On the irreducibility of a truncated binomial expansion

Transcription

1 On the rreducblty of a truncated bnomal expanson by Mchael Flaseta, Angel Kumchev and Dmtr V. Pasechnk 1 Introducton For postve ntegers k and n wth k n 1, defne P n,k (x = =0 ( n x. In the case that k = n 1, the polynomal P n,k (x takes the form P n,n 1 (x = (x + 1 n x n. If n s not a prme, P n,n 1 (x s reducble over Q. If n = p s prme, the polynomal P n,n 1 (x = P p,p 1 (x s rreducble as Esensten s crteron apples to the recprocal polynomal x p 1 P p,p 1 (1/x. Ths note concerns the rreducblty of P n,k (x n the case where 1 k n 2. Computatons for n 100 suggest that n ths case P n,k (x s always rreducble. We wll not be able to establsh ths but nstead gve some results whch gve further evdence that these polynomals are rreducble. The problem arose durng the 2004 MSRI program on Topologcal aspects of real algebrac geometry n the context of work of Inna Scherbak n her nvestgatons of the Schubert calculus n Grassmannans. She had observed that the roots of any gven P n,k (x are smple. Ths follows from the dentty P n,k (x (x + 1 P n,k (x n 1 = k She then asked whether, for a fxed postve nteger n, the varous n(n 1/2 roots of P n,k (x for 1 k n 1 are dstnct. Inna Scherbak [6] tred to answer ths queston usng methods from representaton theory, but to the best of our knowledge the problem s stll open. We wll not resolve the problem, but our methods mply that for each postve nteger n, almost all of the roots are dstnct. In other words, the number of dstnct roots s n 2 /2 as n tends to nfnty. Before closng ths ntroducton, we menton that these same polynomals have recently arsen n the context of work by Iossf V. Ostrovsk [4]. In partcular, he fnds a soluton to a problem posed by Alexandre Eremenko on the dstrbuton of the zeroes of P n,k (x as k and n tend to nfnty wth k/n approachng a lmt α (0, 1. 2 The Results Our methods apply to a wder class of polynomals than the P n,k (x s alone, so we begn by recastng the problem n a more general settng. For a and b nonnegatve ntegers wth a b, the dentty a ( ( b b 1 ( 1 = ( 1 a (1 a =0 =0 x k. s easly establshed by nducton on a. We deduce that ( n ( ( n P n,k (x 1 = (x 1 = ( 1 x = Mathematcs Subect Classfcaton: 12E05 (11C08, 11R09, 14M15, 26C10. The frst author was supported by the Natonal Scence Foundaton durng research for ths paper. The thrd author was supported by NWO grant Parts of the work were completed whle the thrd author was supported by MSRI and by DFG grant SCHN-503/2-1, whle he held postons at MSRI, resp. at CS Dept., Un. Frankfurt. =0

2 = = =0 = =0 k ( =0 ( 1 x = =0 = =0 ( n ( 1 x = where the last equalty makes use of (1. For 0 k, we defne c = 1 k 1 k ( 1 x ( 1 k x, ( 1 k = ( 1k n(n 1 (n + 1(n 1 (n k + 1(n k!(k! so that P n,k (x 1 = k =0 c x. We are nterested n the rreducblty of P n,k (x. A necessary and suffcent condton for P n,k (x to be rreducble s for P n,k (x 1 to be rreducble, so we restrct our attenton to establshng rreducblty results for the polynomals k =0 c x. For our results, we consder F n,k (x = a c x, (2 =0 where a 0, a 1,..., a k denote ntegers, each havng all of ts prme factors k. In partcular, none of the a are zero. Observe that f F n,k (x s rreducble for all such a, then necessarly P n,k (x s rreducble smply by choosng each a = 1. Another nterestng choce for a s a = ( 1 k!(k!. As F n,k (x s rreducble f and only f ((n k 1!/n! x k F n,k (1/x s rreducble, the rreducblty of F n,k (x wll mply the rreducblty of 1 n k + x n k x 2 n k xk n. In partcular, f n = k + 1, these polynomals take a nce form. It s possble to show, stll wth n = k + 1, that these polynomals are rreducble for every postve nteger k. The dea s to use Newton polygons wth respect to two dstnct prmes n the nterval ((k + 1/2, k + 1]. For k 10, t s known that such prmes exst (cf. [5]. As ths s not the focus of the current paper, we omt the detals. Let N be a postve nteger. The number of ntegral pars (n, k wth 1 n N and 1 k n 2 s n N (n 2 N 2 2. Our frst result s that the number of possble reducble polynomals F n,k (x wth n N and 1 k n 2 s small by comparson. More precsely, we show the followng. Theorem 1. Let ε > 0, and let N be a postve nteger. For each ntegral par (n, k wth 1 n N and 1 k n 2, consder the set S(n, k of all polynomals of the form (2 where a 0, a 1,..., a k denote arbtrary ntegers, each havng all of ts prme factors k. The number of such pars (n, k for whch there exsts a polynomal f(x S(n, k that s reducble s O(N 23/18+ε. Under the assumpton of the Lndelöf hypothess, a result of Gang Yu [9] can be used to mprove our estmate for the number of exceptonal pars (n, k to O(N 1+ε. A further mprovement to O(N log 3 N s possble under the Remann hypothess by a classcal result of Atle Selberg [7]. Based on the man result n [1], one can easly modfy our approach to show that for each postve nteger n, there are at most O(n dfferent postve ntegers k n 2 for whch S(n, k contans a reducble polynomal. Ths mples the remark n the ntroducton that for fxed n, the number of dstnct roots of P n,k (x for k n 2 s n 2 /2 as n tends to nfnty. Our second result s an explct crteron for the rreducblty of F n,k (x. Theorem 2. If there s a prme p > k that exactly dvdes n(n k, then F n,k (x s rreducble for every choce of ntegers a 0, a 1,..., a k wth each havng all of ts prme factors k. 2

