View of On Fixed Divisors of Forms in Many Variables, I

ON FIXED DIVISORS OF FORMS IN MANY VARIABLES, I

A. SCHINZEL (In memory of Trygve Nagell)

Abstract

LetDd,r be the maximal fixed divisor of a primitive form of degreedinr variables overZ. A formula is given forDd,2and estimates forDd,rforr >2. As a consequence, a question of Nagell raised in 1919 is completely answered.

LetKbe a finite extension ofQand forf ∈K[x¹, . . . , xr] letC(f )andD(f ) be the highest common ideal factor of the coefficients off and of the values of fforx∈^Zr, respectively. Polynomialsf withC(f )=1 are called primitive.

For a prime idealᒍand an idealᑾofK, let ord_ᒍᑾbe the exponent with which ᒍoccurs in the factorization of ᑾ. T. Nagell has proved ([5], p. 16) that for everyf ∈^Z[x¹, . . . , xr] of degreed

(1) D(f )|d!C(f ).

This result is implicit in [4]. An easy generalization is contained in

Theorem1. For every finite extensionKofQand for everyf ∈K[x¹, . . . , xr]of degreed(1)holds.

Put

Sd,r = {F ∈^Z[x¹, . . . , xr],of degreed, homogeneous and primitive}, S_d,r⁰ = {F ∈Sd,r,splitting overC},

S_d,r¹ = {F ∈Sd,r,splitting overZ}.

It follows from (1) that the following definitions are correct:

Dd,r = max

f∈Sd,rD(f ), D_d,r¹ = max

f∈S_d,r¹ D(f ).

ForK=^Q,D(F )is identified with its positive generator. We shall prove

Received 14 October 2011, in final form 20 August 2012.

Theorem2.For allF ∈S_d,r⁰ and for all primesp ord_pD(F )≤ord_p

p

(p^r−1−1)d p^r −1

!

.

Theorem3.For all positive integersdandr >1and for all primesp ord_pD_d,¹2≥ord_p

p

d p+1

!

,

ord_pD¹_d,r ≥(p^r−1−1)q^r−1ord_p((pq)!)+ord_p

p

d−(p^r−1)q^r p+1

!

, whereq =

r

d p^r−1

.

Corollary1.For all positive integersdand for all primesp ord_pDd,2=ord_p

p

d p+1

!

=ord_pD_d,¹2.

Corollary2.The least integerd, sayd2(n), such thatn! |Dd,2is4for n=3and3 ⁿ₂

, otherwise.

The corollary answers a question asked by Nagell [5]. He has proved that d2(2)=3,d2(3)=4,d2(4)=d2(5)=4,d2(n)≤2n−1. The last result has been anticipated by Hermite (see [2], p. 266).

Corollary3.For all integersd ≥3andr ≥2and for all primesp < d ord_pD¹_d,r =d

1

p−1− 1 p^r−1

+d^r−1r O r

p + logd rlogp +1

, where the constant in theO-symbol is absolute andd^r−1r can be omitted for r =2.

Corollary4.For all integersr ≥2

logDd,r =dlogd+O(d) uniformly inr.

Theorem4.

(i) For all positive integersdandr

Dd,r |(d−1)!

(ii) For all integersd ≥4,rd =d−ord2

2 ^d₃

!

−1andr ≥rd, Dd,r =Dd,rd.

Corollary5.For all positive integersd≤6andr ≥2 Dd,r =Dd,2.

Corollary6.D9¹,3=D¹9,2. Corollary 1 suggests the following

Conjecture.For all positive integersdandr Dd,r =D¹_d,r.

This is true ford <9 and eachr, see Remark after the proof of Corollary 5.

Theorem5.Letdr(n)be the least integerdsuch thatn! |Dd,r. Then for allr the limitlr =lim_r→∞^dr_n⁽ⁿ⁾ exists and satisfieslr ≤ ²₂^rr⁻−¹2. If Conjecture is true we have equality.

