Divisibility of Trinomials by Irreducible Polynomials over F2
aa r X i v : . [ m a t h . R A ] J a n Divisibility of Trinomials by IrreduciblePolynomials over F a Ryul Kim, b Wolfram Koepf a Faculty of Mathematics and Mechanics,
Kim Il Sung
University, D.P.R Korea b Department of Mathematics, University of Kassel, Kassel, Germany a ryul [email protected], b [email protected] Abstract
Irreducible trinomials of given degree n over F do not always exist andin the cases that there is no irreducible trinomial of degree n it may be effec-tive to use trinomials with an irreducible factor of degree n . In this paper weconsider some conditions under which irreducible polynomials divide trino-mials over F . A condition for divisibility of self-reciprocal trinomials byirreducible polynomials over F is established. And we extend Welch’s cri-terion for testing if an irreducible polynomial divides trinomials x m + x s + 1 to the trinomials x am + x bs + 1 . Mathematics Subject Classification:
Keywords:
Trinomial, Self-reciprocal polynomial, Finite field
Irreducible and primitive trinomials over finite fields are of interest both in theory andpractice. We restrict our attention to polynomials over a binary field F . Sparse poly-nomials such as trinomials are commonly used to perform arithmetic in extension fieldsof finite fields since they provide a fast modular reduction but unfortunately irreducibletrinomials of given degree n over F do not always exist. Swan’s theorem [7] rules out n ≡ and also most n ≡ ± . In the cases that there is no irreducibletrinomial of given degree n , one can always use irreducible polynomials with more thanthree nonzero terms like pentanomials. But it may be more effective to use (reducible)trinomials with irreducible or primitive factors of degree n . In 1994, Tromp, Zhang andZhao [8] asked the following question: given an integer n , do there exist integers m, k such that G = gcd (cid:0) x m + x k + 1 , x n − − (cid:1) is a primitive polynomial of degree n over F ? They verified that the answer is yes for n up to 171 and conjectured that the answer is always yes. Blake, Gao and Lambert [1] con-firmed the conjecture for n ≤ and they also relaxed the condition slightly and asked: do there exist integers m, k such that G has a primitive factor of degree n ? Motivatedby [1], Brent and Zimmermann [2] defined an almost primitive (irreducible) trinomialwhich is the trinomial with a primitive (irreducible) factor of given degree n and theyproposed the algorithms for finding almost primitive (irreducible) trinomials. Doche [4]called these trinomials (almost irreducible trinomials) as redundant trinomials and gave aprecise comparison of running times between redundant trinomials and irreducible pen-tanomials over finite fields of characteristic 2. In [5] it was given a positive answer tothe latter question and the authors developed the theory of irreducible polynomials whichdo, or do not, divide trinomials over F . They considered some families of polynomialswith prime order p > that do not divide trinomials. To know which irreducible poly-nomials divide trinomials over F is of interest in many applications such as generationof pseudo-random sequences. In this paper we consider some conditions under which agiven irreducible polynomial divides trinomials over F . We prove a condition for a givenirreducible polynomial to divide self-reciprocal trinomials.Welchs criterion is a clever one for testing if an irreducible polynomial divides trino-mials over F . We give a refinement of a necessary condition for divisibility of trinomials x am + x bs + 1 by a given irreducible polynomial ([3]) and extend Welch’s criterion to thistype of trinomials. In this section we consider divisibility of self-reciprocal trinomials by given irreduciblepolynomials. Let q be a prime power. For a polynomial f ( x ) of degree n over finite field F q a reciprocal of f ( x ) is the polynomial f ∗ ( x ) of degree n over F q given by f ∗ ( x ) = x n f (1 /x ) and a polynomial f ( x ) is called self - reciprocal if f ∗ ( x ) = f ( x ) . Numerousresults are known concerning self-reciprocal irreducible polynomials over finite fields. In[9], it was studied in detail the order of self-reciprocal irreducible polynomials over finitefields. Let f ∈ F q [ x ] be a nonzero polynomial with f (0) = 0 . The least positive integer e for which f divides x e − is called the