The Parity of the Number of Irreducible Factors for Some Pentanomials

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1 The Party of the Nuber of Irreducble Factors for Soe Pentanoals Wolfra Koepf 1, Ryul K 1 Departent of Matheatcs Unversty of Kassel, Kassel, F. R. Gerany Faculty of Matheatcs and Mechancs K Il Sung Unversty, Pyongyang, D.P.R.Korea Aprl 5, 008 Abstract It s well known that Stckelberger-Swan theore s very portant for deternng reducblty of polynoals over a bnary feld. Usng ths theore t was deterned the party of the nuber of rreducble factors for soe knds of polynoals over a bnary feld, for nstance, trnoals, tetranoals, self-recprocal polynoals and so on. We dscuss ths proble for type II pentanoals naely x + x n+ + x n+1 + x n + 1 F x. Such pentanoals can be used for effcent pleentng ultplcaton n fnte felds of characterstc two. Based on the coputaton of dscrnant of these pentanoals wth nteger coeffcents, t wll be characterzed the party of the nuber of rreducble factors over F and be establshed the necessary condtons for the exstence of ths knd of rreducble pentanoals. Keywords: Fnte feld, Irreducble polynoals, Type II pentanoals 1 Introducton For purpose of effcently pleentng feld arthetc n fnte feld F, t s desrable to take an rreducble equally spaced polynoalesp of degree over F of the for x kd + x k 1d + + x d + x d + 1 where kd =, or an rreducble polynoal over F of degree and low weght such as trnoal, pentanoal and so on. Unfortunately, rreducble equally spaced polynoals are very rare and an rreducble trnoal also do not always exst for a gven degree over F. On the other hand, now a conjecture that there exsts an rreducble pentanoal of degree over F for each 4 reans open, but there exsts no known value of 4, for whch an rreducble pentanoal does not exst. In 8, type I and type II pentanoals 1

2 over F were defned as follows: Type I : x + x n+1 + x n + x + 1, where n / 1 Type II : x + x n+ + x n+1 + x n + 1, where 1 n / 1 The authors presented the parallel ultplers for eleents of F defned by these types of pentanoals. In 7, t was proposed another ultpler for eleents of F defned by a type II pentanoal wth better te and gate coplexty copared to ones n 8. Though type I and type II rreducble pentanoals are abundant, they do not exst for each gven degree. Stckelberger-Swan theore s an portant tool for deternng reducblty of a polynoal over a fnte feld. Usng ths theore, Swan proved n 196 that trnoals x 8k +x +1 F x wth 8k > have an even nuber of rreducble factors and so can not be rreducble. Many researchers used ths theore for deternng the party of the nuber of rreducble factors of several knds of polynoals over F.1, 3 5 For applyng the Stckelberger-Swan theore, t s essental to copute a dscrnant of a polynoal wth the nteger coeffcents odulo 8. In ths paper we copute the dscrnant of onc lft for the type II pentanoals of even degrees over F to the ntegers and characterze the party of the nuber of rreducble factors for these pentanoals. Our an results are theore 3-6 that show the condtons for type II pentanoals to have an even nuber of rreducble factors over F. Usng our results one can fnd soe fales of type II pentanoals that are reducble over F. Prelnares Let K be a feld, and let f x = a x x K x where x 0, x 1,, x are the roots of f x n a certan extenson of K. Then a dscrnant Df of f s defned as follows: Df = a <jx x j 1 The followng theore, called the Stckelberger-Swan theore, s very portant for deternng reducblty of a polynoal over F. Theore 1 9 Suppose that the polynoal f x F x of degree has no repeated roots and let r be a nuber of rreducble factors of f x over F. Let F x Z x be any onc lft of f x to the ntegers. Then r od f and only f D F 1 od 8. Ths theore asserts that by coputng the dscrnant of F x Z x odulo 8, one can deterne the party of the nuber of rreducble factors of the polynoal f x over F. For nstance, f s even and D F 1 od 8, then f x has an even nuber of rreducble factors over F and therefore t s reducble over F. Now we scrbe the prelnary results on the dscrnant and the resultant. Let f x be the sae as above and let g x = b n 1 x y j K x, where y 0, y 1,, y n 1 are the roots of g x n a certan extenson of K. The resultant

