution in rational
integers b1, ···, hn -1
andhn( =F 0)
with ab-solute values at most B and
B’, respectively.
Using
this result one can deduce results related to thep-adic analogues
ofthe theorems of
Thue-Siegel-Roth
and Baker asdeveloped by Mahler, Coates, Sprindzuk
and others.(See
forexample [13, 14, 6, 20, 21].)
Wenote that one of the related
conjectures
at the end of Mahler’s book[14, Appendix III]
wasproved by
Straus, but that this result has not beenpublished.
We prove hereTHEOREM 6. Let 0 8 1. Let a and b be
positive integers
such thata
b ~ a+a1-~. Furthermore, letp,
r, a1, a2,b1, b2 be p ositive integers
such that a = al a2, b =
bi b2, P(al b1) ~
p,03C9(a1 b1) ~
r. Then thereexists an e.c. constant il =
il(p,
r,8)
> 0 such thatPROOF. Put A = max
(a2, b2).
It suffices to prove A > a’’ -1. First weassume A ~
2. Let al= pii ... prr,
andbl
=p03B211 ··· p03B2rr
beprime
fac-torizations
of al
andb1 respectively. Hence,
We have |03B2j-03B1j| ~ log b/log 2 for j = 1, ..., r. We applythe above mentio-
ned result of Baker to the linear form at the
right
hand sideof (11).
We takeB =
log b/log
2 and.B’ = 1.Hence,
there exists an e.c. constantC2
=C2 (p,r)
such that
Since
b ~ 2a,
it follows thatOn the other
hand,
The combination of these
inequalities gives
or,
equivalently,
Let 1
> 0 be chosen so small thatThen a2 ~ 2A1/~. Hence A ~
a’’ .Now we consider the
remaining
case A = 1. We see from Theorem 3 that there exists an e.c. constantC3 - C3(p, r, 9)
> 0 such that aC3.
We
choose q
> 0 so small that(12)
is fulfilledand,
moreover,Hence, A = 1 > C~3 -1 > a~-1.
Thiscompletes
theproof
of thetheorem.
8. It is easy to see that Theorem 6 is no
longer valid,
if wereplace (10)
9.
Finally
wegive
anotherapplication
of the result of Baker mentioned in 7. Wintner communicated thefollowing problem
to Erdôsorally
[7, p. 218].
Does there exist an infinite sequence of
primes
p 1 p2 ···such that
if n1 n2 ··· is
the set of allintegers composed
of thep’s
then
limi~~(ni+1-ni)
= oo?Here we solve the
conjecture
in the affirmative and prove even much more:THEOREM 7. Let 0 ~ 1. There exists an
infinite
sequenceof primes
Pl p2 ... such that
if ni
n2 ... isthe set of all integers
com-posed of
theseprimes
thenPROOF. We construct a sequence pl p2 ... with the
required
prop- ertyby
induction. Let p1 = 3. It is obvious that(13)
holds for powers of 3. Now assume that we have a sequence p1 ... p, such thatfor all
integers a,
b with a b andcomposed
of theseprimes. Let p
bea
prime,
p > pr.Suppose
that there existintegers a
and bcomposed of the primes p1,...,pr, p
such that 0 b - aa1-~.
Let a= p03B111 ···
p03B1rrp03B1
and b =Pol ... p03B2rrp03B2
beprime
factorizations of a andb,
respec-tively.
Then11 1= P
because of our inductionhypothesis applied
toalp’
and
b/p03B2.
We haveFurthermore, |03B2j-03B1j| ~ log b ~ log
2afor j
=1,···, r and |03B2-03B1| ~ log b/log p ~ log (2a)/log p.
Weapply
Baker’s result(see 7)
with d =1,
n =
r+ 1, A’ - pr, A
=p, b = 814, B
=log (2a)
and B’ =log (2a)l
log p.
Hence, there exists an e.c. constantC4
=C4(p, r)
such thatIt follows from
(15)
and(16)
thatThis
implies
Thus there exist e.c. constants
C5 - CS(p, r, 8)
andC6 = C6(8)
suchthat x ~ CS log (C6 x),
where x =log a/log p. Hence,
there exists an e.c.constant
C7 = C7(p, r, 8)
such thatSo we
obtain a ~ pC7 and b ~
2a ~pC7,
whence03B1, 03B2 ~ C7
and 03B1j,03B2j ~ C7 log p for j = 1, ···, r.
Let T be a
large integer.
We know from[17,
formula(3.8)]
that thenumber of
primes
in the interval[T/2, T]
islarger
than3T/(10 log T)
for T ~ 41. For each
prime p
in the interval[T/2, T we apply
the aboveargument.
Hence,
implies
that a,j8
~C7
and03B1j, 03B2j ~ C7 log
Tfor j
=1, - - -,
r. The num-ber of
possible
choices for 03B11, ···, a,. , a,03B21, ···, f3r, f3
is at most(C7+ 1)2r+2(log T) 2r.
Assume that all theseintegers
are fixed. Then1
b/a 1 +a-8 implies
thatwhere
C8
=p03B11-03B211 ··· p03B1r-03B2rr.
We recall that 03B1 ~03B2.
First supposep
> oc.Then
Hence,
p is contained in a fixed intervalof length
Secondly
supposefi
11. Then from(18)
Hence, p
is contained in a fixed interval oflength
It turns out that in both cases the number of
primes
p for which(17)
with fixed 03B11, ···, ar, a,
Pl, - - -, fi,, fl
ispossible
does not exceedTa-~.
Since
2a ~ b ~ p ~ T/2,
we see thatTa-~ ~ 4T1-~.
We concludethat the total number of
primes
in the interval[T/2, T]
for whichintegers
a and b
subject
to(17)
can be found is at mostWe have to exclude these
primes. However,
there are more than3T/(10 log T ) primes
in this interval. Forsufficiently large
Tthe number ofprimes
in[T /2, T is
greater than(19)
and we cantake pr + 1
out of theremaining
set.Doing
so everypair
ofintegers
a and b with a b andcomposed of p1, ···, pr+1
satisfies b - a >al - ~. Now
theproof
has beencompleted by
induction.10. REMARKS.
(i)
It follows from the aboveproof
that forevery C
with 0 9 1 it is
possible effectively
togive
a sequencePl, P 2’ ...
such that there exists a sequence p 1 p2 ... with the
required
property and with
Pj/2 ~ pj ~ Pj
for allj. (ii)
0 ~ 1. It followsfrom Theorem 2 that there does not exist a constant
C9 = C9(~)
suchthat Theorem 7 is valid if
(13)
isreplaced by
theinequality
It is
interesting
to compare this with Theorem 1.REFERENCES A. BAKER
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