The Number of Solutions of φ (x) = m

Annals of Mathematics

The Number of Solutions of φ(x) = mAuthor(s): Kevin FordSource: Annals of Mathematics, Second Series, Vol. 150, No. 1 (Jul., 1999), pp. 283-311Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/121103 .

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Annals of Mathematics, 150 (1999), 283-311

The number of solutions of q (x) = m

By KEVIN FORD

Abstract

An old conjecture of Sierpiniski asserts that for every integer k ) 2, there is a number m for which the equation ((x) = m has exactly k solutions. Here q is Euler's totient function. In 1961, Schinzel deduced this conjecture from his

Hypothesis H. The purpose of this paper is to present an unconditional proof of Sierpiniski's conjecture. The proof uses many results from sieve theory, in particular the famous theorem of Chen.

1. Introduction

Of fundamental importance in the theory of numbers is Euler's totient function b(n). Two famous unsolved problems concern the possible values of the function A(m), the number of solutions of 4(x) = m, also called the

multiplicity of m. Carmichael's Conjecture ([1],[2]) states that for every m, the equation b(x) = m has either no solutions x or at least two solutions. In other words, no totient can have multiplicity 1. Although this remains an open problem, very large lower bounds for a possible counterexample are relatively easy to obtain, the latest being 1010?1 ([10, Th. 6]). Recently, the author has also shown that either Carmichael's conjecture is true or a positive proportion of all values m of Euler's function give A(m) = 1 ([10, Th. 2]). In the 1950's, Sierpiniski conjectured that all multiplicities k > 2 are possible (see [9] and

[16]), and in 1961 Schinzel [17] deduced this conjecture from his well-known

Hypothesis H [18].

SCHINZEL'S HYPOTHESIS H. Suppose fi(n),... ,fk(n) are irreducible,

integer-valued polynomials (for integral n) with positive leading coefficients. Also suppose that for every prime q, q \ fi(n) ... fk(n) for some n. Then the numbers fi (n),..., fk(n) are simultaneously prime for infinitely many positive integers n.

Hypothesis H is a generalization of Dickson's prime k-tuples conjecture [7], which covers the special case when each fi is linear. Schinzel's construction of



KEVIN FORD

a number with multiplicity k requires that as many as k distinct polynomials with degrees as large as 3k simultaneously represent primes. Using a different, iterative argument, the author ([10, Th. 9]) deduced Sierpinski's Conjecture from the Prime k-tuples Conjecture, the argument requiring only three linear

polynomials to be simultaneously prime. The proof is quite simple, and we

reproduce it here.

LEMMA 1.1. Suppose A(m) = k and p is a prime satisfying

(i) p > 2m + 1, (ii) 2p + 1 and 2mp + 1 are prime, (iii) dp + 1 is composite for all d I 2m except d = 2 and d = 2m.

Then A(2mp) = k + 2.

Proof. Suppose /(xl) = .= (xk) = m and ((x) = 2mp. Condition (i) implies p t x, hence p J (q - 1) for some prime q dividing x. Since (q - 1) m 2mp, we have q = dp + 1for some d 2m. We have q > 2p, so q2 x and thus

X(x) = (q - l)0(x/q). By conditions (ii) and (iii), either q = 2p + 1 or

q = 2mp+1. In the former case, O(x/q) = m, which has solutions x = (2p+1)xi (1 < i < k). In the latter case, o(x/q) = 1, which has solutions x = q and x = 2q. D

Now suppose A(m) = k and let dl,... ,dj be the divisors of 2m with 3 < di < 2m. Let pl,... ,pj be distinct primes satisfying pi > di for each i.

Using the Chinese Remainder Theorem, let a mod b denote the intersection of the residue classes -d~1 modpi (1 < i < j). Then for every h and i, (a + bh)di + 1 is divisible by pi, hence composite for large enough h. The Prime k-tuples Conjecture implies that there are infinitely many numbers h so that p = a + hb, and 2p + 1 and 2mp + 1 are simultaneously prime. By Lemma 1.1, A(2mp) = k + 2 for each such p. Starting with A(1) = 2 and A(2) = 3, Sierpiniski's Conjecture follows by induction on k.

Although Hypothesis H has not been proved in even the simplest case of two linear polynomials (generalized twin primes), sieve methods have shown the conclusion to hold if the numbers fl (n),... , fk (n) are allowed to be primes or "almost primes" (nonprimes with few prime factors). See [13] for specific statements. Recently, the author and S. Konyagin [11] employed the theory of almost primes to prove that A(m) = k is solvable for every even k. Although falling short of a complete proof of Sierpinski's Conjecture, the method did provide a solution to the corresponding problem for the sum of divisors function o(n). It is proved ([11], Theorem 1) that for every k > 0, there is a number m for which a(x) = m has exactly k solutions, settling another conjecture of Sierpiniski. Curiously, the barrier to solving A(m) = k for odd k by this method is the probable nonexistence of a number with A(m) = 1.

284

THE NUMBER OF SOLUTIONS OF b(X)=M

Combining a different application of almost primes with the iterative ap- proach from [10], we now provide an unconditional proof of Sierpiiski's Con-

jecture.

THEOREM 1. For each integer k 2, there is an integer m with A(m) = k.

Our proof of Theorem 1 is by induction on k. We start with A(10) = 2, A(2) = 3 and for the induction step, we show:

THEOREM 2. Suppose A(m) = k and m is even. Then there is a number m' such that A(mm') = k + 2.

Combining Theorem 1 with Theorem 2 of [10] gives a strong result on the distribution of totients (values taken by b(n)) with a given multiplicity.

COROLLARY 3. For each k 2, a positive proportion of all totients have

multiplicity k. Specifically, if V(x) is the number of totients < x, and Vk(x) is the number of totients < x with multiplicity k, then for each k > 2,

liminf Vk( >0. x->oo V(x)

At the center of the proof of Theorem 2 is a famous theorem of Chen ([3], [4]).

LEMMA 1.2 (Chen's theorem). For each even natural number m and x > xo(m), there are ? x/log2x primes s E (x/2,x], s = 1 (mod m) such that (s - 1)/m has at most two prime factors, each of which exceeds x1/10

Remark. A proof of Chen's Theorem also appears in Chapter 11 of [13]. What Chen actually proved is that there are > x/ log2 x primes p < x with

p + 2 having at most two prime factors. Trivial modifications to the proof give Lemma 1.2.

HYPOTHESIS S(m). For some constants c > 0 and xo, if x > xo then the

number of primes s E (x/2, x] for which (s - 1)/m is also prime is > cx/ log2 x.

We suppose q(a) = m has the k solutions al,... , ok. The proof is then broken into two parts: (I) assuming Hypothesis S(m) is true; (II) assuming Hypothesis S(m) is false. For Part I we show that for many such primes s, A(mq(q - 1)) = k + 2, where q = (s - 1)/m and 0(n) = mq(q - 1) has the k + 2 solutions n = q2ai (i = 1,..., k), n = q(mq + 1) and n = 2q(mq +. 1)

(the "trivial" solutions). For Part II, Chen's Theorem tells us that for an unbounded set of x,

there are x/ log2x primes s E (x/2, x] with s = 1 + mpq, where p and

285

KEVIN FORD

q are primes and x1/10 < p < q ?< x9/10. We show that for many such s, A(mpq(p - l)(q - 1)) = k + 2, the solutions of

(1.1) q(n) = mpq(p - l)(q- 1)

being only the "trivial" ones n = p2q2ai (i = 1,... , k), n = pq(mpq + 1) and n - 2pq(mpq + 1). Proving this is more difficult than Part I, but a simple heuristic indicates that such s are likely. The basis for the heuristic (and the actual proof) is an extension of the classical result of Hardy and Ramanujan [14] that most integers n < x have about loglogx prime factors. Erdos [8] combined their method with Brun's sieve to show that for most primes p < x, p - 1 has about log log x prime factors. One can further refine these ideas (e.g. Lemma 2.8 of [10] or Lemma 4.3 below) to show that for most primes p < x, the prime factors of p- 1 are nicely distributed. This means that p- 1 has about log log b - log log a prime factors in the interval [a, b] for 1 < a < b < x. Call such primes "normal". If p and q are both normal primes and x is large, then w = p(p - )q(q - l)m has about 2loglogx prime factors. Fix t and

suppose n has t distinct prime factors ri,... rt, different from p and q. For simplicity suppose n is square-free. Then

4)(n) = (r- 1) . (rt- 1) = w.