Theorem6.For all positive integersd ≥3⁴·2³=648andr = ^−1+√₂^8d+1 Dd,r ≡0 mod d−3(2d)^3/4!.

Corollary7.Forr ≥ ^−1+√₂^8d+1

we have

logDd,r =dlogd−d+O(d^3/4logd) uniformly inr.

Proof of Theorem 1. Since xⁿ is a linear combination of _x

j

(j = 0, . . . , n) with integral coefficients, it follows that

(2) f =

i∈I_d

αⁱ x¹

i1

. . .

xr

ir

,

whereI_d = {i= [i1, . . . , ir] :i1+ · · · +ir ≤d},αi ∈K. We shall show by induction onkthat

(3) D(f )|αⁱ for i∈I_k.

Sinceα0 = f (0), (3) holds fork = 0. Assume that it holds fork and let j1+ · · · +jr =k+1. By the inductive assumption

D(f )

i∈I_k

αⁱ x¹

i1

. . .

xr

ir

for all x∈^Zr,

hence, by (2), (4) D(f )

i∈I_d\I_k

αⁱ x1

i¹

. . . xr

ir

for all x∈^Zr.

Since fori∈I_d\I_kwe havei¹+ · · · +ir ≥k+1=j¹+ · · · +jr we obtain eitheris =js for alls≤roris > jsfor at least ones ≤r. Therefore,

i∈I_d\I_k

αⁱ j¹

i1

. . .

jr

ir

=α^j

and, by (4),D(f )|αj, which completes the inductive proof of (3). Now, D(f )g.c.d.

i∈I_d αⁱ g.c.d.

i∈I_d

d i1, . . . , ir

αⁱ

d! g.c.d.

i∈I_d

αⁱ i1!. . . ir!

d!C(f ).

Remark. In the same way one can prove that all values of a polynomial f ∈ K[x1, . . . xr] atx ∈ ^Zr belong to an ideal ᑾ ofK, if and only iff =

i∈Iai

_x1

i1

. . ._xr

ir

, whereai∈ᑾfor alli∈I.

For the proof of Theorem 2 we need two lemmas.

Lemma1.For every primepand positive integern ord_p(n!)= n−sp(n)

p−1 , wheresp(n)is the sum of digits ofnin the basep.

Proof. See [1], pp. 54–55.

Lemma 2.For every finite field F_q, whereq = p^f (p prime) and every vectorv∈^Fr_q\ {0}there exist at mostp^r−1−1vectorsx∈^Fr_p\ {0}such that

(5) vx=0.

Proof. Letw1, . . . , wf be a basis ofF_q overF_p and letv = _f

i=1v_iwi, v_i ∈ ^F_p^r. Sincev = 0we havev_j =0for at least onej ≤ f. Moreover, the

equation (5) givesv_ix =0 for alli ≤f, hence the number in question does not exceed the number of non-zero solutions ofv_jx=0, which isp^r−1−1.

Proof of Theorem2. LetKbe a splitting field ofF and F =

d i=1

Li,

whereLi ∈K[x¹, . . . , xr] are linear forms. Letᒍbe a prime ideal ofKand letπbe an element ofKsuch that ord_ᒍπ =1. SinceC(F )=1, multiplying Li by a suitable power ofπ we may achieve

(6) ord_ᒍC(Li)=0 (1≤i≤d).

Let normᒍ=q and forv∈^Fr_q\ {0}

N^v= {i≤d : Li ≡v¹x¹+ · · · +vrxr modᒍ}.

By Lemma 2

x∈^Fp^r\{0}

v∈^Fr_q\{0} vx=0

|Nv| =

v∈^Frp\{0}

|Nv|

x∈^F_p^r\{0} vx=0

1

≤(p^r−1−1)

v∈^Frq\{0}

|N^v| =(p^r−1−1)d.

It follows that there existsx⁰∈^Zr,x⁰≡0modᒍsuch that denoting byx⁰the residue class ofx⁰modpwe have

(7) s(x⁰):=

v∈^Frq\{0} vx⁰=0

|Nv| ≤

(p^r−1−1)d p^r −1

.

