that in Conecture 1 we havefèxè=èx n+1, 1è=èx, 1è so that f 0 èxè = nxn+1, èn +1èx n +1 èx,1è 2 : Higher derivatives of fèxè as in Conecture 2 take a

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1 CLASSES OF POLNOMIALS HAVING ONL ONE NONíCCLOTOMIC IRREDUCIBLE FACTOR A. Borisov, M. Filaseta*, T.. Lam**, and O. Trifonov 1. Introduction In 1986, during the problem session at the West Coast Number Theory Conference, the second author stated the following: Conecture 1. Let n be an integer 2, and let fèxè =1+x+x 2 +æææ+x n. Then f 0 èxè is irreducible over the rationals. He noted then that the conecture is true if n = p, 1 2orif n= p r where p is a prime and r a positive integer. Calculations showed the conecture also held for n 100. Recently, in a study of more general polynomials, the ærst author ë2ë obtained further irreducibility results for f èxè; in particular, he established irreducibility in the case that n + 1 is a squarefree number 3 and in the case that n =2p,1 where p is prime. The third author independently observed that f èkè èxè is Eisenstein if n = p,1 for every integer k 2 ë1;n,1ë and, based on some further computations, conectured: Conecture 2. Let n and k be integers with n 2 and 1 k n, 1, and let fèxè = 1+x+x 2 +æææ+x n. Then f èkè èxè is irreducible over the rationals. In 1991, again during the problem session at the West Coast Number Theory Conference, Jeæ Lagarias mentioned a class of polynomials associated with some work of Eugene Gutkin ë5ë concerning billiards. Eugene Gutkin was interested in showing that the polynomials had no roots in common other than from obvious cyclotomic factors. As a consequence, Jeæ Lagarias made the following conecture attributed to Eugene Gutkin: Conecture 3. Let n be an integer 4, and let pèxè =èn,1èèx n+1, 1è, èn + 1èèx n, xè: Then pèxè is èx,1è 3 times an irreducible polynomial if n is even and pèxè is èx,1è 3 èx+1è times an irreducible polynomial if n is odd. In this paper, we explain some approaches to these three conectures. The connection between Conectures 3 and the two previous conectures is more transparent if one observes *The second author was supported by NSF Grant DMS and NSA Grant MDA **The third author was supported by NSA Grant MDA Typeset by AMS-TE 1

2 that in Conecture 1 we havefèxè=èx n+1, 1è=èx, 1è so that f 0 èxè = nxn+1, èn +1èx n +1 èx,1è 2 : Higher derivatives of fèxè as in Conecture 2 take a similar form. We are able to show that Conectures 1 and 3 hold for almost all n and that Conecture 2 holds for most choices of n and k. More precisely, we establish each of the following theorems. Theorem 1. Let "é0. For all but Oèt è1=3è+" è positive integers n t, the derivative of the polynomial fèxè =1+x+x 2 +æææ+x n is irreducible. Theorem 2. Fix a positive integer k. For all but oètè positive integers n t, the kth derivative of the polynomial fèxè =1+x+x 2 +æææ+x n is irreducible. Theorem 3. Fix a positive integer m. There is an N such that if n is a positive integer N and fèxè =1+x+x 2 +æææ+x n, then the polynomial f èn,mè èxè is irreducible. Theorem 4. Let "é0.for all but Oèt è4=5è+" è positive integers n t, the polynomial pèxè =èn,1èèx n+1, 1è, èn + 1èèx n, xè; is such that pèxè is èx, 1è 3 times an irreducible polynomial if n is even and pèxè is èx, 1è 3 èx +1ètimes an irreducible polynomial if n is odd. In Theorem 2, our arguments give Oèt log log t= log tè in place of oètè. We would be interested in an upper bound of the type Oèt è for some 2 è0; 1è that is independent of k. Our arguments suggest that such aexists, but we have been unable to establish this. The rest of the paper is organized as follows. In the next section, we give a proof of Theorem 3. The proofs of the remaining theorems above that we will present here rely on the location of the p-adic zeroes of the polynomials. Section 3 establishes some preliminary results based on these zeroes. As noted at the end of that section, these preliminary results can be extended to handle certain other classes of polynomials where almost all polynomials in the class have one non-cyclotomic irreducible factor. In the remaining sections of the paper, we give proofs of each of the remaining theorems based on these preliminary results. Acknowledgment: The authors express their gratitude to Andrze Schinzel who encouraged the ærst three authors to correspond with one another in matters related to this research. They also express their gratitude to Charles Nicol for early remarks concerning this work. 2. A Proof of Theorem 3 and Further Remarks Consider fèxè as in Theorem 3. If m = 1, then f èn,mè èxè is linear and, hence, irreducible for every integer n 1. If m = 2, then f èn,mè èxè is quadratic and it is a simple matter to show that this quadratic has imaginary roots. Thus, in this case, f èn,mè èxè is irreducible for every integer n 2. It is of some interest to continue by considering the cubics one obtains in Theorem 3 by setting m = 3. The proof we will present for Theorem 2

3 3 is eæective so that in theory it is possible to determine for a æxed m what polynomials of the form f èn,mè èxè are reducible. We will demonstrate this at the end of the section by showing that for m = 3 the cubic f èn,mè èxè is irreducible for every integer n 4. We turn now to the proof of Theorem 3. Observe that f èn,mè èxè = n =n,m è,1è æææè,n+m+1èx,n+m = We set k = n, m and consider the polynomial F k èxè = xm f èkè è1=xè = k! m k + m, = m, =0 m =0 m =0 èn,èèn,,1è æææèm,+1èx m, : èk + m, èèk + m,, 1è æææèm,+1è x k! m x k + = x m, : =0 It suæces now to show that if k is suæciently large, then the polynomial F k èxè is irreducible. For a prime p and an integer a, we deæne P èaè = p èaè=ewhere p e a. We deæne the n Newton polygon of a polynomial F èxè = =0 a x as the lower convex hull of the points è; èa èè ècf. ë3ë, ë6ë, ë15ëè. We consider the Newton polygon of a polynomial F èxè. Let the lattice points along the edges be èx 0 ;y 0 è;èx 1 ;y 1 è;:::;èx s ;y s è with 0 = x 0 éx 1 éæææ é x s = deg F èxè. Then the degree of any irreducible factor of F èxè èover Zëxëè must be some sum of the diæerences x 1,x 0 ;x 2,x 1 ;:::;x s,x s,1. In other words, if r is the degree of an irreducible factor P of F èxè, then there are integers 1 ;:::; t with 1 1 é 2 éæææ é t s t such that r = i=1 èx i, x i,1è. The next result is due to Sylvester ë13ë and was ærst used to obtain irreducibility results by I.Schur ë12ë. It is a generalization of Bertrand's postulate that for every integer m 1, there is a prime in the interval èm; 2më ètake k = mè. Lemma 1. Let m and k be positive integers with m k. Then there is a prime p k +1 which divides one of the numbers m +1;m+2;:::;m+k. We will also use an eæective version of Thue's theorem èit follows with a little modiæcation from Theorem 4.1 in ë1ë; also see ë11ëè. Lemma 2. Let a, b, and d be integers with d 6= 0. Let q be a positive integer 3. Then there are ænitely many integer pairs èx; yè for which ax q, by q = d. Furthermore, these pairs can eæectively be determined. The following is a combinatorial lemma and follows directly from è5.26è of ë4ë. Lemma 3. Let m and k be positive integers. Let F k èxè be as in the theorem. Then m k + m +1 F k èx +1è= x m, : =0 3