For a given factorization w = d... dt, the probability that d1 + 1,... , dt + 1 are all prime is about ((logdi) ... (log dt))-1. The critical case is t = 2. Since

p and q are normal, w has about 2 log log y prime factors < y, and thus has about (log y)2lg2 divisors < y. By partial summation,

2E (log d1)(log d2) < (log X)21og2-2

w=dl d2 di> 2

the right side representing the probability that (1.1) has a nontrivial solution for a given pair (p, q). Similarly, it can be shown that

E (log dl) (log dt) < (log )21g2-2 w=dl .-dt

di 2

the bulk of the sum coming from taking d1,... , dt-2 all small. The biggest practical problem with these arguments is the fact that b(n)

is quadratic in q for Part I and quadratic in both p and q for Part II. Thus, when certain sieve estimates are employed, the bounds have factors of the form

p z I D p~<z v p

286

THE NUMBER OF SOLUTIONS OF q(X)=M

where (D) is the Legendre symbol. Heuristically, such products should be

0(1) on average, and showing this for a wide range of the parameters D and z requires strong results on the distribution of primes in arithmetic progressions (see (2.1) and Lemmas 2.6 - 2.8 and 4.2 below).

In the next section, various lemmas needed in the proof of both parts are presented. Section 3 is devoted to the proof in Part I. Section 4 contains further lemmas needed for Part II, and the argument counting nontrivial solutions of

(1.1) with (n, pq) > 1. The last section deals with the hardest case, namely counting solutions of (1.1) with (n,pq) = 1.

The following notation will be used throughout. Let P+(n) denote the

largest prime factor of n and P-(n) denote the smallest prime factor of n. The functions w(n) and Q(n) count the number of distinct prime factors of n, respectively the number of prime factors counted according to multiplicity. Let Q(n, U, T) denote the number of prime factors p of n such that U - symbols may depend on m, but not on any other parameter unless indicated. The symbols p, q, and w will always denote primes. The symbol m will always refer to the number in Theorem 2. The cardinality of a set S will be denoted as #S or IS1.

2. Preliminary lemmas

Aside from Chen's Theorem, the other sieve results we require all follow from a basic sieve inequality called the "Fundamental Lemma" of the combi- natorial sieve (Theorem 2.5 of [13]).

LEMMA 2.1. Suppose F is a polynomial of degree h > 1 with integer coefficients and positive leading coefficient. Denote by Q(w) the number of solutions of F(n) - 0 (mod z). For 2 6 z < x,

#{n s x : P-(F(n)) > z} = x f ( - ())

X {1 + Oh (e/x + e-u(1g u-1og log3u-logh-2)) }

where u = log x/ log z.

Proof. If g(w) = w for some prime z <4 z, then the left side is zero. Otherwise the lemma follows from [13, Th. 2.6] and following remarks. O

287

KEVIN FORD

Remark. The error term is Oh(1), and this will suffice for most of our applications. Also, for future reference, we note that for w t 2a,

(2.1) #{n:an2+bn+c=O (modw)} =1+ ( 4 ).

LEMMA 2.2. Suppose

F(n) = (ain + b1) ... (ahn + bh),

where the ai are positive integers and the bi are integers. Suppose

E:= al ah I (aibj -ajbi) O. l i<j<h

If 2 < z < x and logE s log3 x, then

#n x?: P-(F(n)) )> } ?<h (log )(E) ()

<<h x(log log x)h(log z)-h.

Proof. In the notation of Lemma 2.1, we have p(w) = h when w B E; therefore the number of n in question is

?- hXJ7 <(h x (E)) (log z)-h. ( ) (0(E) w <z

wfE

The last part of the lemma follows from the inequality E/q(E) < log log E. [2

LEMMA 2.3. Let V(x) be the number of distinct values of d(n) which are < x. Then for every e > 0,

) (logx)l-6

Proof. This is the main theorem of [8]. E2

LEMMA 2.4. The number of n < x divisible by a number s satisfying s > exp{(loglog X)2} and P+(s) s10/loglog x is < x/log x.

Proof. For exp{(log log x)2} < y < x, let G(y) be the number of s < y with P+(s) < s10/loglogx. Let n denote a number with no prime factor exceeding

288

W = y10/loglogx and set a = 1 - 20log lgx. Then log y 00

n() ) <2 n (log x)20 1 (1 -p ) n(y n=l p W

<< (log )20 exp (

By the prime number theorem, pW dt i200 ' du

S p- (1 + o(1)) --l (1 +o(1)) d EpW a 1+1 t log t J( log ( (l y)-1 u

(1 + o(1))(e200 + loglogy) < 2loglogx

for large x. Thus G(y) << y/(logx)18 and the lemma follows by partial summation. For a survey of results on the distribution of numbers without large prime factors, see [15]. D2

The next four lemmas are needed to bound the product P(D, z) mentioned in the introduction.

LEMMA 2.5. When (a, n) = 1, n is odd and square-free,

(aj+b) 1

(j,n)=l

When (ab, n) = 1 and n is square-free,

n n jn () (aj+b) = 1.

Proof. Denote by f(n; a, b) the first sum in the lemma. The desired bound follows from the identities f(n; a, b) = _() for prime n, and

f(nln2; a, b) = f(nl; an2, b)f(n2; anl, b)

when (a,nin2) = (ni,n2) = 1. The second assertion follows in a similar

fashion. Here we adopt the convention that (a) = 1 for all a. C

LEMMA 2.6. For some absolute constant yo, if y > yo, 1 0 c 2 2, E > 1, and D E Z, then

(i -1 D))C cw(4n) (n) (D ii 7 Y ^ Y^y " ^n n

wz y n ylog log y ufE (n,2E)=l

P+(n)<y

289

290 KEVIN FORD

Proof. If D = 0, the left side is 1 and the right side is 4, so suppose D = 0. For 1 < c < 2 and x > -1/3, (1 + x)C < (1 + cx)(1 + 3x2); thus

-1 -I D 9

1

3 1i (i

zi7y wv 3 ( 4

3ww y w{E F,E

-49 - (A w >3

3

,) P+(n)<y (n,2E)=l

cA

3.92 2

P+(n)<y (n,2E)=l

n<ylog log y

c(nl)pu(n) D 3.92 n n

Since cw(n) < r(n), by Lemma 2.4 and partial summation, we obtain

cw(n)

n n>ylog log y

P+(n) < y dl d2 >ylog log y

P+(did2),y

1

did2 d2>y(log logy)/2

P+ (d2) Y

log d2 d2 < E

d2 >y(log log y)/2

P+ (d2) .< d/ log log d2

< (logy)-8.

