How Many Primes Can Divide the Values of a Polynomial?

HOW MANY PRIMES CAN DIVIDE THE VALUES OF A POLYNOMIAL?

FLORIAN LUCA AND PAUL POLLACK

For Andrzej Schinzel on his seventy-ﬁfth birthday

Abstract. Let F (T ) Z[T ] be a nonconstant polynomial. We prove a result concerning the maximal2 order of ⌦(F (n)), where ⌦( ) denotes the total number of prime factors (counting multiplicity). In the· case when F has only simple roots, the result asserts that ⌦(F (n)) 1 lim sup = , n log n log ` !1 where ` is the least prime for which F has a zero in the `-adic integers Z`. This extends investigations of Erd˝osand Nicolas, who treated the case F (T )=T (T + 1).

1. Introduction

Let ⌦(n):= pk n 1denotethenumberof(positive)primefactorsofn, | counted with multiplicity. The study of statistical properties of ⌦(n)wasa P major impetus for the development of probabilistic number theory in the first half of the twentieth century. It is a simple consequence of elementary prime number theory that ⌦(n)behaveslikeloglogn on average. Hardy and Ramanujan [HR17] showed that this average behavior is typical by demonstrating that ⌦(n) log log n as n along a set of asymptotic ⇠ !1 density 1. This result was refined in celebrated work of Erd˝os and Kac [EK40], who proved that ⌦(n)possessesaGaussiandistributionwithmean and variance log log n. In contrast with the depth of these authors’ work, the minimal and maximal orders of ⌦(n)aretrivialtodetermine:⌦(n)=1 log n for prime n, while ⌦(n)= log 2 when n is a power of 2. Let F (T ) Z[T ] be a fixed polynomial. One can ask for the average, 2 normal, minimal, and maximal orders of ⌦(F (n)). The first two questions have been satisfactorily resolved (see, e.g., [Hal56] for an analogue of the Erd˝os–Kac theorem in this context). The third question is in general very dicult; to take a famous example, we expect that if F (T )=T (T +2), then ⌦(F (n)) = 2 infinitely often, but we are still far from proving this. Probably the following is true:

2010 Mathematics Subject Classiﬁcation. Primary 11N32; Secondary 11N37, 11J68. Key words and phrases. Hypothesis H, subspace theorem, number of prime factors. 1 2F.LUCAANDP.POLLACK

Hypothesis H0. Let F (T ) be a nonzero polynomial with integer coecients.

Let D := gcdn Z F (n) be the greatest ﬁxed divisor of F . Then 2 { } lim inf ⌦(F (n)) = r + ⌦(D), n !1 where r is the number of monic irreducible factors of F in Q[T ], counted with multiplicity.

We remark that it is easy to compute D = D(F )foragivenF . Indeed, setting d := deg F , a theorem of Hensel [Hen96] asserts that

D(F )=gcd(F (0),F(1),...,F(d)).

Moreover, Pólya [Pol15] has shown (cf. [Bha00]) that if the coecients of F do not all share a common factor > 1, then D(F ) d!. | We show in 2 that Hypothesis H0 is equivalent to Schinzel’s well-known § Hypothesis H [SS58] concerning simultaneous prime values of polynomials. This equivalence, and its proof, are similar in flavor to Schinzel’s own argu- ment [Sch62] that his Hypothesis H implies a conjecture of Bunyakovsky. The primary purpose of this note is to give a satisfactory answer to the remaining, fourth question: What is the maximal order of ⌦(F (n))? Erd˝os and Nicolas [EN80] considered this problem when F (T )=T (T +1),so that ⌦(F (n)) = ⌦(n)+⌦(n +1).Trivially,⌦(F (n)) 2 log (n+1) . But these  log 2 authors showed [EN80, Théorème 3] that in fact, log n ⌦(n(n +1)) (1 + o(1)) (as n ).  log 2 !1 We show that ⌦(F (n)) always has maximal order C log n for some positive constant C. More precisely, factor F as

k (1.1) F (T )= Cont(F ) F (T )ei , ± i i=1 Y where Cont(F )isthecontent of F (the greatest common divisor of its coecients) and the Fi are distinct irreducible elements of Z[T ], each with positive leading term. For each 1 i k, we let ` denote the least prime `   i for which Fi has a root in the `-adic integers Z` (equivalently, for which Fi has a zero modulo every power of `). To see that ì is well-defined, first note that a straightforward variant of Euclid’s proof of the infinitude of primes shows that Fi has a root modulo ` for infinitely many primes ` (see, for example, [Sch12, pp. 40–41]). For each such ` not dividing the resultant of