However, fori /∈

v∈^Frq\{0},vx⁰=0N^vwe haveLi(x⁰)≡0 mod_ᒍ, hence ord_ᒍD(F (px+x⁰))=ord_ᒍD

v∈^Frq\{0} vx⁰=0

i∈Nv

Li(px+x⁰)

.

Now for i in question, by (6), ord_ᒍC(Li(px +x⁰)) ≤ ord_ᒍp. Hence by Theorem 1

ord_ᒍD(F )≤ord_ᒍD(F (px+x⁰))≤s(x⁰)ord_ᒍp+ord_ᒍ((s(x⁰))!)

=ord_ᒍ((ps(x⁰))!)

and the inequality ord_p

p

(p^r−1−1)d p^r−1

!

≥ord_pD(F ) follows from (7).

For the proof of Theorem 3 we need again two lemmas

Lemma3. For alld andr and all primitive formsF ∈ ^Z[x¹, . . . , xr] of degreed

D(F )|Dd,r. If, moreover,F splits overZ, then

D(F )|D_d,r¹ .

Proof. We shall prove the first part of the lemma; the proof of the second part is analogous. Assuming the contrary we infer the existence of a primep such that

ord_pD(F ) >ord_pDd,r.

LetDd,r =D(F0), whereF0∈^Z[x1, . . . , xr] is a primitive form of degreed. By the Chinese remainder theorem there exists a formF1 ∈^Z[x1, . . . , xr] of degreedsatisfying the congruences

F1≡F modp^{ordpD(F )}, F1≡F0modDd,r/p^ordpDd,r.

We have F1 = cF2, whereF2is primitive and(c, pDd,r) = 1. Now, by the congruences above

Dd,rp^{ordpD(F )−ordpDd,r} |D(F²), henceD(F2) > Dd,r, contrary to the definition ofDd,r.

Lemma4.For all primespand positive integersdthe form of degreed

Fpd(x, y)=

d−

p+1d

−1

i=0

(x−iy)

d p+1

−1

j=0

(y−jpx) is primitive and satisfies

(8) ord_pD(Fpd)≥ep =ord_p

p d

p+1

!

.

Proof. The formFpd is primitive, since each factor is primitive. We con- sider three cases

y ≡0 modp, (9)

y ≡0≡x, (10)

y ≡0≡xmodp.

(11)

In the case (9) there exists an integerzsuch that x≡zymodp^ep. Hence

d−

p+1d

−1

i=0

(x−iy)≡y^d−

d p+1

d−

p+1d

−1

i=0

(z−i)

≡y^d−

d p+1

d− d

p+1

! z

d− _p+^d₁

modp^ep. Now,

d− d

p+1

≥p d

p+1

, hence

ord_pFpd(x, y)≥min

ord_p

d− d

p+1

!

, ep

≥ep. In the case (10) we havey =pt,t ∈ ^Zand there exists an integerusuch that

t ≡ux modp^ep−

d p+1

. Hence

d p+1

−1

j=0

(y−jpx)=p

d p+1

d

p+1

−1

j=0

(t −j x)

≡p

d p+1

d p+1

! u

p+d1

modp^ep and

ord_pFpd(x, y)≥ d

p+1

+min

ord_p d

p+1

!

, ep− d

p+1

.

Sincep _p+^d₁

and _p+^d₁

have the same sum of digits in the basep, by Lemma 1 the right-hand side equalsep.

In the case (11) we have ord_pFpd(x, y)≥d > d−1

p−1 ≥ord_p(d!)≥ord_p

p d

p+1

!

=ep.

Thus in each case (8) holds.

Proof of Theorem3. Forr = 2 the theorem is contained in Lemma 4.

Forr ≥2 consider the form splitting overZ

F0=

pq−1 a¹=0

. . .

pq−1

ar=0 p(a1,...,ar)

(a1x1+ · · · +arxr).