3 R f, g of f x and g x s n 1 Rf, g = 1 n b The resultant has the followng propertes. f y j = a n Lea 1 4, 9 1 If g = fq + r, Rf, g = Rf, r If c s constant, Rf, c = c = Rc, f 3 Rx, g = g 0, Rf, x = f 0 4 Rf 1 f, g = Rf 1, grf, g, Rf, g 1 g = Rf, g 1 Rf, g g x There s an portant relaton between the dscrnant and the resultant such as Df = 1 / R f, f 3 Now we focus to a onc polynoal f x = x + a 1 x + + a = x x over K. It s well known that the coeffcents a k of f x are all the eleentary syetrc polynoals of x : a k = 1 k x 1 x x k 0 1< < k < for 1 k <. Snce each a k K, we have that S x 0, x 1,, x K for any syetrc polynoal S K x 0, x 1,, x. For any ntegers p, q and k, 0 k, let S k,p = 1,, k =0 j l p x 1 x p k, S k,p = x 1,, k =0 1< < < k p 1 x p k, S p,q =, j x p xq j where j l eans that f j l, then j l. We denote S 1,p = S 1,p sply as S p and put S 0,p = S 0,p = 1. Then we have the followng lea easly. Lea 1 S 0 = S 1,0 = S 1,0 = S p,q = S p S q S p+q 3 S k,p = k! S k,p P roof. It s trval fro ther defnton. Newton s forula s very useful n coputng the dscrnant of a polynoal. Theore 6, Theore 1.75 Let fx, S p and x 0, x 1,, x be as above. Then for any p 1, S p + S p 1 a 1 + S p a + + S p n+1 a n 1 + n S p na n = 0 4 where n = n p,. 3

4 3 The party of the nuber of rreducble factors n case of n even In ths secton we consder the party of the nuber of rreducble factors of the followng pentanoal f x = x + x n+ + x n+1 + x n + 1 F x 5 where both of and n are assued to be even and suppose that n <. The case of n odd s ore coplex and t wll be consdered n the next secton. Let F x Z x be the onc lft of f x to the ntegers whch has all ts coeffcents equal to 0 or 1. The dervatve of F x s F x = x + n + x n+1 + n + 1x n + nx n 1 = = x n 1 + n + x + n + 1x + n Let G x = + n + x + n + 1x + n. Then by lea 1 and the equaton 3, we have DF = 1 / R F, F = 1 / R F, x n 1 R F, G = 1 / F 0 n 1 R F, G = 1 / R F, G By, we can wrte R F, G as R F, G = u + vx + wx + r where u =, v = n +, w = n + 1, r = n and x 0, x 1,, x are the roots of F x. Usng x = 1 and the fact that u, v and r are even, we expand the above expresson to get R F, G = w + uw + vw + w r <j + uw r j 1 x + v w <j x 1 x 1 j + uvw j + u w <j x x j + w r 1 1 j 1 x j x 1 1 x 1 j + vw r x x 1 j + S x 0, x 1, x j where S x 0, x 1, x s an nteger congruent to 0 odulo 8. Therefore R F, G w + uw S n 1 + u w S, n 1 + vw S v w S,1 + w rs 1 + w r S, 1 + uvw S n 1,1 + + uw rs n 1, 1 + vw rs 1, 1 od 8 6 Dvde to two cases. Case 1. n + equvalently n 1 4

