Some irreducibility and indecomposability results for truncated binomial polynomials of small degree

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1 Proc. Idia Acad. Sci. (Math. Sci. Vol. 127, No. 1, February 2017, pp DOI /s Some irreducibility ad idecomposability results for trucated biomial polyomials of small degree ARTŪRAS DUBICKAS ad JONAS ŠIURYS Departmet of Mathematics ad Iformatics, Vilius Uiversity, Naugarduo 24, LT Vilius, Lithuaia * Correspodig author. arturas.dubicas@mif.vu.lt; joas.siurys@mif.vu.lt MS received 4 Jue 2015; revised 4 September 2015 Abstract. I this paper, we show that the trucated biomial polyomials defied by P, (x = ( j j=0 x j are irreducible for each 6adevery + 2. Uder the same assumptio +2, we also show that the polyomial P, caot be expressed as a compositio P, (x = g(h(x with g Q[x] of degree at least 2 ad a quadratic polyomial h Q[x]. Fially, we show that for 2adm, + 1 the roots of the polyomial P m, caot be obtaied from the roots of P,,wherem =, bya liear map. Keywords. Trucated biomial expasio; irreducibility; idecomposability Mathematics Subject Classificatio. Primary: 12E10; Secodary: 12D10, 12E Itroductio ad mai results For positive itegers, let P, (x := j=0 ( x j = j ( + 0 ( x + 1 ( x ( x be a trucated biomial expasio at the -th stage. I particular, P, (x = (1 + x ad P, 1 (x = (1 + x x. The polyomial P, is clearly reducible for each 2, whereas P, 1 is reducible (over Q if ad oly if is a composite iteger. (Ideed, if is a composite iteger the P, 1 (x = (1 + x x is reducible over Q, whereas if is prime the the polyomial P, 1 is irreducible, by Eisestei s criterio. It is easy to see that the polyomial P, is irreducible for = 1 ad also for = 2 ad each 3, sice the roots of P,2 are complex cojugate umbers for 3. Apparetly, the problem of irreducibility of P, was first raised i [9]. It was observed there that P, does ot have multiple roots for >, sice by the idetity P, (x (x + 1P, (x/ = ( 1 x the oly cadidate for a double root is x = 0, which is ot the root of P,. The otrivial cases 3 ad + 2 were first cosidered by c Idia Academy of Scieces 45

2 46 Artūras Dubicas ad Joas Šiurys Filaseta et al [4], ad subsequetly by Khaduja et al [6]. It is ow that the polyomials P, are irreducible for ad for 2 <(+ 1 3 ; see [6]. Also, for each fixed there is a iteger 0 ( such that P, is irreducible for each 0 (; see [4]. It seems very liely that for every pair of positive itegers (, satisfyig the polyomial P, (x is irreducible over Q, although this is ot proved for a specific (except for the two trivial above metioed cases = 1 ad = 2. Now, we show the followig: Theorem 1.1. For positive itegers 6 ad + 2, the polyomial P, (x is irreducible over Q. For example, for = 3 oe ca easily see that the statemet of the theorem is equivalet to the assertio that the diophatie equatio m 3 m 2 + ( 1 ( 1( 2 m = has o solutios i itegers 5 ad m. Aother questio ivestigated i [4] which is still ope is whether for a fixed, two polyomials P, ad P,l ca have a commo root for > > l. The regios to which the roots of the polyomials P, belog to ad the curves to which they coverge as / teds to a limit i [0, 1] have bee ivestigated by Jaso ad Norfol [5] ad by Ostrovsii [8]. Earlier, Szegő [11] showed that the roots of the trucated expoetial fuctio x j j=0 j! ted to the (ow called Szegő curve {z C : ze 1 z = 1 ad z 1}. The roots of the trucated expoetial series were also ivestigated i [1], [2], [7]. I fact, the roots of polyomials P, (x/( ted to the Szegő curve whe ad / ted to ifiity; see [5]. Similar results for the polyomials arisig from the series (1 x trucated after terms were obtaied by She ad Strag i [10]. The ext result shows that the roots of P m, caot be obtaied from the roots of P,, where = m, by a liear trasformatio: Theorem 1.2. Assume that m,, are three distict itegers satisfyig 2 ad m, + 1. The, there do ot exist a,b R such that the idetity P m, (x = P, (ax + b holds for all x R. This, combied with the mai theorem of [3], implies that uder assumptio of irreducibility of P 1, 1 ad P m 1,l 1, the equatio P, (x = P m,l (y has oly fiitely may solutios i itegers x,y whe (, = (m, l ad, l 2 but ot both, l are equal to 2. The questio of irreducibility of trucated biomial polyomials is related to their idecomposability. Recall that a polyomial f(x Q[x] is called decomposable over Q, if it ca be writte as f(x= g(h(x with some g, h Q[x] of degree at least 2 ad idecomposable otherwise. Obviously, each decomposable polyomial should be of degree at