As the left side in the lemma is > (logy)-C, the desired bound follows for sufficiently large y. In particular, the sum in the lemma is positive and > (log y)-. 1

LEMMA 2.7. Suppose y > yo, U > y 10(loglogy)2 A, B and R are positive integers with R even and log(ABR) < log3 U. Suppose D(s) = A(A - Bs), D(s) = As(As - B) or D(s) = A + Bs. Uniformly in A, B, R and 1 < c < 2,

II sU <prim

s prime

( 1

1-I ICU

D(s)c U(loglogU)l+c

C D ( ) logU

S ~n s,s U w pri

s,sR+1 prime

(1 1

(D(s)Vc < U(loglogU)4+c w JJ (log U)2

n>ylog log y

P+(n) 2y

ca(n)

n

_^ ))

,(n) (n) (D

n \n J

1 1 d2 < di

di <d2

THE NUMBER OF SOLUTIONS OF /(X)=M

Proof. Let z = U1/loglogU. If D(s) = A(A - Bs) or D(s) = As(As - B), set E = AB; otherwise set E = B. By imposing the addition condition that w { E, we see that the products on the left are only increased when

D(s) = A(A - Bs) or D(s) = As(As - B), and in the case D(s) = A + Bs the

product is changed by a factor

i (I - 1 (A)) << (log log B)c < (loglog U).

=IB

We will also replace the conditions that "s prime" and "s, sR + 1 prime", with P-(s) > z, respectively P-(s(sR + 1)) > z. This will eliminate at most z values of s from the sums on the left, reducing the sums by at most

O(z(logy)2) = O(U6). Let Yi denote the set of s z and let 92 denote the s z. By Lemma 2.6,

cw(n) u(n) (D(s)) n n S LE I (i! (DA)))

w\E

<4 E

SEIl n<ylog log y

(n,2E)=l P+(n)<y

cw(n) 2(n) D (s)

nx<yg l logy SEy

(n,2E)=l

Denote by S(?; n, b) the number of elements of the set Y which lie in the residue class b mod n. By the sieve fundamental lemma (Lemma 2.1), and the fact that n < z1/5, uniformly in (b, n) = 1 we have

S(Yl; n, b) = (1 + ((log U)-10))

where n log log U

11(n) log U

wCi7n

For each n, Lemma 2.5 gives

Z (Ds)) n

SG~l

= S(YI;n,b) ( (b)

(b,n)=l

n=( ( n )- O(U(og U)-10) (b,n)=l

U log log U 1 U

log U q(n) log10U'

291

KEVIN FORD

and the first inequality follows. The second inequality follows in the same manner by summing over s E 92 and using

S(Y2; n, b) = U (1 + O((log U)-10)),

where W2 =0 if (n, 1 + bR) > 1 and otherwise

W (1 2 ) I (1 1-) 2 ( nR )2 (loglogU)2 W2 \ /11\^7 I(U< nR) log2 U

w4Uz wz w{unR w|R,win

LEMMA 2.8. For some absolute constant b > 0, the following holds. For each Q > 3, there is a number dQ >A (logQ)A for every A > 0, so that whenever ID I Q, dQ { 8D, and log z < exp{(log Q)b},

H 1-1-)) << 1. Z7 Ir7/ Qlog log Q <ZX \ z\

Furthermore, 16 { dQ and p2 f dQ for all odd primes p.

Remark. From now on dQ will refer to the number in this lemma. The implied constant in dQ >A (log Q)A is ineffective for A > 2. With more work, the set of exceptional D may be reduced to those with f(D) = f(dQ), where

f(n) is the product of the odd primes which divide n to an odd power. The upper limit of z may also be extended easily.

Proof. We may assume without loss of generality that D is square-free. If IDI < 2 the result follows from the prime number theorem for arithmetic progressions, since (D) then depends only on w modulo 8. Next suppose IDI > 3. The desired bound follows from a version of the prime number theorem for arithmetic progressions deducible from the Landau-Page theorem and a zero density result of Gallagher (see [6] and [12]). The number dQ is the order of a real, primitive Dirichlet character whose corresponding L- function has a real zero /3 > 1 - co/log Q. If co > 0 is a sufficiently small absolute constant, the Landau-Page theorem asserts that there is at most one such primitive character of order < 8Q. Furthermore, if this character exists, Siegel's Theorem implies that dQ >A (logQ)A for any A > 0. Let 7r(y; q, a) denote the number of primes p < y, p = a (mod q). For n < 8Q not divisible by dQ, (a,n) = 1 and y > Qo : QloglogQ, Theorem 7 of [12] implies

r(y; n, a) = (n) logy (1 + O((log Q)Y))

292

THE NUMBER OF SOLUTIONS OF b(X)=M

for some absolute constant b, 0 Qo,

E 1_ log logz - log logQo loglog z ~ + (0 (n) (log Q)b) q$(n)

QoQ<z(Z w-a (mod n)

Write D = D1D2, where D2 is the largest odd positive divisor of D (i.e. D1 E {-2, -1,1, 2}). Since D2 > 3, quadratic reciprocity implies

(D)- (1_)(( (D2-1)/4 (D1) ( ) , ' ( ) ?~

where iw E {-1, 1} depends only on w modulo 8. Therefore, for numbers ie E{-1,1},

Qo<w,z e=1,3,5,7 l1a(8D2 Qo<wzz a=e (mod 8) w-a (mod 8D2)

log log g l og log Qo . 8a'e log log z

b(8D2) e D2 + (log Q)b' e a =l

Here we have used the fact that ( 2) 0= if (a, D2) > 1. The error term is 0(1) by hypothesis, and the sum on a' is zero, since as a' runs from 1 to D2, 8a' runs over a complete system of residues modulo D2. The lemma now follows from the fact that 1 + h = exp{h + O(h2)} whenever -? < h < .

3. Proof of Theorem 2, Part I

LEMMA 3.1. If A(m) = k and S(m) is true, then A(mm') = k + 2 for some m'.

Proof. Suppose q(al) = = q(ak) = m. Suppose x is sufficiently large, so that x > 2ai for every i and there are >> x/ log2 x primes q E (x/2, x] such that qm + 1 is prime. For each such q, the equation O(n) = q(q - )m has at least k + 2 trivial solutions, namely n = q2ai (i = 1,... ,k), n = q(qm + 1) and trivial n = 2q(qm + 1). Let B(x) denote the number of such q giving rise to additional solutions. If 4(n) = q(q - l)m, then clearly q3 i n. If q2 I n, then

qf(n/q2) = m, which implies n is one of the trivial solutions. If q I n, q2 t n, then q(n/q) = qm. There is a prime r I (n/q) with q I (r - 1), i.e. r = dq + 1 for some d I m. If d = m then n is one of the trivial solutions mpq + 1 or

2(mpq + 1). By Lemma 2.2, the number of q with dq + 1 for some other d is

O(x(log x)-3). The last case, counting solutions with (n, q) = 1, is codified in a lemma, since we will need the argument in Part II. It follows from the next

293

KEVIN FORD

lemma that the number of q with q(n) = q(q - l)m for some n with (n, q) = 1

is O(x(logx)-2'l). Therefore, B(x) = 0(x(logx)-2'1) and hence for some q, A(q(q- l)m) = k + 2. D

LEMMA 3.2. Suppose b = 1 or b is a prime less than yl0. Let N(y) be the number of primes q c (y/2, y] satisfying (i) qbm + 1 is prime, and (ii) for some n with (n,q) = 1, b(n) = mq(q - 1). Then

N(y) < y(log y)2-1.