5 Lea 3 If n + then R F, G 1, = n + 1 3n n, = n n, n + 4 < < n + od 8 7 P roof. All coeffcents of the polynoal F x are 0 except a n = a n 1 = a n = a = 1 and by Newton s forula t thus satsfes S 1 = = S n 3 = 0. Applyng Newton s forula and lea to F x and ts recprocal x F x 1, we can copute each ter n 6 as follows. S 1 = 0, S 1 = a = 0, {, n = 4 S = S 1 a 1 + a = a = 0, n > 4 {, n = S = S 1 a + a = a = 0, n > S,1 = 1 { S 1, n = 4 1 S = 0, n > 4 S, 1 = 1 { S 1, n = 1 S = 0, n > S 1, 1 = S 1 S 1 S 0 = S n = S n 3 a S 1 a n 3 + n a n = n S n 1 = S n a S 1 a n + n 1 a n 1 = n 1 { S4 = S S n = a + 4a 4 = S 4 =, n = 4 n, n > 4 S n 1 = S n + S n 1 + S n + n 1 a n 1 { S n + n 1 + n n 1, = n + = S n + n 1 + n, < n + 1 od 8, = 6, n = n + 1 od 8, = n +, n > 1 od 8, = n + 4, n > 3 n 1 od 8, n + 4 < < n + S, n 1 = 1 S n 1 S n 1 0 od 8, = 6, n = 1 nn + 1 od 8, = n +, n > 1 od 8, = n + 4, n > 1 n 1 n 4 od 8, n + 4 < < n + {, = n + 4 S n 1,1 = S n 1 S 1 S n = n, > n + 4 S n 1, 1 = S n 1 S 1 S n = n By substtuton of above values to 6 we have 7. 5

6 Theore 3 If n +, the pentanoal 5 has an even nuber of the rreducble factors over F f and only f one of the followng condtons holds : 1 = n + = n + 4, n 0, od 8 3 n + 4 < < n + and a 0 od 8, or b od 8, n, 6 od 8, or c 6 od 8, n 0, 4 od 8. P roof. The asserton of the theore s followed by applyng Theore 1 to DF = 1 / R F, G 1, = n + 1 n+4n+3/ 1 3n n, = n / n 1 3, n + 4 < < n + od 8 Corollary 1 Let n+ and suppose that the pentanoal 5 s rreducble over F. 1 If n 0 od 8, then, 4 od 8 If n od 8, then 4, 6 od 8 3 If n 4 od 8, then ether, 4 od 8 or = n If n 6 od 8, then ether 4, 6 od 8 or = n + 4. Case. > n + equvalently n 1 > Lea 4 If > n +, then R F, G n + { 4, n = 0, n > od 8 8 P roof. In ths case by Newton s forula we get S n 1 = S n + S n 1 + S n + S n = S n + S n 1 + S n because S n = 0, for n < n. Snce > n + and n s even, a 1 = a = 0 and therefore S 1 = 0, S = 0, S,1 = 1 S 1 S = 0, {, n = S 1 = a = 0, S = a = 0, n > S 1, 1 =, S, 1 = 1 { S 1, n = 1 S = 0, n > S n = n, S n 1 = n 1, S n = n S n 1 = n + n 1 + n = 3 n 1 S, n 1 = 1 S 1 n 1 S n 1 = n 1 n 4 S n 1,1 = S n 1 S 1 S n = n S n 1, 1 = S n 1 S 1 S n = n. 6

7 Fro theses values we obtan R F, G n + 1 n + 1 n n + 1 n 1 n 4 + { n + 1 n, n = 0, n > + nn + n n n + 1 n n + n + 1 n { 4, n = n + 0, n > Theore 4 If > n +, the pentanoal 5 has an even nuber of the rreducble factors over F f and only f one of the followng condtons holds : 1 n =, 4, 6 od 8 n > and a 0 od 8, or b od 8, n, 6 od 8, or c 6 od 8, n 0, 4 od 8. P roof. We can obtan easly that f n =, then D F = 1 / RF, G 1 / / od 8 and f n >, then D F 1 / n od 8 We obtan the result by applyng theore 1 to the above congruence. Corollary Let > n+ and suppose that the pentanoal 5 s rreducble over F. 1 If n 0, 4 od 8, then, 4 od 8. If n =, then 0, od 8. 3 If n od 8 and n,then 4, 6 od 8. 4 If n 6 od 8, then 4, 6 od 8. 4 The party of the nuber of rreducble factors n case of n odd Ths secton deals wth the type II pentanoals f x = x + x n+ + x n+1 + x n + 1 F x, 1 n 1 9 n the case of n odd. Let F and G be sae as before. The slar consderaton as prevous secton akes us to get R F, G = v + r + uv + v w x 1 + ur + wr x + u v <j j 7