3 Some irreducibility ad idecomposability results 47 least 4. I [3], we observed that for = + 1 the idetity P, (x = (1 + x x = g(x 2 + x holds with g(x := ( 1/2 j=1 (( 2 2 cos ( 2πj x + 1 Z[x], (1.1 so P, 1 is decomposable for each odd 5. It seems very liely that this is the oly case whe P,, where + 1 is decomposable. We coclude with the followig result towards the idecomposability of P, which shows its idecomposability i case + 2 with a quadratic polyomial h: Theorem 1.3. For 4 ad + 1, the idetity P, (x = g(ax 2 + bx + c holds for all x R ad some a,b,c Q, a = 0, ad g Q[x], degg>1, if ad oly if is a eve umber ad = + 1. The proofs of these results come from differet sources ad are give i the ext three sectios. First, i 2 we preset some calculatios which imply Theorem 1.1 through Lemma 2.1. The, i 3 we shall give a self cotaied proof of Theorem 1.2. Fially, i 4 we will prove Theorem Proof of Theorem 1.1 For each 2, set P( := p p, where the product is tae over the primes p. Lemma 2.1. Suppose that 2 3 ad P, is reducible over Q. The at least oe of the followig holds: (a for some d 3 which is a divisor of, there exist a,b,x,y N satisfyig a,b P( d 1 ad = ax d, = by d ; (b is eve, ad there exist a,b,x,y N satisfyig a,b P( 2 ad = a x 1 + 1, = b y 1 1. The proof of the lemma is give i [4]. As we already observed i the Itroductio, the theorem is trivial for = 1 ad = 2. Now, we shall prove Theorem 1.1 for other small values of, amely, for 3 6. The algorithm is based o Lemma 2.1. Sice = 3, = 4or = 5 have oly oe divisor d 3, we eed to chec the coditio (a with d = 3, d = 4 ad d = 5, respectively.

4 48 Artūras Dubicas ad Joas Šiurys We give all the solutios of the correspodig Thue equatio i the followig three tables. ax d by d =, where d = = 3, 4, 5 ad a,b satisfy coditio (a of the lemma. (The algorithms for the explicit determiatio of all solutios of a give Thue equatio are based o Baer s theory of liear forms i logarithms; see [12]. It turs out that there are ot may solutios. I each case we verify that the correspodig polyomial P,, where = ax d 5, is irreducible over Q. Case = 3, d = 3. a b x y Is P, irreducible? True True True True True Case = 4, d = 4. a b x y Is P, irreducible? True True True True True Case = 5, d = 5. a b x y Is P, irreducible? True True True True True True True True True True True True True

5 Some irreducibility ad idecomposability results 49 This proves the theorem for = 3 ad = 5. For = 4, we eed to chec coditio (bofthelemma: Coditio (b for = 4. a b x y Is P, irreducible? True True True True This proves the theorem for = 4. It remais to prove the theorem for = 6. First we chec coditio (b: Coditio (b for = 6. a b x y Is P, irreducible? True True True True True True True True True True True True True True Also, for = 6, the coditio (a should be checed twice: for d = 3 ad for d = 6. Case = 6, d = 3. a b x y Is P, irreducible? True True True True True True True True True True True True True True True True True True

