2 MICHAEL FILASETA proved the analogous result for irreducible cubics. Nair [9] has shown that in the case of an irreducible polynomial f(x) of degree

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1 SQUAREFREE VALUES OF POLNOMIALS Michael Filaseta*. Introduction. The purpose of this paper is to present some results related to squarefree values of polynomials. For f(x) 2 Z[x] with f(x) 6 0; we dene N f = gcd(f(m);m 2 Z): For computational reasons it is worth noting that N f = gcd(f(m);m2f0;; :::; ng) where n denotes the degree of f(x): This observation is due to Hensel (cf. [, p. 334]) and follows in a fairly direct manner after using Lagrange's interpolation formula to deduce that f(m) = n j=0 (,) n,j m j m, j, n, j f(j); where m is any integer > n:we will be interested in estimating the number of polynomials f(x) for which there exists an integer m such that f(m) is squarefree. This property should hold for all polynomials f(x) for which N f is squarefree. However, this seems to be very dicult to establish. Nagel [8] showed that if f(x) 2 Z[x] is an irreducible quadratic and N f is squarefree, then f(m) is squarefree for innitely many integers m: Erd}os [2] *Research was supported in part by the NSF under grant number DMS Typeset by AMS-TE

2 2 MICHAEL FILASETA proved the analogous result for irreducible cubics. Nair [9] has shown that in the case of an irreducible polynomial f(x) of degree n; one may obtain a similar theorem for k-free values of f(x) provided that k (, )n: Of related interest are the papers of Hooley 2 [5], Nair [0], and Huxley and Nair [6]. The problem of determining whether there exists a polynomial f(x) 2 Z[x] of degree 4 for which there are innitely many integers m such that f(m) is squarefree is open. Our interest is in the simpler problem of showing that many polynomials take on at least one squarefree value. If one can show that (i) every polynomial f(x) 2 Z[x] with N f squarefree is such that f(m) is squarefree for at least one integer m; then it will follow that (ii) every polynomial f(x) 2 Z[x] with N f squarefree is such that f(m) is squarefree for infinitely many integers m (cf. the proof of Theorem 2 in [3]). In fact, (i) implies that (iii) every polynomial f(x) 2 Z[x] is such that f(m)=n f is squarefree for innitely many integers m: Our goal is to show the weaker result that almost all polynomials f(x) with N f squarefree take on at least one squarefree value. To clarify our results, we dene S n (N) =ff(x)= n j=0 a j x j 2 Z[x]:ja j jn for j =0;;:::;ng: Thus, js n (N)j = (2[N] +) n+ : We say that almost all polynomials f(x) have a certain property P if for every non-negative integer n; jff(x) 2 S n (N) :f(x) satises P gj () lim N! js n (N)j =: Results associated with almost all polynomials go back to van der Waerden [2]. He showed that for almost all polynomials f(x) the associated Galois group is the symmetric

3 SQUAREFREE VALUES OF POLNOMIALS 3 group on n letters where n = deg f(x): In particular, this implies that almost all polynomials are irreducible. A proof of this latter fact can be found in Polya and Szeg}o [, p. 56]. Other related results can be found in Gallagher [4] and the author's [3]. We make a brief historic remark on the phrase \almost all" in this context. Van der Waerden's Algebra I includes a comment on his result above [3, p. 204]. The German edition states that the Galois group is the symmetric group for asymptotically \00%" of the polynomials rather than using a German equivalent for \almost all." This led to a mistranslation in the English edition [4, p. 200] where a statement is made asserting that the Galois group is the symmetric group for \all" polynomials. The earliest editions of van der Waerden's Algebra I do not refer to his result above. At times we will restrict our attention to polynomials f(x) for which N f is squarefree. An almost all result for such f(x) will mean that () holds with S n (N) replaced by ff(x) 2 S n (N) :N f squarefreeg: We will prove Theorem. Almost all polynomials f(x) with N f squarefree are such that f(m) is squarefree for some integer m: Theorem 2. Almost all polynomials f(x) are such that there is an integer m for which f(m)=n f is squarefree. We will actually prove stronger results (see section 3). As a consequence of the stronger results, we note that almost all polynomials f(x) = P n j=0 a jx j are such that f(m)=n f is squarefree for some positive integer m (max 0jn fja j jg) ; where (x) isany function which tends to innity with x: We end this section by asking whether analogous results hold when one considers values