8 + u r <j + uv w v k r k + k=1 + j + v w <j + uwr 1 1 uv k r k 1 k=1 u v k r k 3 + k=1 v k wr k 1 + k=1 v k w r k 3 + k=1 uv k wr k 3 + k=1 x 1 j 1< <j k x j 1 x j k x j 1,,j k j 1< <j k 1<,j 1< <j k 1, j 1,,j k x x j 1,,j k j 1< <j k 1<,j 1< <j k 1, j 1,,j k 1,j 1< <j k 1, j 1,,j k n x j 1 x j k x n 1 j 1 x j k x 1 x 1 j + w r x x j <j 1 x 1 x j 1 x j k x x x 1 j 1 x j k n x 1 x x j 1 x j k + S x 0, x 1,, x where S x 0, x 1, x s an nteger congruent to 0 odulo 8. Now denote respectvely S, V, W, X, Y, Z the last 6 sus of above equaton, and we rewrte t as R F, G = v + r + uv + v w + u r <j + uv w x 1 + ur + wr x + u v <j j + v w <j + uwr 1 1 x 1 + S + V + W + X + Y + Z. And we use the notatons n secton to get j x 1 x 1 j + w r x x j <j 1 x 1 R F, G v + r + uv S n + ur S n + v w S 1 + wr S 1 + u v S, n + u r S, n + v w S, 1 + w r S,1 + uv w S n, 1 + uwr S n,1 + S + V + W + X + Y + Z od 8 It s necessary to copute the values of S, V, W, X, Y and Z odulo 8 to characterze R F, G. 8

9 Let V k = W k = X k = x j 1,,j k j 1< <j k 1<,j 1< <j k 1, j 1,,j k n x j 1 x j k, 1 k x n 1 x j 1 x j k, 1 k 3 x xj 1 x j k, 1 k j 1,,j k j 1< <j k Y k = Z k = 1<,j 1< <j k 1, j 1,,j k 1,j 1< <j k 1, j 1,,j k x x x 1 j 1 x j k, 1 k 3 n x 1 x x j 1 x j k, 1 k 3 For a fxed nteger, 1 <, and for each postve nteger p, let U k,p = Uk,p = p xj 1 x p j k j 1,,j k j l j t and put U 0,p = 1 for any p. Then we obtan the followng leas. Lea 5 U k,p = k 1 j k! k j! xjp S k j,p P roof. Fro defnton of U k,p, we have U k,p = Lea 6 j l j t x p j 1 x p j k kx p j 1,,j k 1 j l j t p xj 1 x p j k 1 = S k,p kx p U k 1,p = S k,p kx p Sk 1,p k 1x p U k,p = S k,p kx p S k 1,p + kk 1x p U k,p = = = S k,p kx p S k 1,p + kk 1x p S k,p k 1 kk 1 x k 1p S1,p 1 x p k = 1 j k! k j! xjp S k j,p S k,p S p = S k+1,p + k 1 j+1 k! k j! S j+1p S k j,p j=1 9

10 P roof. S k,p S p = x p j 1 x p j k j l j t = S k+1,p + = S k+1,p + = S k+1,p + kx p k j=1 x p = kx p U k 1,p + S k+1,p k 1 1 j k 1! k j 1! xjp S k j 1,p 1 j 1 k! k j! xj+1p S k j,p k 1 j+1 k! k j! S j+1p S k j,p j=1 By applyng Newton s forula and these leas we copute S, V, W, X, Y, and Z odulo 8. For splcty we assue n ths secton that and n are large, for exaple, n > 4 and > 4n. Suppose that s not dvdable by 4. If s ultple of 4, then t s slar and spler. 4.1 Coputaton of S Frst we copute the non-zero values of S p by Newton s forula slarly as prevous secton. The followng table shows the non-zero values of S p for p 1, 4 3n. These are needed for below coputaton. p S p p S p n n 3 n 1 3 n 1 n 1 n 1 3 n 3 n n n 3 n 3 n 3 n 1 3 n 1 n n 3 n 3 n n 3 n 3 3 n 3 n 1 4 n n n 1 n 1 4 4n 7 4 4n 7 n n 4 4n 6 5 n 3 n n 4 4n n 5 n 1 n 1 4 4n 4 19 n 1 n n 4 4n n 3 4 4n 5 n 1 3 n n 4 4n 1 4 4n 1 3 3n 5 3 3n 5 4 4n n 3 3n 4 3 3n 4 4 3n 6 4 3n 6 3 3n 3 7 n 1 4 3n n 5 3 3n 3 3n 4 3n n 4 3 3n 1 3 3n 1 4 3n n 3 3 3n n 4 3n 64 3n 3 n 4 3 n 4 4 3n n 1 3 n 3 3 n 3 4 3n 4 3n 3 n 33 n 10