6 50 Artūras Dubicas ad Joas Šiurys Case = 6, d = 6. a b x y Is P, irreducible? True True True True True True True True True True True True True True True True True This completes the proof of the theorem for = 6. The computatio time for 6is quite short (less tha 15 mi. For larger, the computatio time icreases drastically. 3. Proof of Theorem 1.2 Assume that for some a,b R, we have P, (ax + b = j=0 ( (ax + b j = j Clearly, a = 0. The coefficiets for x o both sides must be equal, so ( m a = ( = j=0 ( m x j = P m, (x. (3.1 j m(m 1...(m + 1 ( 1...( + 1. (3.2 Similarly, from the fact that the coefficiets for x 1 must be equal we derive that ( m = 1 ( ( a 1 b + a 1. 1 Multiplyig both sides of this equality by a ad dividig by ( m 1 i view of (3.2 we deduce that a = ( ( m b ( m + 1( ( m 1( ( m 1( = (m + 1b + m (3.3

7 Some irreducibility ad idecomposability results 51 Next, usig the fact that the coefficiets for x 2 i (3.1 must be equal as well, we fid that ( ( ( ( ( m = a 2 b 2 + ( 1a 2 b + a This time, we multiply both sides by a 2 ad divide by ( m 2. Isertig the value a as i (3.2, we fid that a 2 = ( ( ( m 2 b 2 ( m + 2( ( ( 1 ( 1 m b + ( m 2( ( m 2( ( m 2( This yields (after dividig by (m + 1(m + 2 the followig equality: a 2 (m + 1(m + 2 = b2 2 + b ( + 1( + 2. (3.4 By (3.3, we have b + 1/( + 1 = a/(m + 1. Hece, b 2 2b ( = b ( a 2 = (m ( Isertig this ito (3.4 we obtai a 2 (m + 1(m + 2 a 2 2(m = ( + 1( ( Cosequetly, which implies. a 2 (m 2(m (m + 2 = ( 2( ( + 2, a 2 = (m + 12 (m + 2( ( ( + 2(m. (3.5 Now, as a = (a 2 /2, combiig (3.2 ad (3.5 we fid that Therefore, m(m 1...(m + 1 ( 1...( + 1 = (m + 1 (m + 2 /2 ( /2 ( + 1 ( + 2 /2 (m /2. m(m 1...(m + 1(m /2 ( 1...( + 1( /2 (m + 1 (m + 2 /2 = ( + 1 ( + 2 /2.

8 52 Artūras Dubicas ad Joas Šiurys Settig M := m + 1, N := + 1 ad G (x := (1 x /2 (1 + x 1 /2 (1 + 2x(1 + 3x...(1 + ( 1x, (3.6 i view of m...(m + 1(m /2 (1 (m + 1 (m + 2 /2 = + 1 ( ( M 1 /2 M M M + 1 ( = 1 1 /2 ( /2 ( ( = G (1/M, M M M M we must have G (1/M = G (1/N. (3.7 It remais to show that (3.7 does ot hold for M>N(which ca be assumed without restrictio of geerality, where N = I what follows we will prove this step by step for various 2. For = 2, by (3.6, we have G 2 (x = 1 x. Thus, (3.7 does ot hold i view of M = N. For = 3, G 3 (x = (1 x 3/2 (1 + x 1/2 (1 + 2x,so G 3 (x G 3 (x = 3 2(1 x 1 2(1 + x x = x(5 + 4x (1 x 2 (1 + 2x. Hece, the fuctio G 3 (x i the iterval (0, 1 is decreasig, so (3.7 does ot hold i view of M = N. Suppose ext that 4. The, usig (3.6 we fid that G (x /2 = G (x 1 x / x x ( 1x. (3.8 Now, for 14, we will show that the derivative G (x is positive i the iterval (0, 1/2. Ideed, the fuctio o the right-had side of (3.8 is decreasig for x > 0, sice its derivative, amely /2 (1 x 2 + /2 (1 + x 2 1 (1 + x 2 4 (1 + 2x 2 ( 1 2 (1 + ( 1x 2, is egative i view of 1/(1 + x 2 < 1/(1 x 2. So the smallest value of G (x i the iterval [0, 1/2] is equal to G (1/2. Evaluatig (3.8 at x = 1/2, we fid that G (1/2 G (1/2 = Thus, for 14 we obtai j=2 1 2j 2 + j = 2 + j=2 2j 8 3(2 + j. G (1/2 G (1/2 G 14 (1/2 13 G 14 (1/2 = 2 + 2j 8 = > > 0. 3(2 + j j=2 Therefore, for every 14, the fuctio G (x is positive for 0 < x 1/2, ad, cosequetly, the fuctio G (x is icreasig i the iterval (0, 1/2. As0< 1/M <