Proof. First, we remove from consideration those q for which q - 1 has a divisor s with s > exp{(loglogy)2} and P+(s) < 1/loglogy. By Lemma 2.4, the number of such q is O(y/ log10 y). Consider the equation qf(n) = mq(q - 1). There is a number r n with q r - 1. Say r = 1 + qso and let wo = - (n/r). For some factorization of m as m = m1m2, we have m1 I so, m2 I wo. Fix ml and m2, and let s = so/mi, w = wo/m2, so that sw = q - 1. The number of q in question is at most the number of pairs (s, w) for which y/2 - 1 < sw < y, wm2 is a totient (value taken by X(n)), and the following three numbers are

prime:

(3.1) ws + 1 = q, (ws + l)mb + 1 = qmb + 1, (ws + l)smi + 1 = r.

For brevity write E(c) = exp{(logy)C}. For fixed s, Lemma 2.2 implies that the number of w < y/s with each number in (3.1) prime is

y (log log y)3 s log3(y/s)

Therefore, the number of pairs (s,w) with s ? E(0.89) is O(y(logy)-21). Otherwise w < y/E(0.89). By Lemma 2.1 and (2.1), for fixed w < y/E(O.89), the number of s with each number in (3.1) prime is

(3.2) Y(loglogy)3 - ( 1 (m -4mlw))

w(logy)2.67 IT-

~ 726

where z = E(0.89)1/2. First, suppose that w E (U/2, U] where U is a power of 2 less than 2E(1/2).

Let Q = 8mE(1/2) and recall the definition of dQ (Lemma 2.8). The number of w with dQ | (m2 - 4miw) is clearly

4miU U U

dQ (log Q)2 log z

For the remaining w, Lemma 2.8 implies that

I ( 1 (m- 4mlw))

Qlog log Q z Z

294

THE NUMBER OF SOLUTIONS OF q((X)=M

Hence, since w is a totient, Lemma 2.3 gives

1 1 (m2-4mlw'\ -< 2

(log y)/ logy 2 w (Z V (w w Z J (log U)0'999

U/2<w(U awz

By (3.2), summing on U gives a total of O(y(logy)-2.16) possible pairs (s, w). Lastly, consider E(1/2) < w < y/E(0.89). The product in (3.2) for

w > E(0.48) is O((logy)0.41). By our initial assumption about q, w has a

prime factor > E(0.49). Thus n/r has a prime divisor r2 > E(0.49). We can assume that r n, for the n umber of q divisible by a factor r2(r2 - 1) for such

large r2 if O(y/E(0.49)). Write w* = q(n/(rr2)), so that w = (r2 - l)w*. Fix w* and put r2 E (U/2, U], where U is a power of 2 in [E(0.49), 2y/(sw*)]. By Lemma 2.7,

E 1 II (1 I (ml 2-4mlw*(r2 -11)\) (log logy)2 Z r (r2 II k V - - log U

U/2<r2 U z2 E(0.48)

Summing on U and again using (3.2) and Lemma 2.3, we see that the total number of pairs (s, w) is O(y(logy)-2.25). Thus, for fixed ml, m2, the number of possible q is O(y(logy)-21). Since the number of choices for (ml,m2) is

0(1), the lemma follows. [

4. Part II, the case (n,pq) > 1

We require one more sieve lemma, needed only for a special case of the

proof for Part II.

LEMMA 4.1. Suppose r, s, t, u are positive integers, and x, y, z are positive real numbers satisfying x > y > z > 2 and log(rstu) < log3 x. Let

Wg,h = (rg+ 1)(sh+l1)

and

F(g, h) = (twg,h + 1)(ughwg,h + l)ghwg,h.

Then

(log log x)6 #{x < g < 2x,y < h A 2y: P-(F(g,h)) > z} < xy ( )6

Proof. Let g(z) denote the number of solutions of F(g, h) - 0 (mod w). If Qe() = w2 for some w 5 z then the left side is zero. Otherwise by the sieve

fundamental lemma (Theorem 2.5 of [13]),

#{x < g s 2x, y < h < 2y: P-(F(g, h)) > z} < Zy (1 ~e( )) w 7 z

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KEVIN FORD

We next show that for w t 30rstu,

(4.1) -12 - 2w3/4 < p(w)- 6w < -10 + 2Wz3/4.

From (4.1), the lemma follows at once, since then

H (1 _e(a << -Q(( 30rstu 6 (1 /6

(log log x)6

(log z)6

To prove (4.1), suppose w ~ z, w \ 30rstu. Let M1, M2, M3, M4 denote the set of solutions mod w of gh 0, Wg,h 0, tWg,h + 1 0 and ughwg,h + 1 0, respectively. It is easily verified that M1 n M4 = M2 n M3 = M2 n M4 = 0, IMi n M21 = IM n M31 = 2 and iM3 nM4| l 2. Also, IMil = IM21 = 2w - 1 and M31 = w - 1. By inclusion-exclusion,

(z) = IMi U M2 U M3 U M41 = 5w - 7 + M41 - IM3 n M4I.

Since the number of solutions of q(qR + 1) - a (mod w) is 1 + (1+4aR), we have

i- l((1 +

+ 4ar))

( ( 4sa-lu-l ))

- ( +4ar) (a) (a--4su-1 ) 3 + ----+ .

By Theorem 1 of Davenport [5], we have IEl < 2w3/4 and (4.1) follows. OD

We next combine Lemmas 2.7 and 2.8 into an estimate which has enough generality for our applications.

LEMMA 4.2. Suppose 6 > O, z > 100. Suppose A, B, Q, R, U are positive integers satisfying (logz)6 < logQ < (log z)3, log(ABRU) < log3 , U > 10,

(AB, dQ) < dQ/2 and Q > AU(AU + B). Suppose either D(s) = A(A - sB) or

D(s) = sA(sA - B). Let Y1 denote the set of primes s < U, and 92 the set of primes s < U with sR + 1 prime. If 1 < c < 2 and k E {1,2}, then

/ 1 D(s)\ C U log Q

SI (1 n -)) ?8 (log U) log (log log )4+4c log5 z. 5e^

Z( W (w )) (logU)k logU) sEYk w z<z

Proof. Partition 9i into 9', the set of s C Y for which dQ f AD(s), and /", the remaining s. If D(s) = A(A - sB), then y" is contained in a single

residue class modulo g, where g = dQ/(dQ, AB) > d1Q2. Recall that 16 { dQ

296

THE NUMBER OF SOLUTIONS OF qb(X)=M

and dQ is not divisible by the square of any odd prime. If D(s) = sA(sA - B), then (s, dQ) > 1 for at most w(dQ) < logQ < log3z primes s (Here we use

dQ < 8Q). For other s; dQIA(sA - B) for s lying in a single residue class modulo g = d'/(d'/,A), where d' = dQ/(dQ,A). Here (d',A) < 8, therefore

g > d1/2. Thus, in every case

U U 1" << U- + 1 + log3 << + log3z, g (log Q)8+g

so the contribution to the sum in the lemma from s E Y" is

O(U(log z)-6 + log5 z) = O(U/ log2 U + log5 z).

By Lemma 2.8, if s E Y' and z > Qo := QloglogQ then

1-- () 1.

Qo<wz<z

Set y = U1/(20(loglogU)2) Trivially we have

Iv 1 w haD(s) ve n (I 1 (- D(s ())Q (< < (lo )g ( (loglog3.

Y<w(Qo k z}}w log y log U

Therefore, for any admissible z, Lemma 2.7 implies

(log Q (loglog z)3 ( (1 Ds)) <

log Vj - , w logU sE" /w<min(y,z)

<< ( log) U) (log log )4+4c.

As mentioned in the introduction, for most primes p < x, the prime factors of p - 1 are nicely distributed. We prove a simple result of this kind below. For brevity write E(z) = exp{(log x)Z}.