3 Theorem 2 has a smple proof based on Esensten s crteron. It mples, n partcular, that f n s a prme, then P n,k (x s rreducble for every k {1, 2,..., n 1}. Ths then resolves the problem of Scherbak n the case that n s a prme. Our thrd and fnal result concernng the rreducblty of F n,k (x s as follows. Theorem 3. Let k be a fxed nteger 3. There s an n 0 = n 0 (k such that f n n 0, then F n,k (x s rreducble for every choce of ntegers a 0, a 1,..., a k wth each havng all of ts prme factors k. The value of n 0 (k n ths last result, beng based on the solutons to certan Thue equatons, can be effectvely determned. The result s of added nterest as the proof of Theorem 1 reles on consderng k large. Thus, the proof of Theorem 1 gves no nformaton about the stuaton n Theorem 3, where k s fxed and n s large. In the case that k = 1, the polynomals F n,k (x are lnear and, hence, rreducble. In the case that k = 2, our approach does not apply; but we note that the polynomals P n,2 (x are easly seen to be rreducble for n 3 as P n,2 (x has magnary roots for such n. 3 The Proofs Proof of Theorem 1. Let f(x = k =0 d x Z[x] wth d k d 0 0. In the argument, we wll make use of the Newton polygon of f(x wth respect to a prme p. The Newton polygon of f(x wth respect to p can be defned as the lower part of the convex hull of the ponts (, ν p (d where 0 k and ν p (m s defned to be the nteger r satsfyng p r m and p r+1 m. Thus, the Newton polygon has ts left most endpont beng (0, ν p (d 0 and ts rght most endpont beng (k, ν p (d k. A theorem of Gustave Dumas [2] asserts that, for a fxed prme, the Newton polygon of a product of two polynomals can be obtaned by translatng the edges of the Newton polygons of each of the polynomals. The endponts on the translaton of an edge always occur at lattce ponts. For the proof of Theorem 1, we wll use a specfc consequence of ths result: If the lattce ponts along the edges of the Newton polygon of f(x wth respect to p consst of (0, ν p (d 0, (k, ν p (d k and only one addtonal lattce pont, say at (u, v, then ether f(x s rreducble or t s the product of an rreducble polynomal of degree u tmes an rreducble polynomal of degree k u. We note that the lattce pont (u, v n ths context need not be one of the ponts (, ν p (d (for example, consder f(x = x 2 + 4x + 4 and p = 2. Consder n suffcently large. Let p denote the th prme, and let t be maxmal such that p t < n. Denote by δ(n the dstance from n to p t 1 so that δ(n = n p t 1. Suppose that k satsfes 2δ(n < k < n δ(n. We show that n ths case, the polynomal f(x = F n,k (x s rreducble over Q (ndependent of the choces of a as n the theorem. Frst, we explan why ths mples our result. For the moment, suppose that we have shown that F n,k (x s rreducble over Q for n suffcently large and 2δ(n < k < n δ(n. Let ρ(n = p t 1 where t s defned as above. It follows that the number of pars (n, k as n the theorem for whch there exsts a reducble polynomal f(x S(n, k s δ(n (n ρ(n (n p t 1. n N n N 2<p t<n n N ρ(n=p t 1 Ths last double sum can be handled rather easly by extendng the range on n slghtly (to the least prme that s N. It does not exceed (n p t 1 (p t+1 p t 1 (p t+1 p t 1 (p t+1 p t. 2<p t<n p t<n p t+1 2<p t<n p t<n p t+1 2<p t<n Settng d = p +1 p, we deduce from the arthmetc-geometrc mean nequalty that (p t+1 p t (p t+1 p t 1 = d t (d t + d t 1 = d 2 t + d t d t d2 t d2 t 1. Hence, the number of pars (n, k as n the theorem for whch there exsts a reducble polynomal f(x S(n, k s p t<n d2 t. A theorem of Roger Heath-Brown [3] asserts that p t N d 2 t N 23/18+ε. 3