4 4 MICHAEL FILASETA f(m) with large prime factors rather than squarefree numbers. In particular, is there an absolute constant c> (or even a c> which depends on deg f(x)) such that almost all polynomials f(x) are such that there is a positive integer m and a prime p for which pjf(m) and p>m c? 2. Preliminaries Throughout this section and the next we make use of the notation established in the introduction. We view n as being a xed nonnegative integer so that, in particular, other quantities such asmay depend on n: We will, however, stress when such a dependence is necessary. We reserve p for denoting primes. Lemma. Let >0;and let B = B(N) be a function which increases to innity with N: Suppose further that B(N) =o(n):then there exists N 0 = N 0 (n; ; B) such that if N N 0 ; then the number of pairs (f(x);m) with f(x) 2 S n (N); m2z\[;b];and f(m) squarefree is in the interval (, ) 6 2 (2N)n+ B;( + ) 6 2 (2N)n+ B : Proof. Let 0 > 0: Fix m 0 to be a positive integer satisfying m 0 (= 0 ) + so that if m m 0 ; then m n, + +m+= mn, m, <0 m n : For the moment xmto be an integer in [m 0 ;B]; and consider an integer d such that (2) jdj (, 0 )Nm n :

5 SQUAREFREE VALUES OF POLNOMIALS 5 If a 0 ;a ;:::;a n, are arbitrary integers in [,N;N] and N is suciently large, depending only on 0 ; we get that (3) d,, an, m n, + +a m+a 0 Nm n : We successively choose a 0 ;a ;:::;a n, as above with a 0 d (mod m) and for j 2 f; 2;:::;n,g; a j, d,a 0,,a j, m j, =m j (mod m): Thus, the total number of choices for (a 0 ;a ;:::;a n, )is 2[N] + m n +O() 2N = m n + O n N n, : m n, By (3), we can now nd a unique a n 2 [,N;N] such that d = a n m n + +a m+a 0 : The above steps may bereversed. More specically, given m and d as above, we must have that a 0 ;:::;a n, satisfy the congruences above, and this uniquely determines a n as above. Thus, for m xed in [m 0 ;B]; each integer d satisfying (2) has (2N=m) n +O n, N n, =m n, representations of the form f(m) where f(x) 2 S n (N): We now let m vary over all the positive integers m B: We divide the pairs (f(x);m); where f(x) 2 S n (N) and m B; into 3 sets S ;S 2 ; and S 3 : The set S consists of those (f(x);m) for which d = f(m) is squarefree, m 2 [m 0 ;B]; and (2) holds. The set S 2 consists of those (f(x);m) for which d = f(m) is nonsquarefree, m 2 [m 0 ;B]; and (2) holds. The set S 3 consists of the remaining pairs (f(x);m): Then since for any t>0 the

6 6 MICHAEL FILASETA number of squarefree numbers t is (6= 2 )t + O( p t); we get that js j = m 0 mb 2N m n (6= 2 )(, 0 )(2N)m n + O n (N m)+o n N n+ 2 =(6= 2 )(, 0 )(2N) n+ B + O n, N n+ m 0 + On, N n B 2 + O N n+ 2 B ; js 2 j =, 6,, (, 0 )(2N) n+ B + O 2 n N n+ m 0 + On N n B 2 + O N n+ 2 B ; and js 3 j = (2[N]+) n+ [B],jS j,js 2 j = 0 (2N) n+ B + O n, N n+ m 0 + On, N n B 2 + O N n+ 2 B : Now, js j gives us a lower bound on the number of pairs (f(x);m) with f(m) squarefree and m 2 [;B]: An upper is,, js j + js 3 j < (6= 2 )( + 0 )(2N) n+ B + O n N n+ m 0 + On N n B 2 + O N n+ 2 B : Thus, taking 0 = =2 and N suciently large, the result follows. The proof of Lemma given above is similar to the proof of Lemma in [3]. Lemma asserts that the f(x) 2 S n (N) onaverage take on 6 2Bsquarefree values as x ranges over the positive integers B: We note that this is true despite the fact that a positive proportion of the f(x) 2 S n (N) take onno squarefree values. More specically, observe that N f is divisible by if and only if f(x) x 2 (x, ) 2 (x,(p,)) 2 g(x)+px(x, ) (x,(p,))h(x) (mod );

7 SQUAREFREE VALUES OF POLNOMIALS 7 for some polynomials g(x) and h(x) 2 Z[x]: Thus, if p n +; then f(x) 0 is the only such f(x) modulo ;if(n+)=2 p n; then there are exactly p n,p+ incongruent such f(x) modulo ; and if p n=2; then there are exactly n,3p+2 incongruent such f(x) modulo : A simple application of the sieve of Eratosthenes implies that for N suciently large, the proportion of f(x) 2 S n (N) for which N f is nonsquarefree is asymptotic to, pn=2, p 3p, p (n+)=2pn, p n++p pn+, p 3p = 0:05675 ::::, n+2 Thus, the polynomials f(x) 2 S n (N) which take on at least one squarefree value as x ranges over the positive integers B on average take on(6= 2 )B(:059 :::) squarefree values. This curiosity is due to the size of the coecients of the polynomials under consideration in comparison to B: For f(x) 2 Z[x] and ` 2 Z; we dene (`) = f (`)tobethenumber of incongruent solutions to f(x) 0 (mod `): The next lemma gives some basic properties of (`): Lemma 2. Let f(x) 2 Z[x] of degree n: Then (`) has the following properties: (i) (`) is multiplicative (i.e., if ` and `2 are relatively prime integers, then (``2) = (`)(`2)); (ii) if (p) =p; then either p n or f(x) 0 (mod p); (iii) if (p) <p;then (p) n; (iv) if ( ) >(p);then f(x) hasamultiple root modulo p (i.e., there exist an integer a and a polynomial g(x) such that f(x) (x, a) 2 g(x) (mod p)), (v) if ( ) < ;then ( ) pn;