11 Fro these values we copute S k, for 1 k by applyng lea 6. If 1 k < n 1, we see that S k, = 0. n 3 S n 1, = 1 n 1 1 = 1 n 1 S, = 1 n 3 = 1 n 1! 1! n 1!! S n 1! S n 1, S n 1 S n, = 1 n 3 n 3! S n = 1 n 3 n! S n 1, = 1 n 3 n! n 1! S n 1, S n n n! S n 1 = 1 n n 1! S n, = 1 n 3 n 1! n+1! S n+1, S n n 1 n 1! S n = 1 n 1 n! n 3 n 3 S n 1, = 1! S!, S n n 3 n 1 S n 1 = 1 n 1 S, = 1 1 = 1! =!! S! n 1, S + 1 n 3 n 1! 1! S!, S + 1 1! S n 3 Now we are ready to copute S odulo 8. Throughout ths secton we often use the fact that the quadrate of each odd nteger s congruent to 1 odulo 8. Lea 7 S 1 n+1 4 n + n 1 n n+1 1 od 8 P roof. S = k=1 v k r k S k, v n 1 r +n+1 S n 1, + v r S, + v n r n+ S n, + v n 1 r n+1 S n 1, + v n r n S n, + v n 1 r n+1 S n 1, n + n 1 n n+1 1 n 1 n + 1 n nn + + nn nn + 1 n+1 4 n + n 1 n n+1 1 od 8! 11

12 4. Coputaton of V Lea 8 For any k, 1 k, V k = k 1 j 1 k j! xj S k j, P roof. By lea and lea 5, V k = x j 1,,j k j 1< <j k = n x j 1 x j k = 1 k! U k, = x j 1,,j k j 1< <j k j 1 x j k = k 1 j 1 k j! xj S k j, Usng lea 8 and the values of S k, we have non-zero values of V k as follows. k V k k V k n 3 1 n 3 n 3 n 3 n 1 n+1 n n n n 1 n 1 n 3 n 1 n 3 n n 1 Lea 9 V n + 1 od 8 P roof. V = u v k r k 1 V k k=1 n + n 3 n +n+1 1 n 3 n 3 + n + n 1 n +n 1 1 n+1 + n + 1 n 1 1 n + n + n 1 1 n n + n 3 n n+ n + n + n n n+1 n n + n 1 n n n + n + n 3 n n+1 1 n 3 n n + 1 od Coputaton of W We frst deterne W k for each k, 1 k. To do ths we ntroduce another new notaton. For fxed dfferent ntegers 1,, 1 1, <, let Q k = xj 1 x j k 1, j 1,,j k j l j t 1

13 and put Q 0 = 1. We obtan the followng lea by a consderaton slar as lea 5. Lea 10 For any k, P roof. Q k = j 1,,j k j l j t Q k = k 1 j k! k j! xj 1 U k j, xj 1 x j k kx 1 1, j 1,,j k 1 j l j t xj 1 x j k 1 = U k, kx 1 Q k 1 = U k, kx 1 Uk 1, k 1x 1 Q k = U k, kx 1 U k 1, + kk 1x 1 Q k = = U k, kx 1 U k 1, + kk 1x 1 U k, k 1 kk 1 x k 1 1 U 1, x k 1 = 1 j k! k j! xj 1 U k j, Usng lea 5 wth already coputed non-zero values of S k,, we get the values of U k, for any k 0, 1. The result s as follows. 0 k < n 1 = U k, = 1 k k! x k n 1 k < = U k, = 1 k k! k < n = U k, = 1 k k! U n, = n! U n 1, = n 1! n k < n 1 n k+1 k! + 1 k+1 k! x k + x k n 1 x k + x k n x k x + 1 = U k, = 1 k k! x k + x k n 1 + x k x k n + x k n 1 + x k n k < = U k, = 1 k k! x k + x k n 1 + x k x k n + x k n 1 + x k n x k n 1 Now we fnd the expresson of W k. By lea, lea 10 and above equatons, we obtan the followng for any k 1, 3. Snce U k, depends on the ndex k, the expresson for W k s also dfferent accordng to k. For exaple, f k < n, then W k s the su of frst 6 ters n the rght sde of the last equaton. We note the non-zero values of W k for all k 1, 3 n the followng table. 13