9 Some irreducibility ad idecomposability results 53 1/N = 1/( + 1 1/2, we obtai G (1/M < G (1/N. Hece (3.7 does ot hold for each 14. It remais to cosider the remaiig itegers i the rage We split the proof ito three cases, first, 6 13, the = 5 ad, fially, = 4. Observe that for 6, the derivative G (x is positive i the iterval (0, 1/3, sice, arguig as above, we ca coclude this from the iequality G 4 (1/3 = > > 0. G 4 (1/3 So G (x is icreasig i the iterval (0, 1/3. Hece G (1/M = G (1/N caot hold for M>N 3. Thus, it ca oly hold whe N = 2. Let us chec that this is ot the case. Ideed, for = 13, we have G 13 (1/2 >8428 ad G 13 (1/3 <4532, so G 13 (1/M < G 13 (1/2 for M 3. Similarly, we obtai G 12 (1/2 >2085 > 1282 >G 12 (1/3, G 11 (1/2 >555 > 389 >G 11 (1/3, G 10 (1/2 >160 > 127 >G 10 (1/3, G 9 (1/2 >50 > 45 >G 9 (1/3, G 8 (1/2 = 17.5 > 17.3 >G 8 (1/3. As above, this implies G (1/M < G (1/2 for 8 12 ad M 3. This completes the proof that (3.7 does ot hold for For = 7, we have G 7 (1/2 = 35 3/9 = Cosequetly, G 7 (1/3 = > >G 7 (1/2 ad G 7 (1/M G 7 (1/4 = < <G 7 (1/2 for each M 4. Hece, the equality (3.7 does ot hold for = 7. For = 6, we have G 6 (1/2 = 35/12 = This time, G 6 (1/3 = > >G 6 (1/4 = > >G 6 (1/2 ad G 6 (1/M G 6 (1/5 = <G 6 (1/2 for each M 5, so the equality (3.7 does ot hold for = 6. For = 5, usig (3.8 we fid with Maple that the derivative G 5 (x is positive i the iterval (0, 1/4. Also, we have G 5 (1/2 = 5 3/6 = ad G 5 (1/3 = 35 2/27 = To show that oe of the values of G 5 (1/M, where M 4, are equal to oe of the two values above, let us evaluate G 5 (x at x = 1/4, 1/10, 1/11. We fid the values , , , respectively. Sice G 5 (1/4 < G 5 (1/3 ad >G 5 (1/2 >1.4185, this shows that the equality (3.7 does ot hold for = 5. Fially, for = 4, from (3.8 oe ca easily chec with Maple that the derivative G 4 (x is positive i the iterval (0, 1/6. Now, there are four potetial cadidates to which G 4 (1/M ca be equal for some M 6, amely, G 4 (1/2 = 5/6 = , G 4 (1/3 = 10/9 = , G 4 (1/4 = 189/160 =