8 8 MICHAEL FILASETA (vi) if p>nand (p r )=p r for some positive integer r; then f(x) 0 (mod p r ): Proof. Property (i) is an immediate consequence of the Chinese Remainder Theorem. A theorem of Lagrange states that either the number of solutions to the congruence f(x) 0 (mod p) isnor f(x) is identically 0 as a polynomial modulo p: This easily implies (ii) and (iii). Each rootmof f(x) modulo p extends to at most p roots m + kp; where k 2f0;;:::;p,g; modulo : Furthermore, m will extend to exactly root of f(x) modulo unless m isamultiple root of f(x) modulo p (cf. [7, pp ]). Thus, (iv) follows. From the above, if (p) <p;then (v) is a consequence of (iii). Also, if p n; then (v) is immediate since then ( ) pn: Now, suppose that p>nand (p) =p: Then ( ) < implies that f(x) =pg(x) where g(x) is a polynomial in Z[x] which is not identically 0 modulo p: By Lagrange's Theorem, we get that g(x) has deg g(x) =nroots modulo p: Each such rootmof g(x) modulo p corresponds to exactly p incongruent roots of f(x) modulo since f(m + kp) pg(m + kp) 0 (mod ) for each k 2f0;;:::;p,g: Thus, (v) follows. Finally, we just note that the proof of (vi) is similar to the proof of (v). Lemma 3. For B e e ;f(x) 2 Z[x]; and z log log B; the number of positive integers m B for which f(m) is not divisible by for each p z is equal to, (p2 ) (B + O(log B)) : In particular, there exists an absolute constant C > 0 such that the number of positive integers m B for which f(m) is squarefree is, (p2 ) (B + C log B) :

9 SQUAREFREE VALUES OF POLNOMIALS 9 The proof of Lemma 3 is omitted. It is a direct application of the sieve of Eratosthenes. The main idea in the paper is to show that for most f(x) 2 S n (N) the upper bound given above isvery close to the actual number of integers m B for which f(m) is squarefree. This is what is to be expected since the product above converges as z tends to innity. Lemma 4. Let x j 2 (0; ) for j 2f;2;:::;rg: Then r (, x j ), r j= j= x j : The proof of Lemma 4 is easily done by induction since by the conditions on x j ; r, j= x j A (, xr ), r j= x j : Lemma 5. As f(x) ranges over all the incongruent polynomials of degree n modulo ; the average value of f ( ) is : We omit the proof of Lemma 5 as it follows in a fairly straight forward manner by using translation considerations to establish that each of0;;:::;,have an equal probability of being attained as a value of f(m) (mod ): Our next goal is to show that for most f(x) 2 S n (N); if, (p2 ) > 0; then it is not too small. We formulate this in the following manner. Lemma 6. Let > 0; and let N be suciently large (depending on n and ). Let z log log N: Then there exist positive numbers n 0 = n 0 () and 0 = 0 (; n) such that the

10 0 MICHAEL FILASETA number of f(x) 2 S n (N) satisfying (i) pn 2 +n 0, f( ) > 0 and (ii), f ( ) < 0 is (2N) n+ : Proof. Consider the f(x) 2 S n (N) for which (i) holds (where n 0 as well as 0 are for the moment unspecied). Thus, ( ) < for each such f(x) and each prime p n 2 + n 0 : Hence, pn 2 +n 0, f ( ) pn 2 +n 0, p2, = pn 2 +n 0, p,2 : Now, consider any f(x) 2 S n (N): We get from Lemma 2 (ii), (iii), and (iv) that for n 2 + n 0 <; either f ( ) n or f(x) has a multiple root modulo p: Letting c(n; z) = n 2 +n 0 < we see that c(n; z) is greater than the product c(n) =, n ;, n ; p>n 2 +n 0 which is easily seen to converge to a positive quantity. Hence, for each f(x) 2 S n (N); n 2 +n 0 <, f ( ) n 2 +n 0 < c(n) n 2 +n 0 <, n n 2 +n 0 <, f ( ), f ( ) where Q indicates that the product is over those primes p for which f(x) has a multiple root modulo p: We now show that this latter product is not small for most polynomials f(x) 2 S n (N): ;