LEMMA 4.3. Let N , 10 be an integer, put 6 = 1/N and suppose x1/10 <

y < x. The number of primes p < y not satisfying

Q(p- 1, E(j), E((j + 1)6)) < (6 + 62) log log x (O < j N - 1)

is

<<? y(log x)-1-63/3

Proof. Assume that x is sufficiently large with respect to 6. First, the number of p < y for which P+ (p- 1) < yl/lloglogX is O(y/log0 x) by Lemma

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KEVIN FORD

2.4. For the remaining p, let q = P(p - 1) and p - 1 = qa. For fixed a, the number of q < y/a with qa + 1 prime is, by Lemma 2.2,

y log log x y (log log x)3 a log2 (y/a) alog2 x

For each j, let Ij = [E(j6), E((j + 1)6)] and let Bj be the sum of 1/a over a < y with at least (5 + 62) log log x- 1 prime factors in Ij. We have

Bj (1 - +)1-(i6+62) loglogx (1 + -) 2(a,E(jS)'E(j-+6)) a,ax a~x

< 2(log x)-(6+62) log(l+) II -l ) I -1

(P -pIj1 p P'j

? (log X)1+62-(6+2) log(1+6) < (log )1-0453,

and the lemma follows. O

Suppose that Hypothesis S(m) (following Lemma 1.2) is false. By Lemma 1.2, for an unbounded set of x, there are > x/ log2 x primes s E (x/2, x] for which s = 1 + pqm, where p and q are primes and x1/10 < p < x1/2. Denote by sl the set of such pairs of primes (p, q). We will show that for x sufficiently large, there is some such pair (p, q) giving A(mpq(p - l)(q - 1)) = k + 2, with

(4.2) q(n) = mpq(p- 1)(q-1)

having only the trivial solutions n = p2q2ai (i = 1,..., k), n = pq(mpq + 1) and n = 2pq(mpq + 1). Before proceeding, we must cull from vW certain pairs (p, q) possessing undesirable properties. Let F(p) be the number of q with

(p, q) c .W and let G(q) be the number of p such that (p, q) E S. By Lemma 2.2, we have

(4.3) F(p) << , G(q) << plog2 x q log2 x

By (4.3) and the Brun-Titchmarsh inequality, if K is a large enough constant (depending only on m),

#{(p, q) CE : p l(mod 2K) or q = l(mod 2K)} < ?[|1.

Denote by -' the set of remaining pairs. Define v by 2v ] m, 2v+1 { m. If n has h distinct prime factors, then 2h-1 I q(n). Consequently, if (4.2) holds and (p, q) E /', w(n) < C where C = 2K + 1 + v. Let

N = 100([4 log(2C + 2)] + 14 + 2C),

298

(4.4) 6 = 1/N.

THE NUMBER OF SOLUTIONS OF <b(X)=M

For brevity, write Ej = exp{(log x)1-j6} for 0 < j < N- 1 and let EN = 2. Define

(4.5) Qj Ej , d =dQ, (j=2,... ,N).

Let S be the set of (p, q) EE d' satisfying the additional conditions

(4.6) d2 (p - 1)(q - 1) implies d < log4 x,

(4.7) dp + 1, dq + 1, dpq + 1 are composite for each d I m,

except for mpq + 1,

(4.8) p(p- 1) +1, q(q - 1) + ,pq(p- 1) + 1,pq(q - 1) 1, are all composite,

(4.9) (m(p- 1)(q- ), dj) < d4/2 (j = 2,... ,N),

(4.10) (q -)

and

(4.11) Q(p- 1, Ej+,1 Ej) (6 + 62) log log x (0 j N- 1)

|(q - 1, Ej+l, Ej) < ($ + 62) log log (0 < j N - 1).

First, the number of (p, q) failing (4.6) does not exceed

E E nd2 < xlogx (d2) < x(logx)-29. d>log4 x ald2 n<x,aln d>log4 x

Since m is fixed, by Lemma 2.2 and (4.3), the number of pairs failing (4.7) is

O(x/ log3 x). For (4.8), given q the number of p with pq(q - 1) + 1 prime is

O(x(log log x)3/(q log3 x)) by Lemma 2.2. Thus the number of pairs (p, q) with

pq(q - 1) + 1 prime is O(x(log x)-2 99). Likewise the number of pairs satisfying pq(p - 1) + 1 prime is O(x(logx)-299). Also, by Lemma 2.2 and the prime number theorem for arithmetic progressions, for each q the number of p with

p(p- 1) + 1 prime is

x 1 -3 x < log3 I ( P ))

- < 3 qlog 3x k</ p qlog3 x

Thus the number of pairs (p, q) with p(p- 1) + 1 prime is O(x/ log3 x), and the same bound holds for the number of pairs with q(q- 1) + 1 prime. To deal with

(4.9), for each j let Vj denote the set of numbers a < dj with (a, dj) > d1/4/m. Then

lVl S= b 0(n/d) < n/d

dIdj,d>djl/4/m dldj,d>d./4/m

= , d < (dj)mdj3/4 < d4/5

dldj,d<md3/4

299

By Lemma 2.8, dj > (log Qj)40/6 > (log )40 and for x large, m < d/100 for

every j. Therefore, the number of pairs (p, q) failing (4.9) is at most

N-1 N-1

E (E( E p+ E -)AKE i Vl ( d; +1) - + << .IVj1 dj

j=l aEV, xl/lo<pxl/2 Xl/2<q x9/10 j=1 p=l+a (mod dj) q=l+a (mod dj)

x

(log X)3

The number of pairs failing (4.10) is trivially O(x9/10). By (4.3), Lemma 4.3 and partial summation, the number of pairs failing (4.11) is at most

F(p) + G(q) <<og)2+63/3 (log x)2+)3/3 ' badp bad q

Thus, for x large,

(4.12) 1> 1' ? x/ log2x.

We will show that for most pairs (p, q) E M, the equation (4.2) has only the trivial solutions. We break this argument into two parts, the second of which is the most delicate and is proved in the next section.

LEMMA 4.4. The number of pairs (p, q) E X for which (4.2) has a nontrivial solution n with (n,pq) > 1 is O(x(logx)-2'1).

Proof. Consider a generic solution (n,p, q) of (4.2) with (n,pq) > 1. By the definition of X, p3 f n and q3 { n. If (n,p2q2) = p2q2, then b(n/p2q2) =

m, which has the k trivial solutions n = p2q2ai (i = 1,... ,k). Next, if

(n, p2q2) = p2q, then q(n/p2q) = mq. This implies that there is some prime r I n with ql(r - l)lmq, which is impossible by (4.7). Similarly, (n,p2q2) = pq2 is impossible. If (n,p2q2) = pq, then q(n/pq) = mpq, implying n has a prime divisor of the form dp + 1, dq + 1 or dpq + 1 for some d [m. By (4.7), n = pq(mpq + 1) or n = 2pq(mpq + 1). If (n,p2q2) = p2, then qn(n/p2) mq(q - 1). By Lemma 3.2 (with b = p), for fixed p the number of possible q is O((x/p)(logx)-21); therefore the total number of possible pairs (p,q) is

O(x(logx)-2'1). The same argument handles the case (n,p2q2) = q2. Finally, suppose (n, p2q2) = p (the symmetric case (n, p2q2) = q is dealt

with the same way). Then q((n') = mpq(q - 1), where n' = n/p, and there are two cases: (i) there is a prime r n' with pq (r - 1); (ii) there are primes ri, r2 dividing n' with p I (r - 1) and q (r2 - 1).