4 Fix a positive integer m. By the comments at the beginning of this section, we may suppose that m 3 èand do soè. If F k èxè is reducible, then it has a factor with degree in the interval ë1;m=2ë. It suæces therefore to show that for each ` 2 ë1;m=2ë, there are only ænitely many k for which F k èxè has a factor of degree `. Fix an integer ` 2 ë1;m=2ë, and suppose F k èxè has a factor gèxè inzëxë of degree `. Deæne q = m in the case that ` =1. Otherwise, deæne q as the largest prime divisor of mèm,1è æææèm,`+1è. Since m,` `, we deduce from Lemma 1 that q ` +1. Observe that our choice of q guarantees that q 3. Let t 2f0;1;:::;`,1g such that q divides m, t. Suppose now that pémis a prime dividing k + t + 1 èif no such p exists, we can skip this partè. Let r be the positive integer such that p r èk + t + 1è. We claim that q divides r. For t +1m,we deduce from k + èk + èèk +, 1è æææèk+1è =! that p, which isém, divides the numerator of this last expression but not its denominator. In fact, p r must exactly divide the numerator. On the, other æ hand, one easily deduces from k+t pémétand pèk + t + 1è that p does not divide t. Hence, the Newton polygon of F k èxè with respect to the prime p has as its left-most edge the line segment with endpoints è0;rè and èm, t; 0è. Recall that ` t +1. Since F k èxè has the factor gèxè of degree `, it follows that there must be two lattice points, say èa; bè and èc; dè with céa, on the left-most edge of the Newton polygon of F k èxè with c, a `. On the other hand, by considering the slope of the left-most edge, we see that d, b c, a = r m, t =è èm, tèd, b =èc,aèr: The deænition of q implies c, a `éq. Thus, q and c, a are relatively prime èin the case that q is a prime, this is clear; in the case that ` = 1 where we have deæned q = m, this follows since c, a ` = 1 implies c, a = 1è. On the other hand, qèm, tè, so the above equation gives that q divides r as claimed. We now make use of Lemma 3. We consider any prime pémdividing k + m, t +1, and let r be the positive integer such that p r exactly divides k + m, t + 1. Observe that since t `, 1 èm=2è, 1, we havek+m,t+16=k+t+1, sowe are in a diæerent situation than the above. We use an argument similar to the above to show that q divides r in this situation as well. Here, we have k+m+1 èk+m+ 1èèk + mè æææèk+m,+2è = :! The conditions pémand p r exactly divides, k + m æ, t + 1 with r 1 imply that, for every with t +1 m, p r k+m+1 exactly divides. Also, p does not divide k+m+1 æ t. We deduce that the Newton polygon of F k èx + 1è with respect to p contains the line segment with endpoints è0;rè and èm, t; 0è. The same argument asabove gives as before that since F k èxè èand hence F k èx + 1èè has a factor of degree `, q must divide r. 4

5 Let p 1 ;:::;p s denote the distinct primes m. Let T = fp e 1 1 pe 2 2 æææpe s s :0e q,1 for each g: By the above, k + m, t +1=au q and k + t +1=bv q for some integers a and b in T and some integers u and v. We deduce that èu; vè is a solution to the diophantine equation ax q, by q = m, 2t. Note that m, 2t é0 and that q and t only depend on m and `. For each choice of a and b in T,we deduce from Lemma 2 that there are only ænitely many k with k + m, t +1=au q and k + t +1=bv q as above. Since T is a ænite set, there are only ænitely many F k èxè with a factor in Zëxë of degree `. This completes the proof of Theorem 3. We end this section by establishing that the cubics obtained by taking derivatives of fèxè as in Theorem 3 are all irreducible. Theorem 5. Let fèxè =1+x+x 2 +æææ+x n. For every integer n 4, the polynomial f èn,3è èxè is irreducible. As in our arguments above èwith m = 3è, we consider k+1 k+2 F k èxè =x 3 + x k+3 x+ 3 Wewant to show that F k èxè is irreducible for all k 1. In the argument for Theorem 3, we havem=q=3,`= 1, and t =0. We deduce that k +4=au 3 and k +1=bv 3 for some positive integers a, b, u, and v with a and b divisors of 36. Such k are determined from the diophantine equation au 3, bv 3 =3. A simple restriction on a and b that follows directly from au 3, bv 3 = 3 is that either both a and b are divisible by 3 or neither is. Also, since one of k +4 = au 3 and k +1 = bv 3 is odd, at least one of a and b is odd. We show further that only the cases where a and b are both not divisible by 9 are of interest to us èin other words, we need only consider a and b divisors of 12è. If 3 3e+2 exactly divides k + 1 for some non-negative integer e, then the Newton polygon of F k èxè with respect to 3 consists of a line segment with endpoints è0; 3e + 1è and è3; 0è. This segment contains no lattice points other than the endpoints. Hence, F k èxè is irreducible. An analogous argument works when 3 3e+2 exactly divides k +4 by considering F k èx + 1è rather than F k èxè. It follows then that a and b must be divisors of 12. Our next two lemmas appear in ë7ë, Theorem 5 on page 220 and Theorem 6 on page 225. Lemma 4. If dé1, the equation u 3 + dv 3 =1has at most one integer solution with uv 6= 0. If such a solution exists, then necessarily u + v 3p d is the fundamental unit in the ring Zë 3p dë. Lemma 5. The complete set of solutions to the diophantine equation 2u 3, v 3 =3is given by èu; vè =è1;,1è and èu; vè =è4;5è, and the complete set of solutions to the diophantinve equation 4u 3, v 3 =3is given by èu; vè =è1;1è. Lemma 4 will be used to examine solutions to u 3, 2v 3 =1 and u 3, 4v 3 =1: : 5