11 SQUAREFREE VALUES OF POLNOMIALS Let k = k() be a positive integer such that Such a kexists since j=0 j=0 2 j k j k 7 < 0 2e : j=k j 7 = 0 k 7 : Dene t(j) =, n 2 +n 0 2 j for j 2f0;;:::;s+g; where s is chosen so that (n 2 + n 0 ) 2s <z(n 2 +n 0 ) 2s+ : Thus, n 2 +n 0 <, f ( ) s j=0 t(j)<pt(j+), (p2 ) A : Let T = T (n; N) be the set of f(x) 2 S n (N) for which there is a j 2f0;;:::;sg such that f(x) hasamultiple root modulo p for 2 j k primes (t(j);t(j + )]: Also, we dene T 0 = T 0 (n; N) to be the set of f(x) 2 S n (N) for which f ( )= for some prime (n 2 + n 0 ;z]: We show that (4) jt [ T 0 j(2n) n+ and then establish that Q, f( ) 0 for the remaining f(x) 2 S n (N): We deal with T 0 rst. By Lemma 2 (vi), each f(x) 2 T 0 is such that f(x) 0 (mod ) for some prime (n 2 + n 0 ;z]: Note that the number of f(x) 2 S n (N) such that f(x) 0 (mod ) for a given prime p is n+ n+ 2N 2N p + O() = + O 2 n (N n ) :

12 2 MICHAEL FILASETA The choice of z log log N easily implies that the total number of such f(x) 2 T 0 is n 2 +n 0 < n+! 2N + O n (N n ) 2N n+ A + On (N n log log N) p>n 2 +n 0 (2N) n+ p>n 0! + O n (N n log log N) : For n 0 chosen suciently large (depending only on ) we get that jt 0 j(=2)(2n) n+ : We now turn to considering T: We begin by dividing up T into subsets T j which are not necessarily disjoint. For each j 2f0;;:::;sg;we dene T j as the set of f(x) 2 S n (N) such that f(x) hasamultiple root modulo p for 2 j k primes (t(j);t(j+ )]: Fix j; and set w =2 j k: Let p ;:::;p w be w distinct primes in (t(j);t(j+ )]: Dene T j (p ;:::;p w )tobe the set of f(x) 2 T j such that f(x) hasamultiple root modulo p j for each j 2f;:::;wg: Note that each f(x) 2 T j belongs to some set T j (p ;:::;p w ): The number of incongruent polynomials modulo a prime p of degree n which have a multiple root modulo p is equal to the number of incongruent polynomials of the form (x,a) 2 g(x) where a 2f0;;:::;p,g and deg g(x) n, 2: Thus, the number of such polynomials is p n : Thus, the Chinese Remainder Theorem easily gives that the number of incongruent polynomials f(x) modulo p p w of degree n such that f(x) has a multiple root modulo p j for each j 2f;:::;wg is p n p n w : By dividing T j (p ;:::;p w )into these p n p n w congruence classes, we get that jt j (p ;:::;p w )j n+ 2N+ + p n p n w p p : w By the denition of s; we have that (n 2 + n 0 ) 2s <z;so that for n 0 suciently large, w 2 s k<z:also, each p j t(s +)= t(s) 2 z 2 so that p p w z 2z : The choice

13 SQUAREFREE VALUES OF POLNOMIALS 3 z log log N gives that p p w for N suciently large (depending on n). Hence, jt j (p ;:::;p w )j = 0 2N+ + p p w + n+ 2N n+,; 2N n+, p p w C A n+ (2N) n+ p p w n+ p n p n w <e (2N)n+ p p w : Since each polynomial in T j belongs to some T j (p ;:::;p w ) described above, we now get that jt j je(2n) n+ t(j)<pt(j+) p A w e(2n) n+ c w ; where we can take c to be any constant > log 2 provided n 0 is suciently large. Here, we have used that py p = log log y + A + o(); for some absolute constant A: We take c =7=0: We are now ready to complete our estimate for jt j: We get that s 2 j k 7 jt j jt j je(2n) n+ < 0 2 (2N)n+ ; j=0 by our choice of k: The above estimates on jt 0 j and jt j easily imply (4). We now consider Q n 2 +n 0 < j=0, f( ) where f(x) 2 S n (N),T,T 0 : By Lemma 2 (v), we get that for each prime p in the range of the product above, ( ) np: Also, for each j 2f0;;:::;sg; there are fewer than 2 j k primes (t(j);t(j + )] for which f(x) has a multiple root modulo p: Hence, t(j)<pt(j+), f ( ) t(j)<pt(j+) 2 j, n, n k : p t(j)