10 54 Artūras Dubicas ad Joas Šiurys ad G 4 (1/5 = 448/375 = Note also that G 4 (1/6 = 25/21 = Here, G 4 (1/2 <1 <G 4 (1/3 <G 4 (1/4 <G 4 (1/6 <G 4 (1/5. With M 7 icreasig, the values of G 4 (1/M are decreasig. Evaluatig G 4 (1/M at M = 7, 14, 15, we fid the values G 4 (1/7 = 405/343 = < <G 4 (1/4 ad G 4 (1/14 = > >G 4 (1/3 = > >G 4 (1/15 = Thus, oe of the values G 4 (1/M, where 7 M 14 is equal to oe of the G 4 (1/M with M i the rage 2 M 6. Furthermore, for M 15 the values of G 4 (1/M are decreasig (as M is icreasig, so that G 4 (1/3 >G 4 (1/M > G 4 (0 = 1 for each M 15. It follows that G 4 (1/M = G 4 (1/N for M>N 2, amely, (3.7 does ot hold for = 4. This completes the proof of the theorem. Below, we shall give a alterative (etirely differet proof for odd. This proof is due to a referee ad is much shorter. Assume that P m, (x = P, (ax + b for all x R. The P m, (x = ap, (ax + b. Isertig these relatios ito P, (y (y + 1P, (y/ = ( 1 y, where y = ax + b, we obtai P m, (x ax + b + 1 ( 1 P m, a (x = (ax + b. Subtractig it from the idetity P m, (x (x + 1P m, (x/m = ( m 1 x ad usig P m, (x = mp m 1, 1(x, we fid that ( m (b + 1m a ( ( m 1 1 P m 1, 1 (x x+ = x (ax+b. a (3.9 The roots z j, j = 1,...,, o the right-had side of (3.9 satisfy ρζ j z j = az j + b, where ζ j = e 2πij/ ad ρ>0isareal root of ρ = ( m 1 ( / 1. The left-had side of (3.9 has a real root (correspodig to the liear factor ad, by Lemma 4.1 below, 1 complex roots of P m 1, 1, sice 1 is eve. The real root o the right-had side occurs for j =. Thus, the roots of P m 1, 1 are 1 roots z j = b/(ρζ j a, where j = 1,..., 1. Sice P m 1, 1 (x = 1 ( m 1 j=0 j x j, by the Vieta s formulas, the sum over reciprocal roots 1/z j is equal to 1 m. Therefore, which implies 1 m = 1 j=1 1 1 ρζ j a = = ρ z j b b j=1 ( 1a, b (m 1b = ρ + ( 1a. (3.10

11 Some irreducibility ad idecomposability results 55 O replacig x by x/a b/a i the formula P m, (x = P, (ax + b, we obtai P m, (x/a b/a = P, (x. Thus, chagig the roles of ad m, we eed to replace (a, b by (1/a, b/a, ad, as it is easy to see, ρ trasforms ito 1/ρ. Thus, the formula (3.10 traslates ito the formula ( 1( b/a = 1/ρ + ( 1/a, amely, ( 1b = a/ρ + 1. (3.11 As is odd, the umber a is positive, by (3.2. Thus, i view of ρ>0 the right-had sides of (3.10 ad (3.11 are both positive. Cosequetly, b>0 ad b >0, which is impossible. 4. Proof of Theorem 1.3 For = + 1 the idetity of the theorem holds i view of P, 1 (x = g(x 2 + x, where g Z[x] of degree ( 1/2 is give by (1.1. Below, we shall use the followig lemma proved i [3]: Lemma 4.1. Let, N be such that + 1. The P, has exactly oe real root if is odd, ad o real roots if is eve. For coveiece of the reader, we will give the proof of the lemma for odd (as this is the oly part which will be used below. Ideed, if is odd, the P, must have a real root. Assume that >1ad that P, has aother real root. (As we already observed earlier all the roots of P, are simple. Let x 0 <x 1 be two smallest real roots of P, (x. The x 0 <x 1 < 0 ad P, (x > 0 for each x (x 0,x 1. Thus, there exists x 2 (x 0,x 1 such that P, (x 2 = 0. Hece, by the idetity P, (x (x + 1P, (x/ = ( 1 x, we get ( 1 0 > x2 = P,(x 2 >0, a cotradictio. Now, let us show the impossibility of the represetatio P, (x = g(ax 2 + bx + c for Evidetly, 4 must be a eve umber. Hece, by Lemma 4.1, the polyomial P 1, 1 (x = P, (x/ of odd degree 1 has exactly oe real root, say s. Write ax 2 + bx + c = a(x + v 2 + w, where v := b/(2a ad w := c 2 b 2 /(4a. The, P, (x = g(ax 2 + bx + c is equivalet to P, (x v = g(ax 2 + w. (4.1 By taig the derivatives of both sides, we fid that P 1, 1 (x v = 2axg (ax 2 + w. Sice x = 0 is the zero of the right-had side, ad the oly real zero of the left-had side is v + s, wemusthavev = s. Thus, from (4.1 it follows that g(ax 2 + w = P, (x + s = 1 + (x + s + + ( (x + s ( (x + s.