14 k W k n 1 1 n n + 1 n 1 1 n n + 1 n 5 1 n n 33 3n 5 n 3 1 n n + 1 n 1 1 n n n n n 4n n +n 1 1 n n + n 6 + n + 1 +n 1 4 n 1 n 4 n 3 n n 3 1 n + n 5 + 5n n n n 3 n 5 1 n+1 n + n 5 + 5n + 6 Now we derve a forula for coputaton of W odulo 8. Lea 11 W 1 n n+1 P roof. It follows fro the coputaton of 1 n + n+1 n n+1 + n + 3 od 8 3 W = u k=1 odulo 8 based on the above table. 4.4 Coputaton of X, Y, Z v k r k W k The coputatons of X, Y, Z are very slar to ones of V, W, so we ot ther concrete process but pont out only the dfferences and results. For any k, 1 k, X k = x k 1 j 1 k j! xj S k j, k X k k X k n 3 1 n 1 n n 1 1 n 1 1 n+1 n 3 n 1 n 3 n + n 1 n 1 1 n 1 n Lea 1 X 0 od 8 14

15 For any k, 1 k, Y k = 1 1k+1 k + 1 S k k k S j+1 S k j k+1 k n S k n+3 k n k Sj+1 S k j n+ + 1 k+1 k + 1 S k + k + 1 k S j+1 S k j k k + n + 3 S k n k n 1k 1 Sj+1 S k j n k k + n + S k n k n 1 1k 1 Sj+1 S k j n k k + n + 1 S k n+ + 1 k n 1k 1 Sj+1 S k j n k+1 k n S k n 1+ k n k Sj+1 S k j n 1+1 k Y k k Y k n 3 1 n n 1 n n n n + 1 Lea 13 Y 1 n 1 1 n + 1 n + n+1 n n 1 od 8 15

16 Z k = k 1 1 x 1 k! Q k = 1 k+1 k + 1S n+k+1 k + 1 k S n+j S k j k k + 1 k k k n S k+ k n 1 k S n+j S k j n 1+1 k + 1 S k n+1 S n+j S k j k k + n + 3 S k +n+5 k n + 1 k k k + n + S k +n+3 k n k k k + n + 1 S k +n+1 k n + 1 k k k S n+j S k j n +1 S n+j S k j n 1+1 S n+j S k j n+1 k n k n 1 Z k S k + S n+j S k j n 1+1 n 1 1 n 1 1 n + 1 n 5 1 n+1 1 n 3 n n 3 1 n 1 1 n n + 3 n 1 1 n 1 1 n n n 3 +n 1 1 n 1 1 n 1 n 3 n n n n 5 1 n+1 n n + 4n 4 n 3 1 n 1 n n 16