The number of factors in the product is(p^r−1)q^r, hence degF⁰Fp,d−(p^r−1)q^r = d. We haveF⁰Fp,d−(p^r−1)q^r = cF¹, wherec ≡0 modp andF¹ is primitive, thusF1∈S_d,r¹ and by Lemma 4

ord_pD¹_d,r ≥ord_pD(F⁰)+ord_p

p

d−(p^r −1)q^r p+1

!

.

In order to prove that

(12) ord_pD(F0)≥(p^r−1−1)q^r−1ord_p((pq)!) we distinguish two cases:

(13) xj ≡0 modp for at least onej ≤r and

(14) x¹≡ · · · ≡xr ≡0 modp.

In the case (13) we may assume in view of symmetry betweenxj thatxr ≡ 0 modp. Then there exist integersyj such thatxj ≡yjxr modp^d(j < r)and

we obtain F0(x1, . . . , xr)

≡x_r^(pr−1)qr

pq−1 a¹=0

. . .

pq−1

ar=0 p(a1,...,ar)

(a1y1+ · · · +ar−1yr−1+ar)

=x_r^(pr−1)qr

pq−1 a1=0

. . .

pq−1

ar−1=0 p(a1,...,ar−1)

(pq)!

a¹y¹+ · · · +ar−1yr−1+pq−1 pq

·

pq−1 a¹=0

. . .

pq−1

ar−1=0 p|(a1,...,ar−1)

pq−1

ar=0 par

(a1y1+ · · · +ar−1yr−1+ar)modp^d.

Since there are(p^r−1−1)q^r−1vectors [a1, . . . , ar−1]∈ {0, . . . , pq−1}^r such thatp(a¹, . . . , ar−1)we obtain

ord_pF0(x1, . . . , xr)≥min{d, (p^r−1−1)q^r−1ord_p((pq)!)}

=(p^r−1−1)q^r−1ord_p((pq)!), hence (12) follows.

In the case (14) the same inequality is obvious.

Proof of Corollary 1. For r = 2 we have Sd,r = S_d,r⁰ . The upper estimate for ord_pD(F )given in Theorem 2 and the lower estimate for ord_pD_d,¹2

given in Theorem 3 coincide.

Remark. There exists a more direct proof of Corollary 1 using a factoriz- ation ofF over thep-adic field instead of a factorization overC.

Proof of Corollary2. d²(n)is the least non-negative integer such that n!|Dd,2. By Corollary 1 this divisibility is equivalent to

ord_p(n!)≤ord_p

p d

p+1

!

for all primesp < d,

or to

n p

≤ d

p+1

.

The leastdsatisfying this inequality for all primesp < dis maxp

(p+1)

n p

≥3 n

2

Except forn=3 we have the equality.

Proof of Corollary3. We have by Lemma 1 ord_p

p

(p^r−1−1)d p^r−1

!

< p(p^r−1−1)d (p^r −1)(p−1) =

1

p−1 − 1 p^r −1

d.

On the other hand, ford ≥p^r −1

q^r >

r

d p^r −1−1

r

> d p^r−1 −r

d p^r −1

^r−1_r

and ford < p^r −1

0> d p^r−1 −r

d p^r −1

^r−1

r

, thus by Lemma 1

(p^r−1−1)q^r−1ord_p((pq)!)+ord_p

p

d−(p^r −1)q^r p+1

!

≥(p^r−1−1) pq^r

p−1 −(p^r−1−1)q^r−1 logd

logp +1

+ p

p−1· d−(p^r −1)q^r

p+1 − logd logp

≥ pd

p²−1+ (p^r −p²)q^r

p²−1 +d^r−1r O

logd rlogp+1

> d 1

p−1 − 1 p^r −1

+d^r−1r O r

p + logd rlogp +1

.

It is easy to see that forr =2 the factord^r−1r can be omitted.