Fi and Fi0, one sees from Hensel’s lemma that F has a root in Z`. We can now state our main result. HOW MANY PRIMES CAN DIVIDE THE VALUES OF A POLYNOMIAL? 3

Theorem 1. Let F (T ) Z[T ] be a nonconstant polynomial with integer 2 coecients. Suppose that F is written in the form (1.1). For each 1 i k,   define ì as above. Then ⌦(F (n)) e lim sup = C(F ), where C(F ):=max i . n log n 1 i k log ì !1   In particular, if F has only simple roots, then C(F )=1/ log `, where ` is the least prime for which F has a zero in Z`.

Our proof is similar in spirit to that of Erd˝os and Nicolas, but replaces the use of Ridout’s version of Roth’s theorem with an application of the subspace theorem. The questions we have raised make sense also for !(F (n)), where ! counts the number of distinct prime factors. However, as observed already by Erd˝os and Nicolas, here it seems very dicult to prove any nontrivial re- sults about the maximal order. To illustrate the diculties, call the natural number n special if n(n +1)istheproductofthefirst k primes for some k; e.g., n =714isspecial,withk =7.Improvingthetrivialbound !(n(n +1)) lim sup 2 n log n/ log log n  !1 entails showing that there are only finitely many special n. This conjecture, first proposed by Nelson, Penney, and Pomerance [NPP74], seems unattack- able at present. It may be mentioned that in a 2009 preprint, A. D¸abrowski conjectured that there are precisely 28 solutions to the equation

(1.2) N 2 1=pa1 pak , 1 ··· k where N and the ai are positive integers, and pi denotes the ith prime. Note that if n is special, then N =2n +1givesasolutionto(1.2).Theresults of [LN11] imply the truth of D¸abrowski’s conjecture for all k 25.  Notation. Most of our notation is standard or will be explained when needed, so we make only a few brief remarks: We let (without a subscript) |·| denote the usual absolute value on C.Foraprimep and a nonzero rational number x, we write ordp(x)fortheexponentofp in the prime factorization of x. We say that a number n is y-smooth if each prime factor of n is bounded by y, and we deﬁne the y-smooth part of n as its largest y-smooth divisor.

2. Minimal order: The equivalence of Hypotheses H and H0 We begin by recalling the statement of Schinzel’s Hypothesis H [SS58]: 4F.LUCAANDP.POLLACK

Hypothesis H. Suppose that G1(T ),...,Gk(T ) Z[T ] are irreducible over 2 k Q, each with positive leading coecient. Put G := i=1 Gi. Suppose that G has no ﬁxed prime divisor: for every prime p, there is an integer m Q p for which p - G(mp). Then there are inﬁnitely many natural numbers n for which each Gi(n) is prime.

It is clear that Hypothesis H0 implies H: We need only apply H0 with r F = i=1 Gi (assuming, as we may, that the Gi are distinct). So we may focus our energies on showing that H implies H . Q 0 Let F (T ) Z[T ] be a nonzero polynomial for which we wish to establish 2 the conclusion of Hypothesis H0. We can assume that F is nonconstant and

k ei that Cont(F ) = 1. Hence, we may write F = i=1 Fi(T ) , where each F (T ) is nonconstant, irreducible over , and possesses a positive leading i Z Q coecient. Let D =gcdn Z F (n) ; we must prove that 2 { } k