6 We will want uv 6= 0. Integer solutions to these correspond to integer solutions to u 3 + 2è,vè 3 = 1 and u 3 +4è,vè 3 =1. Lemma 4 asserts that there is at most one solution to u 3 +2è,vè 3 = 1 with uv 6= 0. Apparently, this is given by èu; vè =è,1;,1è. Now, we apply Lemma 4 to the second equation. Observe that p 4+2 3p 4 2 is a unit in Zë 3p 4ë. If the fundamental unit in Zë 3p 4ë were of the form u+v 3p 4 with u and v integers satisfying u 3 +4v 3 = 1, then there would be some positive integer t for which, u + v 3 p 4 æ t =5+3 3 p 4+2 3p 4 2 : Expanding the right side and expressing it in terms of the basis f1; 3p 4; 3p 4 2 g, it is easy to see that v will be a divisor of the coeæcient of 3p p 4 and a divisor of the coeæcient of We deduce that v divides both 3 and 2 and, hence, is æ1. Since u 3 +4v 3 =1,we easily obtain a contradiction. Therefore, the fundamental in Zë 3p 4ë cannot be of the form stated in Lemma 4, and we deduce that there are no solutions to u 3 +4è,vè 3 = 1 with uv 6= 0. Lemma 5 is only part of Theorem 6 in ë7, p. 225ë. The ærst sentence of Lemma 5 is stated explicitly. The second sentence follows by considering 4u 3 +è,vè 3 = 3 in Theorem 6. Theorem 6 in ë7ë implies that there is at most one solution to this diophantine equation. Apparently, itisgiven by èu; vè = è1; 1è. Given the restrictions on a and b above, we show next that the only solutions to au 3, bv 3 = 3 with u and v positive arise from one of the following: èiè èa; bè =è4;1è and èu; vè =è1;1è. èiiè èa; bè =è6;3è and èu; vè =è1;1è. èiiiè èa; bè =è2;1è and èu; vè =è4;5è. To simplify matters, we restrict ourselves to a b. If aéband au 3, bv 3 = 3 with u and v positive, then also bè,vè 3, aè,uè 3 =3. Thus, we can make the restriction a b provided we also consider solutions with both u and v negative. Given our restrictions on a and b, we get that there are only six cases to consider. Case 1. èa; bè =è1;1è. Here, we want solutions to u 3, v 3 =3. Since we are considering u and v to have the same sign, we haveuv é 0. Then the factor u 2 + uv + v 2 of u 3, v 3 is 3 with equality if and only if uv =1. We easily deduce that u 3, v 3 = 3 has no solutions in integers u and v with uv é 0. Case 2. èa; bè =è2;1è. Here, we are interested in solutions of 2u 3, v 3 = 3 with uv é 0. We apply Lemma 5 above to obtain the unique solution èu; vè = è4; 5è. Case 3. èa; bè =è4;1è. From Lemma 5, the only solution to 4u 3, v 3 = 3 is èu; vè =è1;1è. Case 4. èa; bè =è3;3è. If 3u 3, 3v 3 = 3, then u 3, v 3 = 1. Since we require uv é 0, the factor u 2 + uv + v 2 of u 3, v 3 is 3 so that u 3, v 3 = 1 has no solutions in integers u and v with uv é 0. 6

7 Case 5. èa; bè =è6;3è. If 6u 3, 3v 3 = 3, then 2u 3, v 3 =1. As noted above, Lemma 4 implies u 3, 2v 3 =1 has only the solution èu; vè =è,1;,1è. Interchanging the roles of u and v and changing the signs of u and v, we deduce that 2u 3, v 3 = 1 has only the solution èu; vè =è1;1è èassuming uv é 0è. Case 6. èa; bè = è12; 3è. If 12u 3, 3v 3 = 3, then 4u 3, v 3 =1. From the comments after Lemma 4 above, it follows that there are no integer solutions to 4u 3, v 3 = 1 with uv 6= 0. We deduce from èiè, èiiè, and èiiiè that we only need consider the three possibilities k+1 = 1, k+1 = 3, and k+1 = 125. One checks the latter two directly to see that F k èxè is irreducible. We are not allowing k = 0 so the ærst possibility does not really arise. This completes the proof of Theorem Preliminary Results For p a prime, we let p represent the p-adic norm on Q and let Q p denote the completion of the rationals with respect to this norm. We denote by p èaè the value of, log a p = log p where we interpret p è0è as 1. Both p and p extend in a natural way to the algebraic closure of Q p.wedrop the subscripts when using p when it is clear what the prime P p under n consideration is. We make use of the Newton polygon of a polynomial fèxè = =0 a x with coeæcients in some extension of Q p ; as in the previous section, this Newton polygon is deæned as the lower convex hull of the points è; èa èè. Throughout the remainder of this paper, we work in an algebraic closure of Q p unless noted otherwise or unless it is clear from the context that we are working in C. As references, we mention the books of Gouvea ë3ë, Koblitz ë6ë, and Weis ë15ë. A lemma we will make use of throughout the remainder of the paper is the following. Lemma 6. Let be an m th p-adic root of unity and 0 and m 0 th p-adic root of unity. Suppose p - mm 0. Then è, 0 è=0. The lemma follows from Lemma 2.12 of ë14ë. It is also easily established by observing that è 0 è,1, 1isarootof P mm 0,1 =0 èx +1è, a monic polynomial with constant term relatively prime to p. We will make particular use of the lemma with 0 = æ1. The next result, an essential ingredient to our arguments for Theorems 1 and 2, is based on the work of the ærst author in ë2ë. Proposition 1. Let wèxè = P n+1 =0 a x 2 Zëxë with a n+1 6=0, and let m and r be integers with mé0,r0,n+1=m+r. Let p be a prime such that pm, pér, and p - a n+1. Write m = p`m 0 where p èm 0 è=0. Suppose that wèxè a n+1 èx m, 1èx r èmod p`è and that, for each 6= 1such that m0 =1,wehave p èwèèè = `. Let wèxè =gèxèhèxèbe a factorization of wèxè in Zëxë. Let A = æ, 1 ; B = æ, 1 ; C = è1, æè and D = è1, æè; æ æ gèæè=0 hèæè=0 gèæè=0 hèæè=0 7

8 where the sums are over the distinct roots of gèxè and hèxè and where we consider A and B only in the case that a 0 6=0. Then A, B, C, and D are rational numbers satisfying: èiè if r =0, then each ofèaè,èbè,ècè, and èdè is positive, èiiè if r é0, p`a 0, and gcdè`; rè =1, then either èaè é 0, ècè é 0, phè0è, and D 6= 0or èbè é 0, èdè é 0, pgè0è, and C 6= 0. Comment: We have deæned A, B, C, and D as sums over distinct roots of gèxè orhèxè. The conclusions of the proposition, however, hold even if any of these sums is taken over the roots counted to their multiplicities. The same proof below, word for word, can be used to establish this. Proof. First, we observe that each ofa,b,c, and D is rational; this follows as each is a symmetric function of the roots of either gèxè or hèxè both of which contain rational coeæcients. Note that the rational values of A, B, C, and D depend only on the coeæcients of gèxè and hèxè. It follows that these values are independent of whether we view the roots æ of gèxè and the roots æ of hèxè as complex numbers or as lying in an algebraic closure of Q p. We begin by determining information about the p-adic location of the zeroes of wèxè. Let be an m 0 th root of unity diæerent from 1. We determine next the Newton polygon of fèxè =wèx+è. Write fèxè = P n+1 =0 b x and observe that b 0 = fè0è = wèè. We deduce that the left-most endpoint of the Newton polygon of f èxè is è0;èwèèèè = è0;`è. Also, the conditions in the lemma imply that there is a vèxè 2 Zëxë for which wèxè = aèx m+r,x r è+p`vèxè where a = a n+1. Note that p - a. It follows that, fèxè =a èx+è m+r,èx+è ræ +p`vèx+è m+r m + r r r, =a = a =0 m+r m + r =0 m+r,, r, x + p`vèx + è x r, + p`vèx + è, r where we interpret æ as zero if é r. We use that èx + yè minfèxè;èyèg with equality when èxè 6= èyè. We deduce èb è min `; m + r r, æ, r æ èmod p`è, and we obtain èb è `. For ér,wehave, r and equality, holds if the minimum is not `. For 1 r, the conditions p`m and pér m+r imply that æ =0. One easily checks that and m + r m + r p u = `, u for 1 u ` `, u if p u ép u+1 and 1 u `, 1: ; 8