In case (i) write r 1 + pqso and wo = -(n'/r), so sowo = m(q - 1). Fix ml and m2, where m = mim2, ml so and m2 wo. Write w = wo/m2 and s = so/m1, so that sw = q- 1. For fixed s < E(3/4) and fixed p, the number

300 KEVIN FORD

of w with sw + 1 = q, psml(sw + 1) + 1 = r and pm(sw + 1) + 1 all prime is

x (log log x)3 < -

3 sp log x

Summing on s I E(3/4) and p yields a total of O(x(log x)-2'24) possible pairs (p,q). The other possibility is that w < q/E(3/4). By Lemma 2.1 (with z = E(0.74)) and (2.1), for fixed w and p, the number of s is

?x (loglogx)3 (1_ (mrp2-4mlpw)) pw (log )2.22 E(O 1-4

w<,E(0.74)

Put p E (U/2, U], where U is a power of 2 between x1/10 and x1/2. By Lemma

2.7, the above product has average O((log log x)2) over such p. Summing over U, it follows that for fixed w, the number of pairs (p, s) is O((x/w)(log x)-2'21). Since wm2 is a totient, Lemma 2.3 implies that the number of pairs (p, q) is

O(x(log x)-2'2). Therefore there are O(x(log x)-2'2) pairs (p, q) counted in case

(i), since there are 0(1) choices for ml and m2. In case (ii), write

r1 = 1 + psi, r2 = 1 + qS2, n' = rir2t, wo = o(t).

Fix mi,m2,m3, where m = mlm2m3, ml s, m2 I s2 and m3 I wo. Let

S3 = si/ml, s4 = s2/m2 and w = wo/m3. The number of possible pairs (p, q) is at most the number of quadruples (s3, s4, w, p) where p is prime, wm3 is a

totient, and the following four numbers are prime:

(4.13) ws3S4 + 1 = q,

ps3ml + 1 = rl,

(W5354 + 1)s4m2 + 1 = r2,

(w3S4 + l)pm + 1 = pqm + 1.

Suppose U is a power of 2, x1/10 E(0.99); 2) those with w < E(0.99), s > E(0.8); 3) those with w < E(0.99) and s < E(0.8).

First, suppose w > E(0.99). By Lemma 2.2, given (s3, s4), the number of

p with ps3ml + 1 prime is O(U(log log x)2/ log2 x) and for each p the number of w with the remaining three numbers in (4.13) all prime is

x(log log x)3 x(log log x)3 US3S4 log3 (x/(US354)) US3S4 (log x)297'

By (4.11), Q(S3s4) 1.01 loglogx, so

-- E < (m)<< (logx)1+l.Ollog2 83,S4 m

m

Q(m) 1.01 log log x

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KEVIN FORD

Therefore, the number of quadruples in the first class is O(x(log x)-22), since there are O(logx) possibilities for U.

Secondly, suppose w < E(0.99) and s3 > E(0.8). Applying Lemma 2.2 twice, for each pair (S4, w) the number of possible (p, S3) with p and the numbers in (4.13) all prime is

x (log log X)4 (log log )4 W84 (log3 X) log2(x/(US4W)) ws4(log x)4 6'

Since wm3 is a totient, by Lemma 2.3 and partial summation we have

W 84 E E << (logx)l'? w

84

Summing over U gives O(x(log x)-2'58) quadruples in the second class.

Finally, suppose S3 < E(0.8) and w < E(0.99). By Lemma 2.2, for each

pair (S3, W), the number of (p, 4) with p, ps3ml + 1 = rl, S3S4W + 1 = q and

(S3S4W + l)pm + 1 all prime is

x(log log X)4 (log log )4

83W l2 log2 xlog /(U3)) S3W log4 x

We have ignored the fact that r2 is prime, because r2 is a quadratic function of s4. As in the previous case, wm3 being a totient implies that the sum of 1/w is < (log x)0-01. Since the sum of 1/s3 is O((log x)0?8), there are O(x(log x)-2'18) quadruples in the third class. The number of triples (ml, m2, m3) is 0(1), so

there are O(x(log x)-2'18) pairs (p, q) counted in case (ii). E]

5. Part II, the case (n,pq)= 1

In this argument we use for the first time the fact that the prime factors of p- 1 and q - 1 are nicely distributed (i.e. the full strength of (4.11)). Recall that Ej = exp{(logx)1-j6} for 0 < j < N- 1 and EN = 2.

We may assume that any prime divisor of n which is > EN-1 divides n to the first power, for the number of (p, q) with r(r - 1) I m(p - l)(q - 1) for some r > EN-1 is O(x/EN-1). Denote by ' the set of remaining pairs (p, q).

Let rl,... , rM be the prime factors of n satisfying

P+(r*) > EN-1, r = ri - ~~i Ti (ri- l,pq)

Clearly 1 < M < C. Let w = q(n/(ri ... rM)), so that

P+(w*) < EN-1, W* = (w, pq)

302

In our new notation, we have

(5.1) r ...rMw* =m(p- 1)(q- 1).

We will partition the solutions of (5.1) according to the value of M (the number of "large" prime factors of n), the size of the largest prime factor of each r*, and the location of the factors p and q among the factors ri - 1 and w. For

convenience, set ro = w+1. Define numbers i' and i" byp I (ri -1), q I (ri,-1). It is possible that i' = i". Let Ij = [Ej,Ej-_) and define numbers ji by P+(ri) E Iji. The primes ri may be ordered so that 1 = jl < j2 * M" The fact that jl = 1 follows from the j = 0 case of (4.11).

By (4.8), it follows that p - 1 ri* and q - 1 ri*,. Thus we may define

(p ^ S,,_ p+ ( (q- -1)ri*,

(5.2) S p( (-l)r ) I= 1*+

(( 1 r* )2 \(p- l, ri*)2 , (q-1 ri*,)2

In other words, S' is the largest prime dividing p- 1 and ri*, to different powers. Define j', j" by S' E Ij,, S" E Ij,,, so that j' and j" measure the rough sizes of S' and S". By (4.11), if j' > 1 then the prime factors of p- 1 which are > E1 also divide ri*, so that ji, = 1. Similarly, if j" > 1 then ji" = 1. Furthermore, if either j' > 1 or j" > 1 then there are at least two values of i with ji = 1. Thus, without loss of generality, if j' > 1 we may assume ' = 1 and if j" > 1, we may assume " = 2.

We count separately the number of (p, q) E O' giving solutions of (5.1) with M, j,... ,jM, i', i"l, j' and j" fixed. The number of choices for these

parameters is 0(1) since it only depends on m. To reap the full benefit of (4.11), we break each factor in (5.1) into N

pieces as follows:

NN N

m(p-1)= n rj, q-1= n Oj, r* = I Qi, (1 i M), j=l j=1 j=1

where 7rj, Oj and Qi,j are composed only of primes in Ij. We can suppose x is large enough so that P+(m) < EN-1, i.e. m I rN. Some of these variables

may equal 1. In particular, for each i we have Qi,j = 1 for j < ji. By the definition of j' and j" we also have

(5.3) 7j = Ql,j (j < j), Oj = e2,j (i < j").

Let

(5 = a={i( 1: jij}l ( 1,... ,N- 1),

bo= I{i 1: (ri-l 1,pq) > 1}1,

bj = I{i ~ 1 : ji = j, (ri - 1,pq) = 1}1, (j = 1, ... , N- 1).