Proof of Corollary4. By (1) we have

logDd,r ≤logd!=dlogd+O(d),

on the other hand, by Corollary 1, logDd,r ≥logDd,2≥

p<d pprime

d p+1

logp

≥d

p<d pprime

logp

p+1 −

p<d pprime

logp=dlogd+O(d).

For the proof of Theorem 4 we need three lemmas.

Lemma5.If, for a primep,

(15) ord_pDd,r+1>ord_pDd,r, then there exists a formF ∈Sd,r+1such that (16) ord_pD(F )=ord_pDd,r+1

and

x1x2. . . xr+1|F.

Proof. Let

(17) Dd,r+1=D(F⁰), when F⁰∈Sd,r+1. We have

F0=

S⊂{1,...,r+1}

FS,

whereFS consists of those monomials ofF in which occur just the variables with indices belonging toS,F∅=0. It follows by induction ons≤r+1 that

(18) p^ordpDd,r+1 |D(Fs)

and

(19) pC(Fs),

whereFs =

{1,...,s}⊂S⊂{1,...,r+1}FS.

Indeed, fors=0, (18) and (19) follow from (17). Assuming that (18) holds for ans≤rand puttingxs+1=0 we find that

(20) p^ordpDd,r+1 |D(Fs−Fs+1),

hence by (18)

p^ordpDd,r+1 |D(Fs+1).

Since the form Fs − Fs+1 depends only on the variables x¹, . . . , xs, xs+2, . . . , xr+1it follows from (15) and (20) that

p|C(Fs−Fs+1), hence by (19)

pC(Fs+1),

which completes the inductive proof of (18) and (19). Applying these formulae fors =r+1 we infer that

p^ordpDd,r+1 |D(Fr+1), pC(Fr+1).

The formF =Fr+1C(Fr+1)⁻¹satisfies the conditions of the lemma.

Lemma6.For all positive integersd andrand for all primesp ord_pDd,r+1≤max

ord_pDd,r,

d−r−1 p−1

.

Proof. If ord_pDd,r+1>ord_pDd,r letF be a form of Lemma 5. We have

F =

i∈I

ai·x1ⁱ¹x2ⁱ². . . x_r+^ir+11,

wherei=[i¹, . . . , ir+1],is (1≤s ≤r+1)are positive integers,Iis a certain finite set andi¹+ · · · +ir+1=d,aⁱ∈^Zfor alli∈I. We have

xⁱ =i! x

i

+fi(x), wherefi ∈^Z[x], degfi < i, hence

F =

i∈I

ai r+1

s=1

is! xs

is

+fi(x1, . . . , xr+1)

wherefⁱ∈^Z[x¹, . . . , xr+1], degfⁱ< d. It follows now from Nagell’s theorem [5, p. 15] and (16) that

(21) p^ordpDd,r+1 |g.c.d.i∈I

ai

r+1

s=1

is!

.

However, by Lemma 1,

ord_p

r+1

s=1

is!=

r+1

s=1

ord_pis!≤

r+1

s=1

is−1

p−1 = d−r−1 p−1 , and sinceF ∈Sd,r+1

ord_pg.c.d.i∈I

ai

r+1

s=1

is!

≤ d−r−1 p−1 , hence we obtain from (21)

ord_pDd,r+1≤

d−r−1 p−1

.

Lemma7.For all primesp < d

(22) ord_p

p d

p+1

!

≥ ord2

2 ^d₃

! p−1

.

Proof. In order to diminish the number of parentheses we agree to perform factorial after multiplication. Forp = 2, (22) becomes equality. For p = 3, d <12 we verify (22) directly. Forp= 3,d = 12k+r, 0 ≤r <12,k ≥1 we have

ord3

3

d 4

!

− ord2

2 ^d₃

! 2

≥ k−ord2k!+2 ^r₄

−ord2

2 ^r₃

! 2

≥ k−ord2k!−1

2 ≥0.

Forp=5,d =6k+r, 0≤r <6,k≥1 we have

ord5

5

d 6

!