9 Furthermore, this last inequality holds also for u = 0 provided is restricted to réép. We deduce that èb p uè=`,ufor 1 u ` and that èb è `, u for p u ép u+1 and 0 u `, 1. Also, èb è 0 for p` én+1. It follows that the Newton polygon of fèxè has left-most edges oining the points è0;`è and èp u ;`,uè for 1 u `. èit is easy to see that the right-most edge is the segment with endpoints èp`; 0è and èn +1;0è, but we will not need this fact.è We use the classical connection between Newton polygons of a polynomial and the p- adic roots of the polynomial. We deduce that fèxè has exactly p roots æ with èæè =1=p and, for each u 2f1;;2;:::;`,1g, exactly p u+1, p u roots æ with èæè =1=èp u+1, p u è. We view these roots as forming ` sets, each set containing roots with equal -values. Note that since p - m 0, p does not ramify in Q p èè. We deduce that the roots in any one set are distinct roots of the same irreducible factor of fèxè over Q p èè. Observe that æ is a root of wèxè if and only if æ, is a root of fèxè. If we view the roots of fèxè in the form æ, and consider the ` sets of roots formed as above, we see that wèxè has ` ëclusters" around of roots with the property that if æ and æ 0 belong to the same cluster, then èæ, è =èæ 0,èé0. Furthermore, the roots in any one of these clusters are distinct roots of the same irreducible factor of wèxè over Q p èè and, hence, of the same irreducible factor of wèxè over Q. In other words, if one root from a cluster is a root of gèxè èor hèxèè, then all the roots from that cluster are roots of gèxè èor hèxè, respectivelyè. The above holds for each 6= 1 satisfying m0 =1. There are m 0, 1 such forming èm 0, 1è æ ` clusters of roots of wèxè. We show next that these are disoint clusters. This is clearly true of clusters formed from the same ; in other words, if æ and æ 0 are roots with èæ, è 6= èæ 0, è, then clearly æ 6= æ 0. Now, suppose æ is in a cluster around and in a cluster around 0 where 6= 0, 6= 1, 0 6=1, m0 = 1, and è 0 è m0 =1. Then it follows that, è 0, èæ æ =, 0 èæ, è, èæ, 0 è æ minf, 0 èæ, è æ ;, èæ, 0 è æ g é0: Lemma 6 implies that è 0, è =0. Since èæ, è é 0 and èè =0,we also deduce èæè = 0. We therefore obtain a contradiction, and we can conclude that the èm 0, 1è æ ` clusters consist of distinct roots. The total number of roots in these èm 0, 1è æ ` clusters is èm 0, 1è æ p`. Since wèxè has m + r = m 0 p` + r roots, we haveyet to account for p` + r roots of wèxè. By considering the Newton polygon of wèxè and using the condition wèxè aèx m, 1èx r èmod p`è, we deduce that wèxè has exactly r roots æ with the property that èæè é 0. Note that the other roots æ of wèxè necessarily satisfy èæè = 0. In a manner similar to the above èbut easierè, we deduce that each of the r roots around 0 does not belong to any of the above clusters of roots. These r roots around 0 form a cluster as before except that we cannot in general deduce that these roots necessarily are roots of the same irreducible factor of wèxè over Q p èè èor over Qè. The condition gcdè`; rè = 1 in èiiè implies that the left-most edge of the Newton polygon of wèxè contains only the lattice points at its endpoints, namely è0;`è and èr; 0è. Since p does not ramify in Q p èè, we deduce that in this case the cluster of r roots around 0 are distinct roots of a single irreducible factor of wèxè over Q p èè. 9

10 We show now that the remaining p` roots of wèxè form a cluster of roots around 1. The argument for roots around 1 is analogous to the case for above èust set = 1è except that we cannot obtain here that èb 0 è=èwè1èè = `. On the other hand, the condition wèxè aèx m, 1èx r èmod p`è implies èb 0 è=èwè1èè `. The argument proceeds as before, and we deduce that there are p` roots æ of wèxè with the property that èæ,1è é 0 èwe could say more, but this is all we will needè. As before, it is easy to argue that these p` roots around 1 are distinct from the roots of wèxè belonging to other clusters. We cannot, however, deduce that these roots are distinct or that they are roots of the same irreducible factor of wèxè over Q p èè èor over Qè. We now apply the information we have established about the location of the zeroes of wèxè. We consider the case that r =0. Then there are no roots in the cluster described above around 0. It follows that the roots of gèxè consist of complete clusters around for some choices of 6= 1 together with possibly some of the p` roots around 1; likewise for hèxè. If C 1 ; C 2 ;:::;C s denote the clusters around 6= 1 which contain roots of gèxè and C 0 denotes the roots in the cluster around 1 that are roots of gèxè, then we deduce that ècè min 0s æ2c è1, æè : Observe that æ æ è1, æè min è1, æè é 0: æ2c 0 æ2c0 For each 2f1;2;:::;sg,we deæne as the m 0 th root of unity such that the roots of C are those around, and we write æ2c è1, æè = æ2c, è1, è, èæ, è æ = C è1, è, æ2c èæ, è: Since C by construction is a multiple of p, we deduce that each of the terms in this last expression has -value é 0. It follows now that ècè é 0. The same argument gives èdè é 0. Since r = 0 and wèxè aèx m, 1èx r èmod p`è, we deduce that a 0 6= 0 so that A and B are deæned. Also, in this case, èæè = 0 for each rootæof gèxè and èæè = 0 for each rootæof hèxè. Deæne 1 ;:::; s as before, and let 0 =1. We use that èaè min 0s min 0s æ2c æ2c æ, 1 æ èæ, è+ æ, æ +, 1 Following along lines similar to our argument that ècè é 0, we deduce that èaè é 0. An analogous argument gives èbè é 0. For èiiè, we have shown that the cluster of r roots around 0 are roots of a single irreducible factor of wèxè over Q p èè. Hence, these r roots are either roots of gèxè or roots of : 10