lim inf ⌦(F (n)) = ⌦(D)+r, where r := ei. n !1 i=1 X It is easy to prove that ⌦(F (n)) ⌦(D)+r for all large n. Indeed, since k k D F (n)= F (n)ei , there is a factorization D = D where each | i=1 i i=1 i e (ordpD )/e D F (n) i . Then with D0 := pd i ie,wehaveD0 F (n)forall i | i Q i p Q i | i 1 i n, and so   Q k k ei ei F (n)= Di0 (Fi(n)/Di0) . i=1 ! i=1 ! Y Y The ﬁrst product contributes at least ⌦(D)primefactors,sinceD = Di e k | D0 i , and the second product contributes at least e = r primes (for i i=1 i Q large n). This gives the desired lower bound on ⌦(F (n)). Q P Turning to the upper bound, let be the set of primes p for which S either p D or p deg F . For each p , choose an integer n so that |  2 S p 1+ordp(D) p - F (np). Choose n0 to satisfy the simultaneous congruences

n n mod p1+ordp(D) for all p . 0 ⌘ p 2 S 1+ordp(D) ˜ ˜ With M := p p ,putFi(T )=Fi(MT + n0), and set F (T )= 2S k F (MT + n ), so that F˜(T )= F˜ (T )ei . Since D M and D F (n ), it 0 Q i=1 i | | 0 is clear that F˜(T )/D Z[T ]. Moreover, F˜(T )/D has no ﬁxed prime divisor: 2 Q Indeed, gcd(F˜(0)/D, M)=gcd(F (n0)/D, M)=1byconstruction,sothat no prime in is a ﬁxed divisor of F˜(T ). Moreover, if p , then the S 62 S reduction of F˜(T )modulop is nonzero and has degree deg F

Since F˜(T )/D has no ﬁxed prime divisor, we have in particular that k ei (2.1) D =Cont(F˜)= Cont(F˜i) . i=1 Y ˜ ˜ Let Gi(T ):=Fi(T )/Cont(Fi). Since each Fi is irreducible over Q, so is each Gi. In Z[T ], we have (referring back to (2.1)) k k G(T ):= G (T ) G (T )ei = F˜(T )/D; i | i i=1 i=1 Y Y as F˜(T )/D has no ﬁxed prime divisor, neither does G(T ). So by Hypothesis

H, there are inﬁnitely many n for which each Gi(n)isprime.Foranysuch n, it is clear that k ei F (Mn+ n0)=F˜(n)=D Gi(n) i=1 Y has precisely ⌦(D)+ i ei = ⌦(D)+r prime factors, as desired.

P 3. Maximal order 3.1. The lower bound. It is simple to prove that ⌦(F (n)) is occasionally at least as large as predicted here: Fix 1 i r so that the ratio e / log `   i i is maximal. Write e = ei and ` = `i (to ease notation). For each natural number j, choose n [`j, 2`j)sothatF (n ) 0(mod`j). Then the n j 2 i j ⌘ j tend to inﬁnity and log(n /2) ⌦(F (n )) e ⌦(F (n )) e j e j . j · i j · log ` This shows that the lim sup considered in Theorem 1 is at least as large as predicted.

3.2. The upper bound. To see that the limsup in Theorem 1 is no larger than predicted, we use a version of Schmidt’s subspace theorem due to Schlickewei. First, some terminology. For a nonzero rational number x,its inﬁnite valuation is x = x . Finite valuations correspond to prime num- | |1 | | bers p, and for such a prime, the p-adic valuation of x is taken to be

ordp(x) x = p . Put = p : p prime .Suchvaluationsare | |p MQ { } [ {1} sometimes called normalized since the product formula

x v =1 holdsforall x Q⇤. | | 2 v 2YMQ Often one extends these valuations to all algebraic numbers. A canonical way to do this is the following. Let K be an algebraic number ﬁeld of degree d over Q. The inﬁnite valuations v of K are in correspondence with the 6F.LUCAANDP.POLLACK

1/d embeddings : K , C. If is real and x K, then x v = (x) , ! 2 | | | | whereas if is complex non-real then x = (x) 2/d. Finite valuations of | |v | | K are in correspondence with prime ideals ⇡ in . More precisely, say ⇡ OK f cvord⇡(x) is a prime ideal in K of norm N / (⇡)=p . Then x v = p , where K Q | | cv = f/d,andord⇡(x)istheexponentatwhichtheprimeideal⇡ appears in the factorization of the fractional ideal x generated by x inside K. OK Put for the set of all valuations of K. Then one checks easily that the MK formula

x v =1 holdsforall x K⇤. | | 2 v 2YMK Let m 2 be given, and let be a ﬁnite subset of containing all S MK the inﬁnite valuations. Assume that for each v we are given a system 2 S of m linearly independent linear forms Lv,i(x)inx =(x1,...,xm)with (j) coecients in K. Using x (1 j d)fortheconjugatesofxi,put i   (j) x := max xi . k k 1 i m | | 1 j d   Then the p-adic subspace theorem of Schlickewei [Sch77] (in the formulation of [Sch91, Theorem 1D, p. 177]) says the following:

Theorem A. For each " > 0, the solutions x m of the inequality 2 OK m " L (x) < x | v,i |v k k v i=1 Y2S Y lie in ﬁnitely many proper subspaces of Km. The proof of the upper-bound assertion of Theorem 1 is based on the following lemma:

Lemma 1. Let F (T ) Z[T ] be a polynomial of degree d 1 with only 2 simple roots. Fix Z>0. For each natural number n for which F (n) =0, 6 write F (n)=UV, where U is the Z-smooth part of F . Given ✏ > 0, we have U>n1+✏ for only ﬁnitely many natural numbers n.

Proof of Lemma 1. We can assume that d 2, otherwise there is nothing to prove. We can also assume that F is monic. To see this, write cd for the leading coecient of F . Replacing F with F if necessary, we can assume d 1 that cd > 0. Then cd F (T )=G(cdT )forsomemonicG of the same degree as F (still with only simple roots), and the lemma holds for F provided that it holds for G. Let K be the splitting ﬁeld of F and write F (T )=(T ✓ )(T ✓ ) (T ✓ ). 1 2 ··· d HOW MANY PRIMES CAN DIVIDE THE VALUES OF A POLYNOMIAL? 7

Let be the set of n such that F (n)=UV, with U>n1+" and Z-smooth. N Since F (T )ismonic,thenumbers✓ ,...,✓ are all in . Put m =2, 1 d OK x =(x ,x ). We shall take x = n ✓ and x = n ✓ .Wetake to be 1 2 1 1 2 2 S the finite subset of consisting of the following valuations: MK (i) all the infinite valuations of K; (ii) all the finite valuations of K sitting above some prime number p Z.  To define the forms L (x)forv and i =1, 2, it is helpful to v,i 2 S introduce the notion of type. Let be the set of finite valuations of .To P S each n we associate a type function f : 1, 2,...,d , as follows: 2 N P ! { } Let ⇡ . For each 1 i d, write 2 P   (n ✓ ) = ⇡ei I , i OK i,⇡ where I is an ideal of coprime to ⇡ and e is a nonnegative integer. i,⇡ OK i We define f(⇡) 1, 2,...,d as that index i for which e is as large as 2 { } i possible, choosing arbitrarily among the possibilities if more than one such i exists. The number of possible types is finite. So to show that is finite, N it suces to show that there are only finitely many n having a fixed type f.

We are now ready to define the forms Lv,i(x). If v is infinite, we take L (x)=x and L (x)=x x . It is clear that they are independent. v,1 1 v,2 1 2 Say that ✓ has degree d d, with minimal polynomial F F .Forlargen, 1 1 | 1 | L (x)= n ✓ ✓ ✓ v,i | 1|v | 1 2| v infiniteY v infiniteY v infiniteY (3.1) N (n ✓ )1/d = F (n)1/d1 n. ⇣ K/Q 1 1 ⇣ (The implied constants may depend on F .) In order to proceed to the finite valuations, observe first that for i =3,...,d,wehaven ✓ = c x + d x , i i 1 i 2 where (ci,di) is the unique solution of the system ci +di =1andci✓1 +di✓2 = ✓ . Observe that d =0,asotherwisec =1and✓ = c ✓ = ✓ . If now v i i 6 i 1 i 1 i 2 P corresponds to a prime ideal ⇡, we then take Lv,1(x)=x1 and Lv,2(x)=x2 if f(⇡) 1, 2 ,andL (x)=x and L = c x + d x if f(⇡)=i 3. 2 { } v,1 1 v,2 i 1 i 2 In all cases, Lv,1(x)andLv,2(x)areindependent.Itremainstocompute L (x) for i =1, 2. Continuing to denote f(⇡)byi,notethatifj = i, | v,i |v 6 then e e , so that ⇡ej divides n ✓ and n ✓ . Thus, it also divides j  i j i ✓ ✓ . Hence, j i ⇡ j=i ej (✓ ✓ ) (F ), 6 | j i | P j=i Y6 where (F )isthediscriminantofF . This shows immediately that f(n) L (x) L (x) n ✓ | |v . | v,1 |v| v,2 |v  | i|v  (F ) | |v 8F.LUCAANDP.POLLACK