14 4 MICHAEL FILASETA Thus, using Lemma 4, n 2 +n 0 <, f ( ) s j=0, 2, n j k t(j) s j=0 2 j kn t(j) =, s j=0 2 j kn (n 2 + n 0 ) 2j > 2 ; provided n 0 is suciently large. We note that we can choose n 0 so that everything above holds and so that n 0 only depends on (and not on n unless, of course, depends on n). For example, by checking the cases n p n 0 and n> p n 0 separately, the last inequality above is easily seen to hold provided that j=0 2 j k n 2j,(=2) 0 < 2 ; which, since k only depended on ; gives a lower bound on n 0 depending only on : Combining the above, we get that for f(x) 2 S n (N), T, T 0 and f(x) satisfying (i),, (p2 ) c(n) 2 pn 2 +n 0 p,2 Thus, the lemma follows by letting 0 be the right-hand side above. A : Lemma 7. Let > 0; and let N be suciently large (depending on n and ). Let z 2 [2; log log N]: Then (5) f(x)2s n (N) 0, f ( ) A A (2N) n+ + O n, N n+ : Proof. For each p z; consider the n+2 incongruent polynomials modulo of degree n; and let w (p);:::;w r (p); where r = r(p) =n+2 ; denote some ordering of the values of f ( )asf(x) ranges over these polynomials. Let p ;:::;p t represent the t = (z)

15 SQUAREFREE VALUES OF POLNOMIALS 5 primes z; and let f (x);:::;f t (x) denote arbitrary polynomials with integral coecients. Then the Chinese Remainder Theorem implies that the number of f(x) 2 S n (N) such that f(x) f j (x) (mod j ) for every j 2f;:::;tg is n+ 2[N] + +O() 2N = p2 t p2 t where we have used that since z log log N; n+ + O n 2N p2 t n ; (6) t (log log N) 2 log log N <N 0 ; where 0 2 (0; ) and N is suciently large (depending on 0 ). For later purposes, we x 0 = minf=2;g: If w 0 j denotes the number of incongruent roots of f j(x) modulo j ; then the contribution of the f(x) f j (x) (mod j ) (for all j 2f;:::;tg) on the left-hand side of (5) is t j=, w0 j j! 2N p2 t n+ + O n Hence, summing over all f(x) 2 S n (N); we get that =, w (p) + + f(x)2s n (N), f( ), w r(p) 2N p2 t Recalling the denition of w j (p) and Lemma 5, we get that Thus, r(p) j= f(x)2s n (N), w j(p) A = r(p), r(p) = 2N p2 t n! n+ + O n n+2 A : 2N p2 t, :, f ( ) 0 0 (2N) n+ + O (2N) n Recalling our choice of 0 = minf=2;g in (6), we get the desired result. n! p AA 2 : :

16 6 MICHAEL FILASETA 3. The Main Theorems We are now ready to prove Theorems and 2 of the introduction. As mentioned there, we will actually be able to prove slightly stronger results. Theorem 3. Let n 2 Z + [f0g; and let B(N) be a function which increases to innity with N: Then the proportion of polynomials f(x) 2 S n (N) with N f squarefree which satisfy that f(m) is squarefree for some integer m 2 [;B] tends to as N tends to innity. Theorem 4. Let n 2 Z + [f0g; and let B(N) be a function which increases to innity with N: Then the proportion of polynomials f(x) 2 S n (N) which satisfy that f(m)=n f is squarefree for some integer m 2 [;B] tends to as N tends to innity. Proof of Theorem 3. We suppose, as we may, that B(N) =o(n) and that N is suciently large (depending on given below and n). Recall the discussion after Lemma and, in particular, that there is a positive proportion of f(x) 2 S n (N) for which N f is squarefree. Alternatively, one may deduce that N f is squarefree for a positive proportion of the f(x) 2 S n (N) as a consequence of Theorem in [3], which stated that for a positive proportion of the f(x) 2 S n (N); there is an integer m for which f(m) is prime. Let > 0: To obtain Theorem 3, we need only prove that if N is suciently large, there are (2N) n+ polynomials f(x) 2 S n (N) with N f squarefree and such that f(m) is nonsquarefree for all integers m 2 [;B]: In fact, for later purposes, we prove something stronger. Using the notation of Lemma 6 with n 0 such that (i) gcd N f ; Q pn 2+n0 = n 0 (=2); we prove that the set T of f(x) 2 S n (N) is squarefree and (ii) f(m) is nonsquarefree for every integer m 2 [;B] satises jt j(2n) n+ (provided N is suciently large). Assume that

17 SQUAREFREE VALUES OF POLNOMIALS 7 jt j >(2N) n+ : Let z = log log B: For each f(x) 2 S n (N); we denote W (f(x)) as the number of integers m 2 [;B] such that f(m) is squarefree. Then Lemma 3 implies that W (f(x)) =, (p2 ) B + E(f(x)); where Thus, using Lemma 7, we get that (7) f(x)2s n (N) E(f(x)) C W (f(x)) = f(x)2s n (N) =, (p2 ) log B:, (p2 ), (2N) n+ B + E ; B + E(f(x)) A with E = f(x)2s n (N) E(f(x)) + O n N n+ 2 B C 2 N n+ log B + N n+ 2 B ; where C 2 = C 2 (n) and we note that E may be negative (so that, in particular, we claim no bound on je j at this point). Note that, p > 2 p, p = 6 2 : 2 Recalling that z = log log B(N); we get that since N and, hence, B(N) are suciently large, 6 < 2, < ; where 0 > 0 is arbitrarily small and possibly depends on and n: Thus, f(x)2s n (N) W (f(x)) = 6 2 (2N)n+ B + E 2 ;