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For each j, aj is the maximum number of i,j which can be > 1. The estimation of the number of solutions of (5.1) proceeds in N steps; at step j we count the possibilities for 7rj, Oj and the ei,j, assuming the variables consisting of smaller primes have been fixed. The possibility that j' > 1 or j" > 1 will force us to express some bounds in terms of products involving the Legendre symbol, and these bounds must be carried over many steps. For convenience, for each j > 1 set

(5.5) Ri,j = r ei,h, Pi = 7rh Tj = h. h j h j h>j

If j' > 1 define

(5.6) DI = Rl - 4(Rl,j - Pj/m), H1 = II (I - ( ))

v(E1

and if j" > 1 define

(5.7) D = R2,j(R2,j- 4Tj) H2 = (1 - ( ))

Note that by (5.3),

(5.8) (2-) - (r -(r* -4(p-1))) (D) _ (r(r -4(q -1)))

Below are the estimates that we will prove:

Step 1. Let w*, P2, T2 and R1,2,... ,RM,2 be given. The number of possibilities for 7rl, 01 and l,J, ..., QM,I is

(log log^ )3+bo+bl+2a (log X)-4-bo-bl+(3+bo+bl+2al)XlX2,

xir -1 X21 --1 X1{ H>X- H> {Hi j' > I Hj > i '

Next, define

yii Z3{ j i2 Step j, 2 < j N- 1. Let w*, Pj+i, Tj+i and Ri,j+ i ... ,RM,j+i be

given. Then

YlE < (log log x) 2b +14i (log x)2 06(1+) log a+26+bj(j6-1)+4j6,

where the sum is over all possible 7rj, Oj and lij,... , M,j, and (j = 1 if j j or j = j" and Cj = 0 otherwise.

304

THE NUMBER OF SOLUTIONS OF (b(X)=M

Step N. We have

YNZN < (logg lox)14(log x)56+26(1+6) log(2M+2) IrNON

where the sum is over w*, lrN, ON, and Qi,N (1 ? i < M).

Using the fact that bo + . + bN-1l = M and combining all N steps, the total number of solutions of (5.1) is

<< x(log log X)2M+45+al (log x)Y,

where N-l

Y = -4 - M + 22 logaj + log(2M +2) + SX,

\j=2 / N-l

X = 16 + bo + bi + 2ai + (jb + 2 + 2 log a)+ 2log(2M + 2). j=2

Since aj < M for each j, the coefficient of 62 is at most 2N log(2M + 2). Also, bo < 2; thus

N-1

X < 2log(2M + 2) + 14 + 2N +2M+ (2loga + jbj). j=1

The number of aj = i is ji+l - ji. Also, jl = 1 and 2 log 2 > 1; thus, if M ) 2 then

N-1 M M-1

(2log aj + jbj) - jZ + 2 5 (ji+i - ji) logi + 2(N - 1 - jM) log M j=l i=1 i=1

M

= 2(N -1) logM + jl + (1- 21og (i 1)) ji i=2

M

2NlogM+ N (1- 2lg (N 1)) i=3

= N(M- 2 + 2 log 2).

The above bound also holds when M = 1, since ai = 1 for all i and bj = 0 for

j > 1. Therefore, by (4.4), we obtain

Y < -4 + 2 log 2 + 6(14 + 2M + 4 log(2M + 2)) < -2.603

for every possible M, i', i", j', j" and ji,... ,jM. In conclusion, the number of solutions of (4.2) with (n,pq) = 1 is O(x(logx)-226). Together with (4.12) and Lemma 4.4, this completes the proof of Part II of Theorem 2.

We now establish the bounds claimed for each of the N steps.

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Proof of Step 1 bound. Let I {i : ji = 1} = {1,2,... ,ai}. Recall (5.5) and for i E I write

(5.9) pi,i = AiBi, Ai = (, Bi = (ii,i, 1).

Then

(5.10) 7rl = Ai, O1 = B iEI icI

and

(5-.11) P-(Q AiBi) ?E1.

We need to count the number of (A1,... , Bai) satisfying (5.11) such that the following 3 + bo + bi numbers are prime:

7liPl/m+ 1 = p,

OiTi + 1 = q,

(5.12) < (7riPi/m + 1)(01Ti + l)m + 1 = pqm + 1,

AiBiR,i(ri - 1,pq) + 1 ri, (i E I),

Rli,i(ri - 1, pq) + 1 = ri, (i = i' ? I or i = i" , I).

Recall that p I (rit - 1) and q J (rii - 1). All variables Ai, Bi occur linearly on the left of the above expressions except in a few cases. If i' c I (i.e. ji' = 1), then Ai, occurs quadratically in the expression for ri, (fourth line in (5.12)), and if i" c I then Bil, occurs quadratically in the expression for ri,, (fourth line of (5.12)). There are 0(1) possibilities for Ii = {i : Ai > 1}, 12 = {i : Bi > 1}. Fix Ii, 12 and note that both I1 and 12 are nonempty and Ii U 12 = I. Let J= 1Il + I21.

Place each variable which is > 1 into a dyadic interval, e.g. Ui < Ai < 2Ui for i E I1, Vi < Bi < 2Vi for i E I2, where each Ui,Vi is a power of 2. For some ordering of the J variables, we use a sieve to bound the possibilities for the first variable with the other variables undetermined. Then with the first variable fixed, we use the sieve to bound the possibilities for the second variable, with the remaining J- 2 variables undetermined, and so on. In most cases the variables may be ordered so as to apply Lemma 2.2 with each. The exceptions are when either Ai/ appears quadratically in the expression for ri, and is isolated (none of the other J - 1 variables appears in the expression for ri/), or when Bi,, has a similar property, or when i' = i" = 1 and Ai/ and Bil are isolated in the expression for ri,. We therefore distinguish three cases: (i) al = i' = i/ = 1; (ii) j' = j" = 1 other than case (i); (iii) j' + j" > 2.

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THE NUMBER OF SOLUTIONS OF b(X)==M

In case (i), we have J = 2, j' = j" = 1, bo = 1, bl = 0, 7rl = A1 and

01 = B1. By Lemma 4.1, the number of pairs A1,B1 satisfying (5.11) such that p = Ai(Pi/m) + 1, q = B1T1 + 1, pqm + 1 and A1BiRi,lpq + 1 = r1 are

prime is O(U1V1 (log log x)6(log El)-6). In case (ii), we first choose the numbers Ai, (if i' > 0) and Bi,, (if i" > 0).

Since j' = j" = 1, none of the primes in (5.12) are determined solely by these choices, so no sieving is required. The remaining variables Ai and Bi occur linearly on the left of the expressions in (5.12), so we may apply Lemma 2.2 repeatedly. The number of solutions counted is

? (niujj ui I ) (loglog x)3+bo+bl+J (log E1)-(3+bo+bl+J) iEIl ieI2

For case (iii), suppose j' > 1, i' = 1, so that 7rl = Q1,1 = A1 and rl =

A1Ri,l(A1iP/m + 1) + 1. By Lemma 2.1 and (5.8), the number of possible A1 is < UlHl(loglogx)2 (logE1)-2. Similarly, if j" > 1 the number of possible B2 is < U2H2 (loglog x)2(log E1)-2. Lemma 2.2 is applied repeatedly to the other variables yielding a total of

? (nu ^ii7j?) (logXlog)3+bo+bl+J(lOg E)-(3+bo+bl+J)xlx2 iel, iEI2

solutions counted. In every case, with II and 12 fixed,

E n n U Vi < ? (log x)-1 Ui ,Vi iEI1 iEI2

and the bound in Step 1 follows.