− ord2

2 ^d₃

! 4

≥ k−ord2k!−ord2

2 ^r₃

! 4

≥ k−ord2k!−1

4 ≥0.

For p = 7, d < 24 we verify (22) directly. For p = 7, d = 24k + r, 0≤r <24,k≥1 we have

ord7

7

d 8

!

− ord2

2 ^d₃

! 6

≥ 3k+6 ^r₈

−ord2k!−ord2

2 ^r₃

! 6

≥ 3k−ord2k!−3

6 ≥0.

Forp < d <3pwe have ord_p

p

d p+1

!

≥ d

p+1

≥1,

ord2

2

d 3

!

≤ord2((2p−2)!)≤2p−3, ord2

2 ^d₃

! p−1

≤1.

For 3p≤d < ^9p−₂³ we have

ord_p

p d

p+1

!

≥ d

p+1

≥2,

ord2

2

d 3

!

≤ord2((3p−3)!)≤3p−4, ord2

2 ^d₃

! p−1

≤2.

For ^9p−₂³ ≤d <6p−3 we have

ord_p

p d

p+1

!

≥ d

p+1

≥3,

ord2

2

d 3

!

≤ord2((4p−4)!)≤4p−5, ord2

2 ^d₃

! p−1

≤3.

Forp≥11,d ≥6p−3 we have ord_p

p

d p+1

!

≥ d

p+1

≥ d−p p+1, ord2

2

d 3

!

≤2 d

3

−1≤ 2 3d−1, hence

ord_p

p d

p+1

!

− ord2

2 ^d₃

! p−1

≥ d−p p+1 −

2 3d−1

p−1

= d(p−5)−3(p²−2p−1) 3(p²−1)

≥ (2p−1)(p−5)−p²+2p+1 p²−1

= p²−9p+6 p²−1 ≥ 7

30.

Proof of Theorem4. (i) We proceed by induction onr. SinceDd,1=1, forr = 1 the assertion holds. Assume thatDd,r | (d−1)!. It for all primes p: ord_pDd,r+1≤max{ord_pDd,r,ord_p((d−1)!)}, we haveDd,r+1|(d−1)!.

Otherwise, by Lemma 5, there exists a primepand a formF ∈ Sd,r+1such that

(23) ord_pD(F )=ord_pDd,r+1>ord_p((d−1)!) and

(24) x¹x². . . xr+1|F.

SinceF ∈ Sd,r+1,F (x1. . . xr,1)is primitive and by (24) of degree at most d−1. Hence by (1)

D(F (x¹. . . xr−1,1)|(d−1)!. contrary to (23).

(ii) We proceed by induction onr ≥rd. Forr =rdthe assertion is obvious.

Assume that it is true for the indexr. IfDd,r+1 > Dd,r, then there exists a primep < dsuch that

ord_pDd,r+1>ord_pDd,r

and since ford ≥4,rd≥2, by Lemma 6 and Corollary 1 d−r−1

p−1

>ord_pDd,r ≥ord_pDd,rd

≥ord_pDd,2=ord_p

p d

p+1

!

. It follows that

ord2

2 ^d₃

! p−1

>ord_p

p d

p+1

!

,

contrary to Lemma 7. ThusDd,r+1= Dd,r and, by the inductive assumption, Dd,r+1=Dd,rd.

Proof of Corollary5. SinceDd,2 ≤Dd,r ≤(d−1)! and ford ≤ 6, d = 5, Dd,2 = (d−1)! we infer thatDd,r = Dd,2. It remains to consider d = 5 and by (ii)r = 3. SinceD⁵,2 = 6,(5−1)! = 24 it suffices to prove thatD5,3≡0 mod 4. Assuming the contrary, by Lemma 5, there exists a form F0∈^Z[x, y, z] such thatxyz|F0

(25) 4|D(F0), 2C(F0).