18 8 MICHAEL FILASETA where E 2 0 (2N) n+ B: On the other hand, Lemma gives us that f(x)2s n (N) W (f(x)) = 6 2 (2N)n+ B + E 3 ; where je 3 j 0 (2N) n+ B: Thus, in fact, je 2 j = je 3 j 0 (2N) n+ B: Recalling how E 2 was obtained, we now get that je j2 0 (2N) n+ B: The importance of this last inequality is that, unlike with the previous inequality one ; wenow are supplied with a lower bound on E : More specically, E,2 0 (2N) n+ B: Recalling the denitions of T and E(f(x)); we get that E(f(x)) =,, f ( ) B for all f(x) 2 T: Thus, f(x)2t E (f(x)) =, f(x)2t, f( ) B: The denition of T easily implies that for each prime p n 2 + n 0 ; f ( )< for all f(x) 2 T: Thus, by Lemma 6, there exists an 00 such that (8), f ( ) 00

19 SQUAREFREE VALUES OF POLNOMIALS 9 for all but at most (=2)(2N) n+ polynomials f(x) 2 T: Since by assumption jt j > (2N) n+ ; there are (=2)(2N) n+ polynomials f(x) 2 T for which (8) holds. Hence, On the other hand, f(x)2s n (N) E(f(x))>0 E (f(x)) C f(x)2t E (f(x)), 2 00 (2N) n+ B: f(x)2s n (N) E(f(x))>0, f ( ) log B C js n (N)j log B C (2N) n+ log B + O n ((2N) n log B) : Thus, recalling the denition of E ; E,, 2 00 (2N) n+ B + O (2N) n+ log B + O n N n+ 2 B : We are still free to choose 0 > 0: We take 0 =( 00 )=5: Then the above contradicts that je j2 0 (2N) n+ B = (2N) n+ B; completing the proof. Proof of Theorem 4. For n = 0; the theorem is clear, so we only consider n : Let 2 (0; ); and let N be suciently large (depending on n and ). Assume that there exists (2N) n+ polynomials f(x) 2 S n (N) such that f(m)=n f is nonsquarefree for every m 2 [;B]: Let T denote the set of such polynomials. By the proof of Theorem 3 and the notation of Lemma 6, the number n 0 = n 0 (=6) is such that jt 2 j(=3)(2n) n+ where T 2 denotes the set of f(x) 2 S n (N) for which (i) gcd N f ; Q pn 2+n0 is squarefree and (ii) f(m) is nonsquarefree for each integer m 2 [;B]: Since increasing the size of n 0 will

20 20 MICHAEL FILASETA only decrease the number of f(x) for which (i) and (ii) hold, we may assume that n 0 7: We do this so that later we may use that jn 0 j 2 < 4 25 : Let T 3 = T, T 2 so that T 3 consists of (2=3)(2N) n+ polynomials f(x) 2 T for which N f is divisible by for some p n 2 + n 0 : Dene 4(n2 2(n + n 0 ) 2 + n 0 ) M = M(n; ) = and B 0 = B 0 (N) = N M B, (2M) n : Using the notation of Lemma 6, dene n = n () =n 0 The proof of Theorem 3 implies that there are 4(2M) n2 +n+2 : 2(2M) n2 +n+2 js n ((2M) n N)j polynomials g(x) 2 S n ((2M) n N) for which (i 0 ) gcd N g ; Q pn 2+n is squarefree and (ii 0 ) g(m) is nonsquarefree for each integer m 2 [;B 0 ((2M) n N)]: We will obtain a contradiction by showing that there are more than (=(2(2M) n2 +n+2 )) js n ((2M) n N)j such g(x) (with even gcd Q N g ; pn 2 +n p = ). We begin by restricting our attention to p n 2 + n 0 : For each such p; let k = k(p) = k(p; n; ) be the minimal positive integer such that p k+ 4(n2 + n 0 ) :