11 hèxè. Suppose the cluster of roots around 0 are roots of hèxè. Then each rootæof gèxè belongs to a cluster around a root of unity so that the arguments above give èaè é 0 and ècè é 0. Since P p - a n+1, the leading coeæcient ofhèxè is not divisible by p and we deduce that èhè0èè = hèæè=0 èæè. Since hèxè has roots from the cluster of roots around 0, we obtain èhè0èè é 0 so that phè0è. If S is the set of r roots clustered around 0, then we consider, æè =r, æ: æ2sè1 æ2s Since èæè é 0 for each æ 2 S, the sum on the right has a positive -value. Since péré0, èrè =0. It follows that,p æ2s è1, æèæ =0. Hence, the arguments in the previous paragraph now imply èdè = 0. In particular, we must have D 6= 0. A similar argument can be used in the case that the cluster of roots around 0 are roots of gèxè. The proposition follows. æ For the proof of Theorem 4, we will make use of three results similar to Proposition 1. They are as follows: Proposition 2. P n+1 Let wèxè = =0 a x 2 Zëxë with a n+1 6=0. Let p be an odd prime such that pèn +1è and p - a n+1. Write n +1= p`m 0 where p èm 0 è=0. Suppose that wèxè a n+1 èx n+1, 1è èmod p`è and that, for each 6= æ1 such that m0 =1,wehave p èwèèè = `. Let wèxè =gèxèhèxèbe a factorization of wèxè in Zëxë. Let A = æ, 1 ; B= æ, 1 ; C 0 = è1, æ 2 è and D 0 = è1, æ 2 è; æ æ gèæè=0 hèæè=0 gèæè=0 hèæè=0 where the sums are over the distinct roots of gèxè and hèxè and where we consider A and B only in the case that a 0 6=0. Then A, B, C 0, and D 0 are rational numbers satisfying èaè é 0, èbè é 0, èc 0 è é 0, and èd 0 è é 0. P n+1 Proposition 3. Let wèxè = =0 a x 2 Zëxë with a n+1 6=0. Let p be an odd prime such that pn and p - a n+1. Write n = p`m 0 where p èm 0 è=0. Suppose that wèxè a n+1 èx n, 1èèx + 1è èmod p`è and that, for each 6= 1 such that m0 = 1, we have p èwèèè = `. Let wèxè = gèxèhèxèbe a factorization of wèxè in Zëxë. Deæne A, B, C 0, and D 0 as in Proposition 2. Then A, B, C 0, and D 0 are rational numbers satisfying èaè é 0, èbè é 0, èc 0 è é 0, and èd 0 è é 0. P n+1 Proposition 4. Let wèxè = =0 a x 2 Zëxë with a n+1 6=0. Suppose wèxè is a reciprocal polynomial so that wèxè = æx n+1 wè1=xè. Let p be an odd prime such that pèn, 1è, pa n+1, and p - a n. Write n, 1 = p`m 0 where p èm 0 è=0. Suppose that wèxè a n èx n,1, 1èx èmod p`è and that, for each 6= æ1 such that m0 =1,wehave p èwèèè = `. Let wèxè =gèxèhèxèbe a factorization of wèxè in Zëxë. Deæne A, B, C 0, and D 0 as above. Then A, B, C 0, and D 0 are rational numbers such that if AB =0, then at least one of èc 0 è é 0 and èd 0 è é 0 holds. Proofs of Propositions 2, 3, and 4 can be given along the lines of the argument presented here for Proposition 1. To aid the reader, we brieæy describe certain aspects of these 11

12 proofs. As in the proof of Proposition 1, the roots of wèxè in each of the above results can be grouped in clusters. In each of Propositions 2, 3, and 4, around each of the m 0, 2 èif m 0 is evenè or m 0, 1 èif m 0 is oddè diæerent satisfying 6= æ1 and m0 = 1, there are p` roots which form various clusters, with each cluster of roots belonging to the same irreducible factor of wèxè and each cluster containing a multiple of p diæerent roots. In the case of Proposition 2, there are p` other roots of wèxè forming a cluster around 1 and, if m 0 is even, p` other roots forming a cluster around,1; each of these clusters contains roots that are not necessarily roots of the same irreducible factor of wèxè. This is suæcient to establish Proposition 2. There are similar clusters of size p` around each of 1 èfor all m 0 è and,1 èif m 0 is evenè in the case of Proposition 4. However, in this case there are two additional roots to account for; one of these two roots æ satisæes èæè é 0 and the other root æ 0 satisæes èæ 0 è é 0. If AB = 0, one can show that the roots æ and æ 0 are either both roots of gèxè or are both roots of hèxè. If the former holds then èd 0 è é 0, and if the latter holds then èc 0 è é 0. In Proposition 3, there is one cluster with p` roots around 1 containing roots that are not necessarily roots of the same irreducible factor of wèxè. There are also p` + 1 roots around,1 èif m 0 is evenè or one such root èif m 0 is oddè forming clusters with the roots in each cluster being roots of the same irreducible factor of wèxè; one cluster contains p + 1 roots èif m 0 is evenè or 1 root èif m 0 is oddè and the remaining clusters contain a multiple of p diæerent roots of wèxè. It follows easily that èaè é 0, èbè é 0, èc 0 è é 0, and èd 0 è é 0. There is a variety of results analogous to the propositions in this section that can be established by similar means. Note that in Proposition 1 we dealt with a sum C of terms of the form 1, æ whereas the remaining propositions dealt with a sum C 0 involving terms of the form 1, æ 2. As will be evident later, C is of value in establishing Theorem 1 as the term 1, æ is 0 when æ is one of the cyclotomic roots of nx n+1, èn +1èx n + 1 èi.e., when æ = 1è, the numerator of f 0 èxè. Similarly, C 0 is helpful in establishing Theorem 4 since 1, æ 2 is 0 when æ is one of the cyclotomic roots of pèxè èi.e., when æ = æ1è. More generally, one can make use of C k = è1, æ k è and D k = è1, æ k è gèæè=0 hèæè=0 in dealing with certain classes of polynomials for which the cyclotomic roots are known to be roots of x k, 1. The proofs presented in the following sections will help illustrate applications of such propositions to the irreducibility of the non-cyclotomic parts of polynomials of a given form. 4. AProof of Theorem 1 Let n 2. We wish to show that nx n+1, èn +1èx n + 1 is èx, 1è 2 times an irreducible polynomial in Zëxë. It suæces to show the same for the reciprocal of nx n+1,èn+1èx n +1, and for this purpose we deæne wèxè = x n+1, èn +1èx+n. We consider n 2 and wèxè = gèxèhèxè where gèxè and hèxè are in Zëxë, deg gèxè 1, deg hèxè 1, and gè1è 6= 0. Note that deg gèxè 1 is possible since the product of the roots of wèxè isæn so that wèxè has a root diæerent from 1. Since wèxè is monic, we may suppose that each of gèxè and hèxè are monic and do so. Our goal is to show hèxè =èx,1è 2. 12

13 We make use of A and B of Proposition 1 but not of C and D. If æ isarootofgèxè, then æ and gè0è=æ are algebraic integers. Also, if æ isarootofhèxè, then æ and hè0è=æ are algebraic integers. Since gè0èhè0è = n, we deduce that nab is a rational integer. We will see momentarily that if B = 0, then hèxè =èx,1è 2. In addition, we show that if B 6= 0, then upper and lower bounds on the value of nab can be obtained which are inconsistent for all but Oèt è1=3è+" è positive integers n t. The proof of Theorem 1 will then be complete. Since èx n+1,1è=èx,1è has distinct roots on the unit circle and since the derivative ofa polynomial has roots inside the convex hull of the roots of the polynomial ècf. ë9, Problem 31 on page 108ëè, the roots of ènx n+1, èn +1èx n +1è=èx,1è 2 have absolute value é 1. It is clear that 1 is a root of wèxè with multiplicity 2. It follows that the remaining roots of wèxè have absolute value é 1. Observe that w 0 èxè only has cyclotomic roots. It follows that the n, 1 roots of wèxè with absolute value é 1 are distinct. Now, we establish that if B = 0, then hèxè =èx,1è 2. We show instead the contrapositive. Suppose hèxè 6= èx,1è 2. Since gè1è 6= 0,èx,1è 2 is a factor of hèxè. The comments above imply that each of gèxè and hèxè must have a root with absolute value é 1. Furthermore, the absolute value of the product of the roots of either of these polynomials exceeds 1. Thus, gè0è and hè0è each has absolute value é 1. Note that gè0è and hè0è must be relatively prime since a common divisor p would divide both gè0èhè0è = n and the coeæcient of x in the product gèxèhèxè, namely n + 1, which is clearly impossible. We apply Proposition 1 with m = n and r =1. We consider ærst a prime divisor p of hè0è. Note then that pm and p - gè0è. We let ` and m 0 be deæned as in the proposition. Since n 0 èmod p`è, we obtain wèxè, x n, 1 æ x èmod p`è. Suppose m0 = 1 and 6= 1. Then n = 1 so that wèè =nè1, è. Since è1, è = 0, we obtain èwèèè = ènè =`. Observe that the conclusions of Proposition 1 èiiè now follow aswè0è = n 6= 0 and r =1 imply the hypotheses in Proposition 1 èiiè hold. Since p - gè0è, we deduce that èaè é 0. On the other hand, A + B = wèæè=0 æ, 1 æ = n +1 n ; where we have used here that the roots of wèxè other than 1 are distinct and that the summand above is 0 when æ = 1 èso that we can consider the sum aboveasasumover roots of wèxè with each root appearing to its multiplicityè. Since pn, wehaveèèn +1è=nè é 0. Since èaè é 0, we obtain B 6= 0. Thus, we can deduce that if B = 0, then hèxè =èx,1è 2. Now, suppose B 6= 0. Since gè1è 6= 0,we still have that gè0è has absolute value é 1. If we repeat the argument in the previous paragraph but this time considering a prime p dividing gè0è èso that the roles of gèxè and hèxè and the roles of A and B are switchedè, we obtain A 6= 0. In addition, we see that for each prime divisor p of n èso p divides hè0è or gè0èè, these arguments give from Proposition 1 èiiè that either èaè é 0orèBèé0. We deduce that at least one of the rational integers gè0èa and hè0èb is a multiple of p. Thus, if pn, then pnab. Next, we show that if pèn + 1è, then p 2 nab. Since we nowhave that AB 6= 0,we will 13