4 Therefore, Theorem 3 follows provded we establsh that F n,k (x s rreducble over Q for n suffcently large and 2δ(n < k < n δ(n. Consder n suffcently large and k an nteger n the nterval (2δ(n, n δ(n. Let p = p t and q = p t 1. Note that both p and q are greater than k. We set u and v to be the postve ntegers satsfyng p = n u and q = n v. Then 1 u < v = δ(n < k/2. Observe that the numerator of c s the product of the ntegers from n k to n nclusve but wth the factor n mssng. Also, the denomnator of c s not dvsble by any prme > k and, n partcular, by p or by q. We look at the Newton polygon of f(x wth respect to p and the Newton polygon of f(x wth respect to q. Note that ν p (n u = 1 and, for each, we have ν p (a = 0. Therefore, the Newton polygon of f(x wth respect to p conssts of two lne segments, one from (0, 1 to (u, 0 and one from (u, 0 to (k, 1. The theorem of Dumas mples that f f(x s reducble, then t must be an rreducble polynomal of degree u tmes an rreducble polynomal of degree k u. Smlarly, by consderng the Newton polygon of f(x wth respect to q, we deduce that f f(x s reducble, then t s an rreducble polynomal of degree v tmes an rreducble polynomal of degree k v. Snce k v > δ(n > u and v u, we deduce that f(x cannot be reducble. Thus, f(x s rreducble. Theorem 1 follows. Proof of Theorem 2. Esensten s crteron apples to x k F n,k (1/x whenever there s a prme p > k that exactly dvdes n (.e., p n and p 2 n. Hence, F n,k (x s rreducble whenever such a prme exsts. Also, F n,k (x tself satsfes Esensten s crteron whenever there s a prme p > k that exactly dvdes n k. Proof of Theorem 3. As n the prevous proofs, we work wth f(x = F n,k (x (where the a are arbtrary ntegers dvsble only by prmes k. Wth k fxed, we consder n large and look at the factorzatons of n and n k. Lemma 1. Let p be a prme > k and e a postve nteger for whch ν p (n = e or ν p (n k = e. Then each rreducble factor of f(x has degree a multple of k/ gcd(k, e. The proof of Lemma 1 follows drectly by consderng the Newton polygon of f(x wth respect to p. It conssts of one edge, and the x-coordnates of the lattce ponts along ths edge wll occur at multples of k/ gcd(k, e. The theorem of Dumas mples that the rreducble factors of f(x must have degrees that are multples of k/ gcd(k, e. Lemma 2. Let n be the largest dvsor of n(n k that s relatvely prme to k!. Wrte n = p e1 1 pe2 2 per r, where the p denote dstnct prmes and the e are postve ntegers. Let Then the degree of each rreducble factor of f(x s a multple of k/d. d = gcd(k, e 1, e 2,..., e r. (3 Note that n the statement of Lemma 2, f n = 1, then d = k. The proof of Lemma 2 makes use of Lemma 1. Suppose m s the degree of an rreducble factor of f(x. Then for each {1, 2,..., r}, Lemma 1 mples there s an nteger b such that me = kb. There are ntegers x for whch Hence, kx 0 + e 1 x 1 + e 2 x e r x r = d. m(d kx 0 = m ( e 1 x 1 + e 2 x e r x r = k ( b1 x 1 + b 2 x b r x r. It follows that md s a multple of k so that m s a multple of k/d as clamed. For the proof of Theorem 3, we defne d as n Lemma 2 and consder three cases: ( d = 1, ( d = 2, and ( d 3. In Case (, Lemma 2 mples f(x s rreducble. Case ( s more dffcult and we return to t shortly. In Case (, there exst postve ntegers a, b, m 1, and m 2 satsfyng n = am d 1, n k = bm d 2, and a and b dvde p k p d 1. (4 4