21 SQUAREFREE VALUES OF POLNOMIALS 2 Note that 2 (0; ) implies that the right-hand side above is>n 2 +n 0 so that p k < 4(n 2 + n 0 )=: Let T 4 be the set of polynomials f(x) 2 T 3 such that p k+ divides N f for at least one prime p n 2 + n 0 : The constant term of each such f(x); being f(0); must be divisible by p k+ : Thus, the number of f(x) 2 T 3 for which p k+ divides N f for a given prime p n 2 + n 0 is 2N + (2N +) n + p k+ 4(n 2 + n 0 ) (2N +)n+ +(2N+) n 3(n 2 + n 0 ) (2N)n+ : Hence, jt 4 j, n 2 +n 0 3(n 2 + n 0 ) (2N)n+ 3 (2N)n+ : Dene T 5 = T 3, T 4 : Thus, jt 5 j(=3)(2n) n+ : For f(x) 2 T 5 ; dene M f =, r= pn 2 +n 0 p r jn f p and P f = M f pjm f p: Q Note that N f = M f Q f where gcd(q f ; pn 2 +n 0 p) = and that P f Mf 2 : By the denition of T 5 ; for each prime p n 2 + n 0 and each f(x) 2 T 5 ; we have that p k+ does not divide M f : This easily implies that eachofm f and P f is M(n; ) for every f(x) 2 T 5 : We now dene a function : T 5! S n ((2M) n N) as follows. For each f(x) 2 T 5 and each prime p n 2 + n 0 ; dene r = r(p; f(x)) to be the nonnegative integer satisfying that p r divides M f and p r+ does not divide M f : In particular, p r+ does not divide N f so that there is an integer a = a(p; f(x)) 2 [;p r+ ] such that f(a) 6 0 (mod p r+ ): Necessarily, f(a) 0 (mod p r ): By the Chinese Remainder Theorem, there is a minimal positive integer b = b(f(x)) such that f(b) is divisible by M f and, for each prime p n 2 + n 0 ;f(b)

22 22 MICHAEL FILASETA is not divisible by pm f : Furthermore, since f(x) 2 T 5 ; Dene b pn 2 +n 0 p r(p;f(x))+ pn 2 +n 0 p k(p)+ g(x) =f(p f x+b)=m f : pn 2 +n 0 p k(p) A 2 M(n; ): Each coecient off(p f x+b) is divisible by M f ; except possibly the constant term f(b): But f(b) 0 (mod M f ); and thus g(x) 2 Z[x]: Furthermore, it is easily veried that each coecient ofg(x) has absolute value N(2M) n : We dene (f(x)) = g(x): Note that M f and P f are uniquely determined by one another; in other words, given M f ; one can determine P f ; and given P f ; one can determine M f : Since there exist M(n; ) possible values for P f and M(n; ) possible values for b; it is easy to see that for each g(x) in the image of ; there are at most M 2 possible f(x) 2 T 5 such that (f(x)) = g(x): In particular, since N is suciently large, j(t 5 )j M 2jT 5j = 3M 2(2N)n+ 3(2 n2 +n )(M n2 +n+2 ) (2(2M)n N) n+ (2M) n2 +n+2 js n ((2M) n N)j : On the other hand, one can check that the denitions of b and g(x) above imply that for g(x) 2 (T 5 ); 0 gcd@ Ng ; pn 2 +n 0 p A =: Recall that by assumption, each f(x) 2 T 5 T is such that f(m)=n f is nonsquarefree for each integer m 2 [;B]: Note that B 0 ((2M) n N) = (B(N)=M), : Now, if m 2 [; (B(N)=M), ] and b is as in the denition of ; then P f m + b is a positive integer

23 SQUAREFREE VALUES OF POLNOMIALS 23 B(N): Also, the denition of M f implies that M f divides N f : We now get that if f(x) 2 T 5 and g(x) =(f(x)); then g(m) =f(p f m+b)=m f is nonsquarefree for each integer m 2 [;B 0 ((2M) n N)]: Thus far, we have shown that there are (2M) n2 +n+2 js n ((2M) n N)j polynomials g(x) 2 S n ((2M) n N) such that gcd(n g ; Q pn 2 +n 0 p) = and (ii 0 ) holds. Let T 0 denote the set of all such g(x): Let T 0 2 denote the set of all g(x) 2 T 0 which also satisfy that gcd(n g ; Q pn 2 +n p)=:it now suces to prove that jt 0 2 j > 2(2M) n2 +n+2 js n ((2M) n N)j : For (n 2 + n 0 ;n 2 +n ]; dene k 0 = k 0 (p) =k 0 (p; n; ) as the minimal positive integer such that p k0 + 4(n2 + n )(2M) n2 + n +2 : Then following the argument which led to an estimate of jt 5 j; we get that there are 2 3(2M) n2 +n+2 js n ((2M) n N)j polynomials g(x) 2 T 0 such that if (n 2 + n 0 ;n 2 +n ] and p r divides N g ; then r k 0 (p): Let T 0 denote the set of all such 3 g(x): Note that T 0 2 T 0 3 : In fact, our goal now istoshow that most of the polynomials in T 0 3 are in T 0 2 : For each g(x) 2 T 0 3 ; let M 0 g = r= n 2 +n 0 <pn 2 +n pjn g p = r= pn 2 +n pjn g p :