14 get the lower bound è1è nab pèn+1è 2 p p : pn We apply Proposition 1 with m = n + 1 and r =0. Thus, p is a prime divisor of m. Again, we let ` and m 0 be deæned as in the proposition. Since n,1 èmod p`è, we obtain wèxè x n+1, 1 èmod p`è. If m0 = 1, then n+1 = 1 so that wèè =èn+ 1èè1, è. If also 6= 1, then è1, è = 0 and we obtain èwèèè = èn +1è=`.Thus, we can apply Proposition 1 èiè. We obtain èaè é 0 and èbè é 0. Therefore, each of the rational integers gè0èa and hè0èb isamultiple of p. It easily follows that the integer nab is divisible by p 2, and we obtain è1è. To obtain an upper bound for nab, we use the following result about the complex zeroes of wèxè. Lemma 7. If n 2 and re i èwith r; 2Rè is a root of wèxè =x n+1, èn +1èx+n, then r, 1 é è5=nè log n. The result is essentially contained in ë2ë and ë8ë. It can be established by observing wèæè =0 implies æ n+1 èn+1èæ,nè2n +1èæ so that æ è2n +1è 1=n logè2n +1è 2 logè2n +1è = exp log n n n n : Observe that since the roots of wèxè other than 1 have absolute value é 1, Lemma 7 implies that for all integers n 2, if re i 6= 1 is a root of wèxè, then 0 ér,1éè5=nè log n. Next, we show that è2è A 10 log n and B 10 log n: Using æ to denote the conugate of æ, we can rearrange the terms in the deænition of A to obtain A = gèæè=0 æ, 1 æ Since gèæè = 0 implies æ isarootofwèxè, we deduce that if æ = re i, then æ æ æ æ, 1 æ = æ r, 1 r : 10 log n : n The ærst inequality in è2è now follows. The second inequality is deduced in an analogous manner. From è2è, we obtain the estimate è3è nab 100nèlog nè 2 : 14

15 Since AB 6= 0,we deduce from è1è and è3è that pèn+1è 2 pn p p 100nèlog nè 2 : Since n t, it follows that pèn+1è p t 1=3 èlog tè 2=3 or pn p t 1=3 èlog tè 2=3 : Theorem 1 is now a consequence of the following Lemma 8. Q Let é0. For n a positive integer, deæne Qènè = pnp. Then for every "é0, the number of n t for which Qènè t is Oèt +" è Proof. Observe that Qènè is always squarefree. For each squarefree number m = p 1 p 2 æææp s t where each p denotes a prime with p 1 ép 2 éæææ ép s, the number of n t for which Qènè =mis is equal to the number of solutions in positive integers x 1 ;x 2 ;:::;x s to x 1 log p 1 + x 2 log p 2 + æææ+x s log p s log t: We consider the p which are p log t ærst. Suppose p k is the largest of these. Clearly k p log t and each x is bounded by 2 log t. Thus, the number of choices for x 1 ;x 2 ;:::;x k is è2 log tè p log t exp, 2 p log t log log t æ. Now, each remaining p satisæes p é p log t so that log p é è1=2è log log t. Hence,, xk+1 + x k+2 + æææ+x s æ log log t 2 x k+1 log p k+1 + x k+2 log p k+2 + æææ+x s log p s log t: Let N denote the greatest integer 2 log t=èlog log tè. Then the number of choices for x k+1 ;x k+2 ;:::;x s is bounded by the number of solutions to x k+1 + x k+2 + æææ+x s N in positive integers x k+1 ;x k+2 ;:::;x s. Equivalently, we seek a bound on the number of solutions to y k+1 + y k+2 + æææ+y s N,s,k in non-negative integers y k+1 ;y k+2 ;:::;y s. Each such solution corresponds to a unique non-negative binary number consisting of N, 1 digits given by y k+1 ones, followed by 1 zero, followed by y k+2 ones, followed by 1 zero, and so on èending with y s onesè. It follows that there are 2 N choices for x k+1 ;x k+2 ;:::;x s as above. Thus, the number of possibilities for the s positive integers x 1 ;x 2 ;:::;x s is, p æ 2 log t exp 2 log t log log t æ 2 2 log t=èlog log tè exp t " : log log t This is a bound on the number of n t for which Qènè =mfor some given squarefree m t. Letting m vary, the lemma follows. æ 15

16 Let n denote a positive integer, and set 5. AProof of Theorem 2 fèxè =1+x+x 2 +æææ+x n : Our goal is to show that for each positive integer k and for most n t, the polynomial f èkè èxè is irreducible. As in the previous section, we will make use of Proposition 1. The main diæculty we will encounter is in showing that the condition èwèèè = ` is satisæed in Proposition 1. Indeed, already for k = 2, it is the case that in many instances èwèèè 6= ` when the other conditions of Proposition 1 hold. Thus, it will become necessary to bound the number of times èwèèè 6= `. For this purpose, we will introduce an auxiliary polynomial uèxè èsee the discussion after Lemma 14è that depends on k and r but not on n and which has the property that èwèèè 6= ` if and only if èuèèè é 0. This allows us to obtain the bound we need on the number of times èwèèè 6= `, and we proceed by applying Proposition 1 as in the previous section. We begin with a lemma which is easily established by induction. The details of the proof are left to the reader. Lemma 9. Let k be a positive integer n, 1. Then è n+1 1 f èkè èxè = è,1è n+1, k èx,1è k+1 n +1, We also make use of =n,k+1 n+1 i=n,k+1 i6= i x +è,1è k+1 k! Lemma 10. Let n and k be positive integers with k n, 1. Then each rootoff èkè èxè has absolute value é 1. Proof. Observe that the roots of fèxè are on the unit circle fz : z =1gand that fèxè has no repeated roots. As in Section 4, we use that the roots of the derivative of a polynomial in Rëxë lie in the convex hull of the roots of the polynomial. It follows that all the derivatives of fèxè have only roots with absolute value é 1. æ Lemma 11. Let n and k be positive integers with k n,1. Let be an integer satisfying n, k +1n+1. Then k n +1, n+1 i=n,k+1 i6= i! : is divisible by k!. Proof. Observe that n+1 i=n,k+1 i6= i = n+1 i=+1 i,1 i=n,k+1 i ; 16