We have for some integersa, b, c, d, e, f

(26) F⁰=xyz(ax²+bxy+cy²+dxz+eyz+f z²) and forx, y, zodd

4|ax²+bxy+cy²+dxz+eyz+f z². However forx, yodd

x²≡y²≡1, xy ≡x+y−1 mod 4, thus

4|a−b+c−d−e+f +(b+d)x+(b+e)y+(d+e)z and, since this holds for allx, y, zodd we have

(27) b≡d≡emod 2

and

(28) a+b+c+d+e+f ≡0 mod 4.

On the other hand, from (25) and (26) forx, y, z = 2,1,1,1,2,1,1,1,2 c+e+f ≡0, a+d+f ≡0, a+b+c≡0 mod 2. It follows from (27) thata ≡c ≡f mod 2, thusb≡d ≡ f ≡ 0 mod 2 and, by (28), 3a ≡0 mod 2,a≡c≡f ≡0 mod 2, contrary to (25).

Remark. We have(D(xyz(x+y)(x+z)(x+y+z)(y+z)(y−z))=48.

SinceD8,2= 2520 it follows by Lemma 3 and Theorem 4 (i) thatD8,r = 7!

for allr ≥3. On the other hand, a complicated computation shows that for all r ≥2,D⁷,r =5!=D⁷,2.

Proof of Corollary6. We have by Theorem 2 and Corollary 1 ord2D¹9,3≤ord2

2

27 7

!

=ord26!=4=ord2D¹9,2,

ord3D¹9,3≤ord3

3

72 26

!

=ord36!=2=ord3D9¹,2. Forp=5,7 we have by Corollary 1

ord2D9¹,2=1=ord_p8!. For the proof of Theorem 5 we need 5 lemmas.

Lemma8.For all primespand all integersr > 1andα >0letd(r, p^α) andd¹(r, p^α)be the leastdsuch thatp^α |Dd,r andp^α |D_d,r¹ , respectively.

We have

d(r, p^α)≤d¹(r, p^α)≤ p²−1 p

α+2logαp logp +1

if α≥p, d(r, p^α)≤d¹(r, p)≤α(p+1) if α < p.

Moreover,

d(r, p^α)≥α(p−1)+2. Proof. By Lemma 1

ord_p((pn)!)≥ pn

p−1 −logpn logp , hence by Corollary 1

ord_pD_d,¹2=ep ≥ p

p²−1d− logd logp −1

and ifα≥p,

d≥ p²−1 p

α+2logαp logp +1

we obtain

ep≥α+2logαp

logp +1− log^p2_p⁻¹

α+2^log_log^αp_p

logp −1≥α.

Thus ord_pDd,2≥αanda fortiori

ord_pDd,r ≥α.

The same is true forα < pprovidedd ≥(p+1)α.

On the other hand, ifp^α |Dd,r, we have by Theorem 4(i) and by Lemma 1 α≤ord_p((d−1)!)≤ d−2

p−1, henced ≥α(p−1)+2.

Lemma9.Ifc >1andp >8/(c−1), thenp^α |n!implies d(r, p^α)≤d¹(r, p^α) < cn.

Proof. Ifα < p, thenn≥αpand by Lemma 8 d(r, p)≤d¹(r, p)≤α(p+1) < α

1+ c−1 8

p < cn.

Ifα ≥p, then again by Lemma 8 d¹(r, p^α)≤ p²−1

p

α+2logαp logp +1

. On the other hand, by Lemma 1

α=ord_pn!< n p−1, hence

d¹(r, p^α)≤ p²−1 p

n

p−1 +2log 2n logp +1

and d¹(r, p^α)

n ≤ p+1

p +2p²−1 p

log 2n

nlogp+ p²−1 pn .

The right hand side is a decreasing function ofnand sincen≥ p², ^d(r,p_n^α) ≤

d¹(r,p^α)

n ≤ ^p+_p¹+6^p2_p⁻3¹ +^p2_p⁻³¹ <1+ ⁸_p < c.

Lemma10.The limit(r, p)=lim_α→∞^d(r,p_α ^α)exists and satisfies(r, p)≥ p−1.