order of f and denoted by ord ( f ) [6]. If f isan irreducible polynomial of degree n over F q and with f (0) = 0 then ord ( f ) is equal tothe order of any root of f in the multiplicative group F ∗ q n and divides q n − . Below weassume all polynomials to be over F . In this case the order of an irreducible polynomialis always odd integer.In [5], it was proved that for prime p > , if there exists a self-reciprocal irreduciblepolynomial of order p then all irreducible polynomials of the same order do not dividetrinomials. In particular, every self-reciprocal irreducible polynomial of prime order > does not divide trinomials. In fact we can easily see that a self-reciprocal irreduciblepolynomial f divides trinomials in F [ x ] if and only if ord ( f ) is a multiple of 3. (Seeexercise 3.93 in [6]).Now consider self-reciprocal trinomials. Self-reciprocal irreducible trinomials over F are only of the form f = x · k + x k + 1 which has order k +1 . Then which irreducible polynomial divides self-reciprocal trinomials? As above mentioned, the order of a self-reciprocal irreducible polynomial which divides self-reciprocal trinomial is a multiple of3. Furthermore, we can say a similar thing about the general irreducible polynomialswhich divide self-reciprocal trinomials. For this we need an auxiliary result. Lemma 1
If an irreducible polynomial f of order e divides a self-reciprocal trinomial x m + x m + 1 , then there exists a unique self-reciprocal trinomial of degree < e which isdivided by f .Proof . Let α be any root of f in a certain extension of F then α m + α m + 1 . Write m = e · q + r, < r < e Note that r is not null. If r < e , then x r + x r + 1 is a desired trinomial. Suppose r > e .( r = e because if r = e , then α m + α m + 1 = α r + α r + 1 = α r + 1 , which isimpossible.) Let r = 2 r − e , then < r − r = e − r < r and α m + α m + 1 = α e (2 q +1)+ r + α eq + r + 1 = α r + α r + 1 On the other hand, ( α − ) m + ( α − ) m + 1 = α − + α − r + 1 = 0 and thus α r + α r − r + 1 = 0 From this we get α r − r = α r , that is, α | r − r | = 1 which means e divides | r − r | . Since | r − r | < e, r = 2 r . Therefore f divides the trinomial x r + x r + 1 . And then wehave also α r = 1( α r = 1) , which implies that e divides r (3 r ) . Since r (2 r ) < e ,we get e = 3 r (3 r ) . If there exists another integer m such that α m + α m + 1 = 0 , m < e then e = 3 m and therefore m = r ( r ) . ✷ Now we are ready to describe the condition for divisibility of self-reciprocal trinomi-als by a given irreducible polynomial.
Theorem 1
Given an irreducible polynomial f over F , f divides self-reciprocal trino-mials if and only if the order of f is a multiple of 3.Proof . Suppose f divides self-reciprocal trinomials. By Lemma 1, f divides self-reciprocaltrinomial x m + x m + 1 with m < e where e is the order of f . Let α be any root of f then α m + α m + 1 = 0 and we get e = 3 m as in the proof of lemma 1. Converselysuppose e = 3 m for a positive integer m . Let α be a root of f then α e = 1 that is α m − α m − α m + α m + 1) . Since α m = 1 , α m + α m + 1 = 0 and thus f divides the trinomial x m + x m + 1 . ✷ Below we show a factorization of an arbitrary self-reciprocal trinomial over F . Theorem 2
For any odd number m , x m + x m + 1 = Y n | m n ∤ m Q n where Q n is the n th cyclotomic polynomial over F .Proof . Suppose n | m, n ∤ m and let f be an irreducible polynomial of order n and α beany root of f in a certain extension of F . Then α n = 1 and therefore α m − α m − α m + α m + 1) = 0 . Since n ∤ m, α m − = 0 and thus α m + α m + 1 = 0 . Therefore f divides the trinomial x m + x m + 1 . Since Q n is a product of all irreducible polynomials of order n , it dividesthe trinomial x m + x m + 1 . From deg ( Q n ) = φ (3 n ) , it is sufficient to show X n | m n ∤ m φ (3 n ) = 2 m Using the formula P d | n φ ( d ) = n , we get X n | m n ∤ m φ (3 n ) = X n | m φ (3 n ) − X n | m φ (3 n )= X n | m φ (3 n ) − X n | m φ (3 n ) = 3 m − m = 2 m. This completes the proof. ✷ Corollary 1 If m is an odd number and m = 3 k · n, ∤ n for a nonnegative integer k ,then the self-reciprocal irreducible trinomial x · k + x k + 1 divides x m + x m + 1 .Proof . The trinomial x · k + x k + 1 divides Q k +1 since it is an irreducible polynomialof order k +1 . Recalling k | m, k +1 ∤ m , we get a desired result from Theorem 2. ✷ We can extend Theorem 2 to any positive degree m.