17 a product of n +1, consecutive integers times a product of k, èn +1,è positive integers. The ærst of these products on the right is therefore divisible by èn+ 1,è! and the second is divisible by èk,èn+ 1,èè!. It follows that k n +1, n+1 i=n,k+1 i6= i = k! èn +1,è!èk, èn +1,èè! is an integer multiple of k!. The lemma follows. æ n+1 i=+1 i,1 i=n,k+1 Lemma 11 is not really necessary for what follows. But it makes matters slightly easier. Note that it follows from Lemma 11 that if æ isarootoff èkè èxè, then 1=æ is an algebraic integer. Lemma 12. Let n and k be positive integers with k n, 1. Let m be an integer satisfying n, k +1mn+1. Set r = n +1,m. Then n+1 è,1è n+k,m k i,k! èmod mè; n +1,m i=n,k+1 i6=m and there is a constant æèk; rè depending only on r and k and independent ofnsuch that n+1 1 è,1è n+k,m k i + k! æèk; rè èmod mè: m n +1,m Proof. Consider the function Observe that i=n,k+1 i6=m F èxè =èx+rèèx + r, 1è æææèx+1èæèx,1èèx, 2è æææèx,èk,rèè: F èmè = n+1 i=n,k+1 i6=m The constant term of F èxè isè,1è k,r r!èk,rè!=è,1è k+m,n,1 èn+1,mè!èk+m,n,1è!., Thus, F èmè è,1è k+m,n,1 k èn +1,mè!èk + m, n, 1è! èmod mè. Writing n+1,mæ as k!=èèn +1,mè!èk + m, n, 1è!è, the ærst congruence in the lemma follows. Observe that è,1è n+k,m, æ, k n+1,m = è,1è r+k,1 k ræ. The above shows that F èxè = f 0 + f 1 x + Gèxèx 2 where è,1è r+k,1, k ræ f0 =,èk!è, f 1 is the coeæcient ofxin F èxè èwhich depends only on r and kè, and Gèxè 2 Zëxë. Since n+1 è,1è n+k,m k i + k! n +1,m i: i=n,k+1 i6=m i 17

18 ræ F èmè+k!, we deduce that the expression on the left-hand side ræ times f1. The lemma is the same as è,1è r+k,1, k of the second congruence is congruent modulo m to è,1è r+k,1, k follows. æ We æx a positive integer k and consider n k +1. Let wèxè beè,1è k,1 =k! times the reciprocal polynomial of the numerator of f èkè èxè in Lemma 9 where k is a positive integer. In other words, we set wèxè =x n n+1 n+1 è,1è n+k, k i x n+1, : k! n +1, This can be rewritten as =n,k+1 wèxè =x n k! k =0 è,1è k+,1 k 0ik i6= i=n,k+1 i6= èn +1,iè x : Note that Lemma 11 implies wèxè 2 Zëxë. Let m be an integer with n, k +1 mn+ 1, and let p be a prime divisor of m with pékèif it existsè. Deæne r, ` and m 0 as in Proposition 1. It follows from the ærst congruence in Lemma 12 that wèxè èx m, 1èx r èmod p`è. The deænition of m implies that 0 n +1,m= r kép. Except for the condition that èwèèè = `, the conditions of Proposition 1 are clearly satisæed. In addition to the condition èwèèè = `, we will want that either both A and B are non-zero or both C and D are non-zero. We address these matters next. Since wèxè èx m, 1èx r èmod p`è, there is a polynomial vèxè inzëxë such that wèxè= èx m,1èx r + p`vèxè. Setting x = where m0 =1,we deduce èwèèè `. We will not be able to prove in general that èwèèè = `, but instead we will show that typically this is the case. Lemma 13. Let uèxè = P s =0 b x 2 Zëxë. Let p be a prime not dividing b s. Then there exist s diæerent numbers such that for some positive integer m 0 relatively prime to p, we have m0 =1and èuèèè é 0. Proof. Let x 1 ;x 2 ;:::;x s be the s not necessarily distinct p-adic roots of uèxè. If is as in the lemma, then è, x i è é 0 for some i 2f1;2;:::;sg. If and 0 are distinct roots of unity as in the lemma and i is such that both è, x i è é 0 and è 0, x i è é 0, then we would have è, 0 è é 0, contradicting Lemma 6. Hence, for each i 2f1;2;:::;sg, there is at most one as in the lemma for which è, x i è é 0. The lemma follows. æ Lemma 14. Let uèxè = P s =0 b x 2 Zëxë with uè1è 6= 0, let z maxfuè1è; b s ; 2g, and let m 0 be a positive integer. Then there is a constant c èdepending only on uèxèè such that there are cm 0 =èlog zè diæerent primes p satisfying gcdèp; m 0 è=1,péz, and there is a for which m0 =1,uèè6=0, and èuèèè é 0. Proof. Let Hèxè be the part of x m0, 1 which is coprime to uèxè; in other words, Hèxè = èx m0,1è= gcdèuèxè;x m0,1è. Let R denote the resultant ofhèxè and uèxè. Then R is a 18

19 non-zero integer which can be expressed as a product of numbers of the form uèè where m0 = 1 and uèè 6= 0. It follows that R is divisible by the product of the primes p in the lemma. If we consider the m 0 complex roots of Hèxè èall with absolute value 1è, we see,p æ s that R is bounded by =0 b m 0. Thus, if P denotes the number of primes p in the lemma, then s m 0 z P =0 b It follows that P m 0 = log z, implying the lemma. æ We describe next the polynomials uèxè that we will use in Lemma 14. We return to our discussion of wèxè and consider 6= 1 for which m0 =1. Since n+1 = n+1,m = r and r k, we can view wèè as a polynomial in of degree k which has, by Lemma 12, each coeæcient divisible by m. We multiply this polynomial by k!=m and use the second congruence in Lemma 12 to deal with the coeæcient of r modulo m. For the remaining coeæcients, observe that 0ik i6=;i6=r èn +1,iè è,1èk,r èk, rè!r! r, : èmod mè: Note that,èk, rè r, r so that this last expression is a rational integer. We deduce now that if k è,1è y 1 èxè =æèk; rèx r +è,1è r,1 èk, rè!r! x ; r, 0k 6=r then èk!=mèwèè, y 1 èè =my 2 èè for some polynomial y 2 èxè 2 Zëxë. The coeæcients of y 1 èxè only depend on r and k, and the coeæcient ofx for 6= r and 0 k in y 1 èxè is clearly non-zero. Recall that wè1è=0. Under the conditions of Proposition 1, è, 1è = 0, so the factor of x, 1inwèxè does not aæect the value of èwèèè. We divide y 1 èxè by the highest power of x, 1 that divides it and call the quotient uèxè. Observe that èwèèè é`if and only if èy 1 èèè é 0 if and only if èuèèè é 0. The advantage of dealing with uèxè over wèxè is that uèxè depends only on k and r and not on n. With k still æxed, we let r = n +1,m vary from 0 to k to obtain k + 1 diæerent polynomials uèxè. The idea now istoshow that in many instances èuèèè = 0. With k and r æxed, we deæne a pair èm 0 ;pè, with m 0 a positive integer and p a prime not dividing m 0,asabad pair èrather than a bad appleè if there is a 6= 1 for which m0 =1 and èuèèè é 0. For té0, we determine an upper bound for the number of bad pairs èm 0 ;pè with p`m 0 t for some positive integer `. The number p`m 0 will correspond to m in Proposition 1. Observe that we do not require that be a primitive m 0 th root of unity. This introduces some complications in bounding the number of bad èm 0 ;pè. For a given m 0, we can use Lemma 14 to bound the number of primes p for which èuèèè é 0, but we must deal with the possibility not covered by Lemma 14 that uèè = 0. We show next that there are at least three choices of r 2f0;1;:::;kg, in the case k 6= 2, for which uèxè has no cyclotomic factors. We will use the following preliminary result. 19