12 56 Artūras Dubicas ad Joas Šiurys Here, the coefficiets of the left-had side, g(ax 2 + w, are all zeros for odd powers of x. Therefore, the coefficiet of the right had side for x 1 must be zero, so that ( ( + s = 0. 1 This yields s = 1/( + 1. Similarly, as the coefficiet for x 3 of the right-had side is zero, we obtai ( ( ( ( ( ( 1 + s( 2 + s 2 + s 3 = This is equivalet to 1 ( + 3( + 2( s ( + 2( + 1 s 2 + 2( s3 6 = 0. Isertig ito this equality s = 1/y, where y := + 1 > 1, ad multiplyig by (y + 2(y + 1y 3 we further obtai y 2 (y + 2y + (y + 2(y (y + 2(y = 0. This leads to y 2 3y + 2 = 0. I view of y>1 the oly possibility is y = + 1 = 2, ad hece = + 1, cotrary to our assumptio + 2. Acowledgemets The authors would lie to tha the referee for givig a alterative proof of Theorem 1.2 i case is odd. Refereces [1] Bucholtz J D, A characterizatio of the expoetial series, Amer. Math. Mothly 73(4 ( [2] Carperter A J, Varga R S ad Waldvogel J, Asymptotics for the zeros of the partial sums for e z, Rocy Moutai J. Math. 21 ( [3] Dubicas A ad Kreso D, Diophatie equatios with trucated biomial polyomials, Idag. Math. 27 ( [4] Filaseta M, Kumchev A ad Pasechi D V, O the irreducibility of a trucated biomial expasio, Rocy Moutai J. Math. 37 ( [5] Jaso S ad Norfol T S, Zeros of sectios of the biomial expasio, Electro. Tras. Numer. Aal. 36 (2009/ [6] Khaduja S K, Khassa R ad Laishram S, Some irreducibility results for trucated biomial expasios, J. Number Theory 131 ( [7] Newma D J ad Rivli T J, The zeros of the partial sums of the expoetial fuctio, J. Approximatio Theory 5 ( ; correctio ibid. 16 ( [8] Ostrovsii I V, O a problem of A. Eremeo, Comput. Methods Fuct. Theory 4 (

13 Some irreducibility ad idecomposability results 57 [9] Scherba I, Itersectios of Schubert varieties ad highest weight vectors i tesor products sl N+1 -represetatios, preprit at arxivmath: v3 (2004 [10] She J ad Strag G, Asymptotic aalysis of Daubechies polyomials, Proc. Amer. Math. Soc. 124 ( [11] Szegő G, Über eie Eigeschaft der Expoetialreihe, Sitzugsber. Berl. Math. Ges. 23 ( [12] Tzaais N ad de Weger B M M, O the practical solutio of the Thue equatio, J. Number Theory 31 ( COMMUNICATING EDITOR: S D Adhiari