17 Lea 14 Z 0 od 8 Now usng above leas we copute the resultant RF, G odulo 8 n case of even, not dvdable by 4 and n odd. Lea 15 Under the above assupton, t satsfes RF, G 1 n + 1 n n+1 n n+1 n + n+1 od 8 P roof. RF, G v + r + uv S n + ur S n + v w S 1 + wr S u v S, n + u r S, n + v w S, 1 + w r S,1 + uv w S n, 1 + uwr S n,1 + S + V + W + X + Y + Z od 8. Snce we assued that n > 6 and > 6n, S 1 = S 1 = S, 1 = S,1 = 0. And by lea Thus we get S n, 1 = S n S 1 S n 3 = 0 S n,1 = S n S 1 S n+1 = 0 RF, G + n + S n + n S n + 1 S n S n + 1 S n S n + S + V + W + X + Y + Z od 8 The asserton s followed by substtutng the correspondng values to the above equaton. Theore 5 If s even, not dvded by 4, and n s odd, then the pentanoal 9 has an even nuber of the rreducble factors over F f and only f one of the followng condtons holds : 1 od 8 and n 3, 7 od 8 6 od 8 and n 1, 5 od 8 P roof. The dscrnant of F s as follows. DF = 1 / RF, G 1 n + 1 n n+1 n n+1 n+1 n + od 8 Thus we search all possble cases, naly od 8, n 1 od 8 = DF 5 od 8 n 3 od 8 = DF 1 od 8 n 5 od 8 = DF 5 od 8 n 7 od 8 = DF 1 od 8 17

18 6 od 8, n 1 od 8 = DF 1 od 8 n 3 od 8 = DF 5 od 8 n 5 od 8 = DF 1 od 8 n 7 od 8 = DF 5 od 8 Therefore we obtan the result of the theore. Fnally we consder the case of, ultple of 4. In ths case we can see drectly W Z 0 od 8. In the sae anner we copute the dscrnant of F but we ot detals and scrbe only the results. S 3 + 4n + 1 n+1 4n + 1 n+1 od 8 V od 8 X 0 od 8 Y 6n + 1 od 8 RF, G n + n n n + S + V + Y 3 + n n+1 4n + 1 n+1 od 8 11 Snce s a ultple of 4, DF = 1 / RF, G = RF, G, so we have the followng theore. Theore 6 If s dvdable by 4, and n s odd, then the pentanoal 9 has an even nuber of the rreducble factors over F f and only f one of the followng condtons holds : 1 0 od 8 and n 1, 5 od 8 4 od 8 and n 3, 7 od 8 Fro theore 5 and theore 6 we obtan the followng corollary. Corollary 3 Let be even and n be odd and suppose that the pentanoal 9 s rreducble over F. 1 If n 1, 5 od 8, then, 4 od 8 If n 3, 7 od 8, then 0, 6 od 8 Note that f f s a polynoal over F wth no repeated root and F s ts onc lft to Z x, then DF 1 or 5 od 8 by Stckelberger s theore.9 As we see fro the proofs of theore 3,4,5 and 6, the dscrnant of the pentanoal F x Z x s always congruent to 1 or 5 odulo 8. Snce Df DF od 8, ths shows that type II pentanoals of even degree have no repeated root. 5 Concluson We have deterned the party of the nuber of rreducble factors for the type II pentanoals of even degrees usng the Stckelberger-Swan theore. Our consderaton s slar wth one n 1 but the coputaton s ore coplex. It wll be publshed elsewhere soe results for type II pentanoals of odd degrees and type I pentanoals. 18

19 References 1 O.Ahad and A.Menezes, Irreducble polynoals over axu weght, Utltas Matheatca, 7007, O.Ahad and A.Menezes, On the nuber of trace-one eleents n polynoal bases for F n, Desgns, Codes and Cryptography, 37005, O.Ahad and G.Vega, On the party of the nuber of rreducble factors of self-recprocal polynoals over fnte felds, Fnte Felds and Ther Applcatons, 14008, A.Bluher, A Swan-lke theore, Fnte Felds and Ther Applcatons, 1006, A.Hales and D.Newhart, Swan s theore for bnary tetranoals, Fnte Felds and Ther Applcatons, 1006, R.Ldl and H.Nederreter, Introducton to fnte felds and ther applcatons, Cabrdge Unversty Press, K.V.Mangpud and R.S. Katt, Montgoery ultpler for a class of specal rreducble pentanoals, Preprnt, F.Rodrguez-Henrquez and C.K. Koc, Parallel ultplers based on specal rreducble pentanoals, IEEE Transactons on Coputers, 5003, R.G.Swan, Factorzaton of polynoals over fnte felds, Pacfc Journal of Matheatcs,1196,

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