20 Lemma 15. For each positive integer k 15, there exist at least two distinct primes in the interval èk=2;k,2ë. Proof. The result was veriæed directly for 15 ké200. Now, suppose k 200. Note that 1:96k=2 ék,2. We show that for each x 100 there is a prime in the interval èx; 1:4xë; the lemma then follows since then there is a prime in the interval èk=2; P 1:4k=2ë and a prime in the interval è1:4k=2; 1:96k=2ë è1:4k=2;k,2ë. Deæne èèxè = pxlog p. We make use of the estimate from Rosser and Schoenfeld ë10ë that x 1, 1 log x éèèxèéx log x for all x 41: To establish there is a prime in èx; 1:4xë for x 100, it suæces therefore to show 1:4x 1, 1 logè1:4xè x log x This is a simple matter to verify; indeed, the inequality above holds for all x 100 follows from the fact that it holds for x = 100. æ Lemma 16. Let k be an integer with k =3or k 5. For each r 2fk,2;k,1;kg, the polynomial uèxè deæned above has no cyclotomic divisors. Proof. We work in the æeld of complex numbers. Fix k and r as in the lemma. By the deænition of uèxè, we know that 1 is not a root of uèxè. We assume now that uèxè has a root which is a root of unity. Then uèè = 0 for some 6= 1 satisfying d = 1 for some positive integer d. We take d minimal and note that d 2. We ustify ærst that d 6= 2. If d = 2, then,1 isarootofuèxè and, hence, also of y 1 èxè. One checks directly that y 1 è,1è 6= 0 in the case that k = 3 and r =1. For the remaining choices of k and r, we use the deænition of y 1 èxè together with the a formula for æèk; rè. In particular, the deænition of y 1 èxè and the choice of r imply that, for k 5, if y 1 è,1è = 0, then æèk; rè é 0. From the proof of Lemma 12, we see that æèk; rè isè,1è r+k,1, k ræ times the coeæcient f1 of x in F èxè =èx+rèèx + r, 1è æææèx+1èæèx,1èèx, 2è æææèx,èk,rèè: When r = k, every coeæcient offèxè is positive and we easily deduce that æèk; rè é 0. Now, suppose r = k,2 and k 5èwehave already dealt with k = 3è. Since the coeæcient of x in the expanded product èx + 2èèx + 1èèx, 1èèx, 2è is zero and its constant term is 4, we see that f 1 is simply 4 times the coeæcient ofxin èx + rèèx + r, 1è æææèx+ 3è. Thus, f 1 é 0, and we conclude that æèk; rè é 0. It remains to consider the case that r = k, 1. One checks directly that in this case f 1 =èk,1è!, k,1 =1 èk, 1è! k,1 =, =2 : èk, 1è! : Since k 3, we obtain æèk; rè =æèk; k,1è é 0. We deduce that uè,1è 6= 0 so that dé2. 20

21 The deænition of uèxè implies that we must also have y 1 èè = 0. Using the deænition of y 1 èxè, we consider the expression y 1 èè=èèk, rè!r! r è as a sum of k + 1 terms and observe that it is an element ofqèè which is an extension of degree èdè over Q. Thus, we can rewrite the expression as a polynomial in f1;;:::; èdè,1 gwith rational coeæcients. Call this polynomial èè. We consider ærst the case that k 15. By Lemma 15, there are two primes in the interval èk=2;k,2ë. Call these primes p 1 and p 2.We show that at least one of these two primes does not divide d. Since is a root of y 1 èxè, a non-zero polynomial of degree k, we deduce that the degree of the minimal polynomial for èin Qëxëè is k. Hence, èdè k. On the other hand, each ofp 1 and p 2 is ék=2. If p 1 and p 2 are both factors of d, then we would have k èdè èp 1 p 2 è=èp 1,1èèp 2, 1è é k 2, 1 2 ; which is easily seen to be impossible for the k under consideration. Thus, either p 1 or p 2 does not divide d. Now, æx p to be a prime in èk=2;k,2ë which does not divide d. Consider 0 = r, p 2f0;1;:::;kg. Observe that for each 2f0;1;:::;kg wehave,pér,kr,ré2p: It follows that in the sum deæning y 1 èxè, the expression r, in the summand is divisible by p if and only if = 0. Since p 2 èk=2;kë, we get that pk!. Since r k, we clearly have that pèk, r + pè!. We obtain k 0 = k! èr, pè!èk, r + pè! 2 Z k =è p does not divide 0 : If we consider the k + 1 non-zero terms in y 1 èxè=èèk, rè!r!x r è, we see that the constant term æèk; rè=èèk, rè!r!è may have denominator divisible by p èand, in fact, does though this is not neededè and the denominator of the coeæcient ofx 0,r =x,p is divisible by p. No other denominators will be divisible by p. We ustify momentarily that,p when expressed as a polynomial in of degree èdè includes a term i with ié0 and with coeæcient not divisible by p. More precisely, we show that,p, b = pgèè is impossible if b 2 Z and Gèxè 2 Zëxë. It will then follow that èè has at least one coeæcient which can be expressed as a rational number èpossibly 0è with denominator not divisible by p plus a non-zero rational number with denominator divisible by p. This coeæcient is clearly non-zero. It follows that èè 6= 0, and we deduce that y 1 èè 6= 0. This is a contradiction. Hence, uèxè does not have a cyclotomic factor for k 15. Assume that there exist b 2 Z and Gèxè 2 Zëxë such that,p, b = pgèè. By applying the automorphisms of Qèè æxing Q, wemay replace in this equation with any primitive root of x d, 1=0. Since dé2, we deduce that there are 1 and 2 primitive roots of x d, 1 = 0 with 1 6=, 2 satisfying,p æ 1, b = pgè 1 è and,p 2, b = pgè 2 è. Subtracting, we obtain,p 2,,p 1 = p Gè 2 è, Gè 1 è. Setting 3 = p 1,p 2 2 Qèè, we easily deduce that N Q èè=q è 3,1è is a multiple of p. Since 3,1 is a root of 1+èx+1è+èx+1è 2 +æææ+èx+1è d,1, 21