24 24 MICHAEL FILASETA Note that with n and xed, so are M and k 0 (p) for each (n 2 + n 0 ;n 2 +n ]: Thus, M 0 g takes on a nite number of distinct values. Let M 0 be one such value of M 0 g : By the denition of n and the proof of Theorem 3, we get that there are (2M) 2(2M) n2 +n+2 n S N n M 0 (2M) n2 +n+2 (M 0 ) js n+ n ((2M) n N)j polynomials h(x) 2 S n ((2M) n N=M 0 ) such that gcd N h ; Q pn 2 +n p = and h(m) is nonsquarefree for each positive integer m B 0 ((2M) n N=M 0 ) B 0 ((2M) n N): We note that we want the above to hold for every choice of M 0 ; and we can do this since N is suciently large and there are only nitely many values of M 0 : Since every prime factor of M 0 is >n 2 +n 0 >n;we get by Lemma 2 (vi) that each g(x) with M 0 g = M 0 satises g(x) 0 (mod M 0 ): But this means that g(x) =M 0 h(x) for some h(x) 2 S n ((2M) n N=M 0 ): The denition of M 0 = M 0 g implies that every such h(x) satises gcd N h ; Q pn 2 +n p : Also, using that gcd Q P f ; n 2 +n 0 <pn 2 +n p = ;one can show from the denition of M f and M 0 g that M f M 0 g divides N f where (f(x)) = g(x): One gets that for h(x) as above, h(m) =f(p f m+b)=(m f M 0 g ) is nonsquarefree for each positive integer m B 0 ((2M) n N=M 0 ): We now get that jt 0 3, T2j 0 (2M) n2 +n+2 (M 0 ) js n+ n ((2M) n N)j = (2M) n2 +n+2 (M 0 ),n, js n ((2M) n N)j ; where P denotes that the sum is over those values of M 0 which are strictly greater than : Since each such M 0 is divisible by some prime p>n 2 +n 0 ;we get that each such M 0 is n 2 + n 0 n 0 : Thus, since n ; (M 0 ),n, jn 0 j 2 ; =

25 SQUAREFREE VALUES OF POLNOMIALS 25 which, by our choice of n 0 7; is < 4=25: Hence, jt 0 3, T 0 2 j 4 25(2M) n2 +n+2 js n ((2M) n N)j ; so that jt 0 2jjT 0 3j,jT 0 3,T 0 2j 38 75(2M) n2 +n+2 js n ((2M) n N)j ; which completes the proof. Before concluding the paper, we note that Theorem 4 and, hence, Theorem 2 can be improved slightly. For f(x) 2 Z[x]; write N f = U f V f ; where V f is the largest squarefree factor of N f : Then one may replace the role of f(m)=n f in the statement of Theorem 4 with f(m)=u f : The proof is essentially the same with the following minor changes. One denes (f(x)) = g(x) where now g(x) =f(p f x+b)=gcd(m f ;U f ): Then g(x) 2 (T 5 ) Q implies that gcd(n g ; pn 2 +n 0 ) is squarefree. One considers, instead of T 0 2 ; the set T 00 2 of g(x) 2 S n ((2M) n N) such that (i 0 ) and (ii 0 ) hold. Since T 0 2 T 00 2 ; the lower bound for jt 0 2j obtained in the proof of Theorem 4 is a lower bound for jt 00 2 j; and the desired improvement follows. References. L. E. Dickson, History of the Theory of Numbers, Vol. I, Chelsea, New ork, P. Erd}os, Arithmetical properties of polynomials, J. London Math. Soc. 28 (953), 46{ M. Filaseta, Prime values of irreducible polynomials, Acta Arith. 50 (988), 33{ P.. Gallagher, The large sieve and probabilistic Galois theory, Proc. Sympos. Pure Math., Vol. 24, Amer. Math. Soc. (973), 9{0.

26 26 MICHAEL FILASETA 5. C. Hooley, On the power free values of polynomials, Mathematika 4 (967), 2{ M. Huxley and M. Nair, Power free values of polynomials, III, Proc. London Math. Soc. 4 (980), 66{ W. J. LeVeque, Fundamentals of Number Theory, Addison-Wesley, Reading, Massachusetts, T. Nagel, Zur Arithmetik der Polynome, Abhandl. Math. Sem. Hamburg (922), 79{ M. Nair, Power free values of polynomials, Mathematika 23 (976), 59{ M. Nair, Power free values of polynomials, II, Proc. London Math. Soc. 38 (979), 353{368.. G. Polya and G. Szeg}o, Problems and Theorems in Analysis II, Springer-Verlag, New ork, B. L. van der Waerden, Die Seltenheit der Reduziblen Gleichungen und der Gleichungen mit Aekt, Monatsh. Math. 43 (936), 33{ B. L. van der Waerden, Algebra I, Springer-Verlag, Berlin, B. L. van der Waerden, Algebra, Vol. I, 7th edition, translated by F. Blum and J. R. Schulenberger, Frederick Ungar Publ. Co., New ork, 970.