Hence,

f(n) v (F ) 1 (3.2) Lv,1(x) v Lv,2(x) v | | | | . | | | |  (F ) v  U ⌧ U v v v Yfinite2S v Yfinite2S | | Thus, putting together (3.1) and (3.2), we have 2 n 1 " L (x) x . | v,i |v ⌧ U ⌧ n" ⌧k k v i=1 Y2S Y By Theorem A, all solutions x are contained in finitely many subspaces 2 of K . In other words, there is a positive integer K and K pairs (C1,D2), ...,(CK ,DK )ofnumbersinK, not both zero, such that each such solution x satisfies C x + D x =0forsome1 i K.Takeani with 1 i K. i 1 i 2     If C =0,thenD =0;thus,x =0andn = ✓ .Similarly,ifD =0,then i i 6 2 2 i x =0andn = ✓ . If neither C nor D vanishes, then (n ✓ )/(n ✓ )= 1 1 i i 1 2 x /x = D /C , which uniquely determines n. So there are only finitely 1 2 i i many possibilities for n, as desired. ⇤ Completion of the proof of Theorem 1. It remains only to prove the upper bound for the lim sup. For an interval I, let us write ⌦I (n):= pk n,p I 1 | 2 for the number of prime power divisors pk of n with p I. 2 P Fix a large real number Z.WriteF in the form (1.1). By the choice of

`i, each prime < `i divides Fi(n) to a bounded power. Hence, for large n,

(3.3) ⌦(F (n)) = ⌦(Cont(F ))+ k

ei ⌦[1,` )(Fi(n)) + ⌦[` ,Z](Fi(n)) + ⌦(Z, )(Fi(n)) i i 1 i=1 X k = O(log n/ log Z)+ e ⌦ (F (n)), i · [ì,Z] i i=1 X where for an interval I we use ⌦I (m)forthenumberofprimefactorsofm in I, multiple factors counted multiply. (As before we suppress the dependence of the implied constants on F .) Let Ui denote the Z-smooth part of Fi(n), so that log Ui (3.4) ⌦[ì,Z](Fi(n)) .  log ì k We now apply Lemma 1 with G(T ):= i=1 Fi(T ). Writing G(n)=UV, where U is the Z-smooth part of G(n), we find that U n1+o(1) as n , Q  !1 and so k log U (3.5) i 1+o(1). log n  i=1 X HOW MANY PRIMES CAN DIVIDE THE VALUES OF A POLYNOMIAL? 9

From (3.3), (3.4), (3.5), and the deﬁnition of C(F )giveninthetheorem statement, k log U e ⌦(F (n)) log n i i + O(log n/ log Z)  log n log ` i=1 i X (C(F )+o(1)) log n + O(log n/ log Z).  Dividing by log n and letting n gives that lim sup ⌦(F (n))/ log n !1  C(F )+O(1/ log Z). Since Z can be taken arbitrarily large, the result follows. ⇤

Acknowledgements

The authors thank Carl Pomerance, Enrique Trevi˜no, and the referee for suggestions which improved the quality of the manuscript. This research was conducted while P. P. was supported by NSF award DMS-0802970.

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The John Knopfmacher Centre for Applicable Analysis and Number Theory, University of the Witwatersrand, P.O. Wits 2050, South Africa Current address: Instituto de Matemáticas, Universidad Nacional Autónoma de México, C.P. 58089, Morelia, Michoacán, México E-mail address: [email protected]

University of British Columbia, Department of Mathematics, 1984 Math- ematics Road, Vancouver, British Columbia V6T 1Z2, Canada E-mail address: [email protected]