Proof. By Lemma 8 we have p²−1

p ≥l(r, p)=lim inf

α→∞

d(r, p^α)

α ≥p−1. For every integernthere exists an integerβnsuch that

d(r, p^βn)≤

l(r, p)+ 1 n

βn.

Iff ∈ Sd,r andp^βn | D(f ), thenp^qβn | D(f^q), wheref^q ∈ Sqd,r. Hence choosing for an arbitrary integerα > 0 an integer q such that(q−1)βn <

α≤qβnwe infer that

d(r, p^α)≤qd(r, p^βn) < qα q−1

l(r, p)+ 1 n

and forα > nβn

d(r, p^α)≤

1+ 1

n l(r, p)+ 1 n

α.

Sincenis arbitrary, it follows that

α→∞lim

d(r, p^α)

α =l(r, p).

Lemma11.The limitlr =lim_n→∞^dr_n⁽ⁿ⁾ exists.

Proof. If lim sup_n→∞^dr_n⁽ⁿ⁾ =1, then since by Theorem 4(i) lim inf

n→∞

dr(n) n ≥1

we have lr = 1. Assume that lim sup_n→∞^dr_n⁽ⁿ⁾ = c > 1. Since dr(n) ≤ d²(n)≤ ³₂nwe havec <∞. We shall prove that

(29) lim

n→∞

dr(n)

n = max

p<_c−1¹⁶

l(r, p) p−1 =:M.

In order to prove (29) it suffices to prove that

(30) lim sup

n→∞

dr(n)

n ≤M

and

(31) lim inf

n→∞

dr(n) n ≥M.

Clearly,

dr(n)=max

p^αnd(r, p^α).

Forp > _c−¹⁶₁ we have by Lemma 9

d(r, p^α) < c+1 2 n, and, since ^c+₂¹ < c,

lim sup

n→∞

dr(n)

n =lim sup

n→∞ max

p^αn p<_c−1¹⁶

d(r, p^α)

n .

Since for everyp < _c−¹⁶1,ε >0 andα > α(ε) d(r, p^α) < (l(r, p)+ε)α while, by Lemma 1

(32) α = n

p−1 +O(logn) we obtain forn > n0(ε)

d(r, p^α) < (M+ε)n, which gives (30).

In order to prove (31) letM = ^l(r,p)_p−1 for a primep. In view of (32)n!|Dd,r

implies for everyε >0 andn > n¹(ε) d≥d(r, p^α)≥(l(r, p)−ε)

n

p−1+O(logn)

≥(M−ε)n+O(logn), which proves (31).

Lemma12.Letd_r¹(n)be the leastd such thatn!|D_d,r¹ . Then

n→∞lim d_r¹(n)

n = 2^r −1 2^r −2. Proof. We shall prove the lemma in two steps

(33) lim sup

n→∞

d_r¹(n)

n ≤ 2^r−1 2^r−2 and

(34) lim inf

n→∞

d_r¹(n)

n ≥ 2^r −1 2^r −2. We have

d_r¹(n)=max

p^αn!d¹(r, p^α).

By Lemma 9 forp >8(2ⁿ−2)we have d¹(r, p^α) < 2^r −1

2^r −2n.

For p < 8(2ⁿ −2), by Corollary 3 for every ε > 0 and n > n(ε), d >

₂r−1 2^r−2+ε

n. ord_pD_d,r¹ =d

1

p−1− 1 p^r−1

+O

d^r−1r logd

>

2^r −1

2^r −2+ε 1

p−1 − 1 p^r −1

+O

n^r−1r logn

> n

p−1 +O(logn)=α.

This shows (33). In order to prove (34) suppose that forε >0 and arbitrarily largen

d <

2^r −1 2^r −2 −ε

n.

Then by Corollary 3 ord2D¹_d,r <

2^r −1

2^r −2−ε 1− 1 2^r−1

n+O

n^r−1r logn

< n−O(logn)=ord2n!.