Proof of bound for step j E [2, N - 1]. All j. There are bj numbers i with

ji = j and (ri - 1,pq) = 1. Recall (5.2) and let I denote the set of such i with the additional property that S' { i,j and S" 1 Qi,j. For each i E I, let Wi = P+(Q,j). Fix the partition of these i into I1, the set of i with Wi i Oj, and 12, the set of i with Wi 7 rj. There are 0(1) such partitions. Also fix

- := {,j, '*? ?*, TQM,j; rj*, j }O. T :== {Ql'j J...j f Q ;7r0}

Here, if j 7 j' and j 7 j", we take Qj = Qii,/Wi if i I and j= Qi,j

otherwise, 7rj* = rj/ liej2 Wi and j* = Oj/ niel Wi. If j = j', we also divide out the factor S' from either 7rj or Oj, as well as the appropriate Qi,j. If j = j" we divide out the factor S" in the same manner. Let

(5.13) t = , .QMj.

and note that by (4.11),

(Q(t) < 26(1 + 6) log log x.

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(5.14)

KEVIN FORD

Our procedure will be to fix r and the Wi (i E I), and sum YjZj/S'S" over S', S" if necessary. Then for each i E I, Wi and Wie jRi,j+l + 1 are prime, so by Lemma 2.2 and partial summation,

(5.15) E << (log log x)2(log x)j-1. Wi

For fixed t, let h(t) be the number of (M + 2)-tuples r with product t. At most aj of the numbers ei,j are > 1, thus h(t) is at most the number of dual factorizations of t into 2 and aj factors, respectively. Actually if j < max(j', j") we can do better, but this bound suffices. Therefore h(t) < (2aj)Q(t). Since 2 , Ij, by (5.14) we have

(5.16) h(t) a26(1+6) log log 2(t)

(5.16) S4 t ogx t5 t t

< (log x)26(1+6) logaj H(1 - 2/p)- pEIj

<< (log x)26(1+6) log aj+26

All that remains is to take care of the sum on S', S".

Proof of bound for step j c [2, N- 1]; j j j', j - j". Here there is no sum of S' or S", so we are done by (5.15) and (5.16).

Proof of bound in step j c [2, N - 1]; j = j' 4 j". Since j' > 0, we have i' = 1. There are three cases to consider: (i) S' I l,j, S' { 7rj; (ii) S' 17rj, S' 1 l,j, S' i,j with ji < j; (iii) S' I rj, S' t Ql,j, S' |i,j with ji = j; for all cases suppose S' E (U, 2U] where U is a power of 2 in Ij. In case (i), (5.6) gives

D = AS'(AS' -4Pj/m),

with A = Ri,jI/S' and Pj//m fixed. Therefore, by (4.5), (4.9) and Lemma 4.2 (with k = 1, c= 1, Q = Qj),

E E H Hi log Qj, (log log Z)4 < (log log X)5(log )

U U<S'<2U U log2 U

In case (ii), (5.6) gives

D1 = Ri,j,(Ri,j, -SB)

with B = 4Pj/(mS') and Ri,j, fixed. As in case (i), by Lemma 4.2 (with k = 1, c = 1, Q = Qj,), we have

UE E S' << (loglogx)5(logX)6. U U<S'<2U

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THE NUMBER OF SOLUTIONS OF 0(X)=M

For case (iii), we also have

D1 = Rl,j,(Rl,j, - SB),

with B = 4Pj/(mS') and Rl,j, fixed. Here we must sum over S' for which S'R + 1 is prime, where R = Ri,j/S' is fixed. By (4.5) and Lemma 4.2 (with k = 2, c = 1), we obtain

z z H1 log Qj, (log log x)4 <(oglogx)5(log) << (log log )0 (log x) -l+J1 U U<S'<2U U

The desired bounds now follow from (5.15) and (5.16). A symmetric argument using (5.7) proves the step j bound when j

j" j'. Proof of bound in step j; j j = j"; S' =7 S". By (5.3), ij-_ = 2. Define

i*, i** by S' i*,j, S" Ii**,j. Put S' E (U',2U'], S" E (U",2U"] where U' and U" are powers of 2 in [Ej, Ej_). By the Cauchy-Schwarz inequality,

1/2 1/2

E VE HiH2 1 H22 H2

U',U" S' S" Ul U// S',S// SI?St/

For each double sum over S', S", we first fix S" and sum over S', then sum on S", bounding each sum with Lemma 4.2, Lemma 2.2 or the prime number theorem, as appropriate. For example, for the sum on S' in the first sum we use Lemma 4.2 (with c = 2) if D1 depends on S'. If i* > 3 then S'R' + 1 is also prime, where R' = Ri*,j/S' is fixed, so we use the k = 2 case of the lemma. Otherwise we use the k = 1 case. If D1 does not depend on S', we use Lemma 2.2 if i* > 3 and the prime number theorem otherwise. For the sum on S", we use Lemma 4.2 only if D1 depends on S" but not on S'. If i** > 3 and i** : i*, then we use the k = 2 case of Lemma 4.2, since S"R" + 1 is prime, where R" = Ri**,jS" is fixed. Otherwise use the k = 1 case. If D1 does not

depend on S", or it depends on both S' and S", we apply either Lemma 2.2, if i** > 3 and i** $ i*, or the prime number theorem, otherwise. In every case we obtain the bound

2 U'U" log4 Qj E H1 ? 3 and i** y7 i*, k1 = 0 otherwise, and k2 = 1 if i* ? 3 and k2 = 0 otherwise. The same bound holds for the sum on H22, so

y HIH2 << (loglog )14(log X)4- (kl+k2)(l1j)

U',U" S',S"

The desired bound again follows at once from (5.15) and (5.16).

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KEVIN FORD

Proof of bound for step j E [2, N- 1]; j = j' j", S' = S". Since

(7rj, Oj) = (1,j, o2,j) = 1, either S' 7, S' I 2,j or S' Oj,, S' Q,j. In either

case, by (5.6) and (5.7), D1 and D2 both depend on S' (recall T and the Wi for i E I are fixed). As before we put S' c (U, 2U] where U is a power of 2 in

[ Ej-1, Ej). By the Cauchy-Schwarz inequality and Lemma 4.2 (with c = 2, k = 1) we obtain

T-TH1H2 T1 (( 1/2 1/2 xi 11 2 H 2 H

Eu -^ u<ls V-^ 1 r2 \SIu< u s 1 2

U U<S'?2U U S' S'

<< (llog log )14(log x)46. Combined with (5.15) and (5.16) gives the desired bound. This concludes the proof of the bounds for steps j = 2 through j = N - 1.

Proof of the bound for step N. Let

= {rN, ON; Qe,N,'", QM,N W*},

where the starred variables (except w*) are the quotients of the corresponding unstarred variables and S', S", as appropriate. As before let t - 7r0N. We proceed as in steps 2 through N-1, except now there are no variables Wi to sum over. The sum over S' and S" is handled exactly as in steps 2 through N-1, but the sum over - of 1/t is handled a bit differently. Since 2 E IN, the procedure in (5.16) will not work. By (4.11), we have Q(t) 4 26(1 + 6) log log x + Qf(m). As in the other steps, given t, the number of dual factorizations of t into M +1 factors and 2 factors is at most (2M + 2)Q(t). We thus have

1 (2M+ 2)2 )1 1 T P+(t)<EN-1

< (log x)6+26(1+6) log(2M+2)

and the bound for step N follows.

Acknowledgements. The author thanks Sergei Konyagin for helpful dis- cussions and the referee for helpful criticisms of the exposition.

UNIVERSITY OF TEXAS AT AUSTIN, AUSTIN, TX

Current address: UNIVERSITY OF SOUTH CAROLINA, COLUMBIA, SC

E-mail address: [email protected]

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(Received August 25, 1997)

(Revised April 28, 1998)

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The Number of Solutions of φ (x) = m

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