5 As d k and k s fxed, there are fntely many choces for d, a and b as n (4. For each such d, a and b, the possble values of n correspond to ax d gven by solutons to the Dophantne equaton ax d by d = k. The above s a Thue equaton, and t s well known that, snce d 3, t has fntely many solutons n ntegers x and y (see [8]. It follows that there are fntely many ntegers n for whch Case ( holds. Hence, for n suffcently large, Case ( cannot occur. We are left wth consderng Case (. For Case (, the defnton of d mples k s even. As we already have k 3, we deduce k 4. Lemma 2 mples that f f(x s reducble, then t factors as a product of two rreducble polynomals each of degree k/2. To fnsh the analyss for Case (, we make use of the followng. Lemma 3. Let f(x be as above wth d, as defned n (3, equal to 2. Let n be the largest dvsor of (n 1(n k + 1 that s relatvely prme to k!. Suppose ν p (n = e where p s a prme > k and e s a postve nteger. If f(x s reducble, then (k 1 e. For the proof of Lemma 3, we agan appeal to the theorem of Dumas. Suppose frst that p (n 1. Snce p > k, we deduce that ν p (n 1 = e. The Newton polygon of f(x wth respect to p conssts of two lne segments, one from (0, e to (1, 0 and one from (1, 0 to (k, e. Let d = gcd(k 1, e. As d = 2, we deduce as above that f(x s a product of two rreducble polynomals of degree k/2. The fact that f(x has ust two rreducble factors mples by the theorem of Dumas that one of these factors has degree that s a multple of (k 1/d and the other has degree that s one more than a multple of (k 1/d. We deduce that there are ntegers m and m such that k 1 d m = k 2 It follows that (k 1/d dvdes 1, whence and k 1 d m + 1 = k 2. k 1 = d = gcd(k 1, e. We deduce that (k 1 e. A smlar argument works n the case that p (n k + 1. To fnsh the analyss for Case (, we use Lemma 3 to deduce that there are postve ntegers a, b, m 3, and m 4 such that n 1 = a m k 1 3, n k + 1 = b m k 1 4, and a and b dvde p k p k 2. (5 As k s fxed, there are fntely many choces for a and b as n (5. For each of these, the possble values of n correspond to a x k determned by solvng the Dophantne equaton a x k 1 b y k 1 = k 2. As k 4, the above s a Thue equaton and has fntely many solutons n ntegers x and y. Thus, there are fntely many ntegers n for whch Case ( holds. Hence, for n suffcently large, we deduce that F n,k (x s rreducble, completng the proof of Theorem 3. References [1] R. C. Baker, G. Harman, and J. Pntz, The dfference between consecutve prmes, II, Proc. London Math. Soc. (3 83 (2001, no. 3, [2] G. Dumas, Sur quelques cas d rréductblté des polynômes à coeffcents ratonnels, Journal de Math. Pures et Appl. 2 (1906, [3] D. R. Heath-Brown, The dfferences between consecutve prmes, III, J. London Math. Soc. 20 (1979,

6 [4] I. V. Ostrovsk, On a problem of A. Eremenko, Comput. Methods Funct. Theory 4 (2004, No. 2, [5] S. Ramanuan. A proof of Bertrand s postulate, Journal of the Indan Math. Soc. 11 (1919, [6] I. Scherbak, Intersectons of Schubert varetes and hghest weght vectors n tensor products sl N+1 - representatons, arxv e-prnt math.rt/ , July [7] A. Selberg, On the normal densty of prmes n small ntervals, and the dfference between consecutve prmes, Arch. Math. Naturvd. 47 (1943, [8] T. N. Shorey and R. Tdeman, Exponental Dophantne Equatons, Cambrdge Tracts n Mathematcs, 87, Cambrdge Unversty Press, Cambrdge, [9] G. Yu, The dfferences between consecutve prmes, Bull. London Math. Soc. 28 (1996, Mathematcs Department Unversty of South Carolna Columba, SC USA Emal: flaseta@math.sc.edu Department of Mathematcs Towson Unversty 8000 York Road Towson, MD USA Emal: akumchev@towson.edu Dept. E & OR Tlburg Unversty, P.O. Box LE Tlburg The Netherlands Emal: d.v.pasechnk@uvt.nl 6