Corollary 2
Suppose that m = 2 k · n, ∤ n . Then x m + x m + 1 = Y n | n n ∤ n Q n k . Proof . Since x m + x m + 1 = (cid:0) x n (cid:1) k + ( x n ) k + 1 = (cid:0) x n + x n + 1 (cid:1) k , the assertion is followed from Theorem 2. ✷ If an irreducible polynomial f of order e divides a trinomial x n + x k + 1 , then for allpositive integer r and s, f divides x n + re + x k + se +1 and it divides at least one trinomial ofdegree < e . Consider a number of trinomials of degree < e which are divided by a givenirreducible polynomial. Denote as N f the number of trinomials of degree < e which aredivided by given irreducible polynomial f of order e . Theorem 3
Let f ( x ) be an irreducible polynomial of order e which divides trinomialsover F . Then N f = 12 deg ( gcd (1 + x e , x ) e )) , where deg means the degree of the polynomial.Proof . Let x e = g ( x ) · g ( x ) · · · · · g t ( x ) be a product of all irreducible polynomials whose orders divide e . Then we get x ) e = g ( x + 1) · g ( x + 1) · · · · · g t ( x + 1) . Let α be a root of f ( x ) then , α, α , · · · , α e − are all roots of g ( x ) , g ( x ) , · · · , g t ( x ) and , α, α , · · · , α e − are all roots of g ( x + 1) , g ( x + 1) , · · · , g t ( x + 1) .From the assumption there exists at least one pair ( i, j ) such that ≤ i, j < e, i = j, α i = α j + 1 . It can be easily seen that the number of such pairs is equal to the num-ber of common roots of x e and x ) e that is the degree of the polynomialgcd (1 + x e , x ) e ) . (Note that gcd (1 + x e , x ) e ) cannot has any mul-tiple root.) Since the different pairs ( i, j ) and ( j, i ) correspond the same trinomial, theresult is true. ✷ Corollary 3
The number of trinomials of degree < k − which are divided by a givenprimitive polynomial of degree k is k − − . In particular it is interesting the case when the number N f is 1. Theorem 4 If N f is 1, then f divides a self-reciprocal trinomial.Proof . Let e be an order of f . From Theorem 1, it is sufficient to prove that e is dividedby 3. Suppose that e is not divided by 3 and f divides a trinomial x n + x k + 1 . Then byTheorem 1 n = 2 k . Let α be a root of f . Then α − is a root of f ∗ , the reciprocal of f .Since f ∗ divides x n + x n − k + 1 , α − n + α − ( n − k ) + 1 = 0 , that is, α e − n + α e − n + k + 1 = 0 . Here < e − n, e − n + k < e, e − n = e − n + k. Therefore f divides the trinomial x e − n + x e − n + k + 1 . Since e is odd, e − n = n . Assumenow e − n = k . We then get α n + k = α e = 1 . Multiplying α k on both sides of the equation α n + α k + 1 = 0 , we have α k + α k + 1 = 0 which says that f divides some self-reciprocal trinomial that contradicts the assumption.Thus f divides two different trinomials x n + x k + 1 and x e − n + x e − n + k + 1 of degree < e , that is, N f ≥ . ✷ x am + x bs + 1 In this section we consider the conditions for divisibility of trinomials x am + x bs + 1 bya given irreducible polynomial over F . Let f be an irreducible polynomial of degree n over F and a and b be positive integers. In [3] it was proved that if there exist positiveintegers m and s such that f divides x am + x bs + 1 , then a and b are not divisible by n − . Below we give a refinement of this result. Theorem 5
Let f be an irreducible polynomial of order e > over F and a and b bepositive integers. If there exist positive integers m and s such that f divides trinomial x am + x bs + 1( am > bs ) , then am, bs and am − bs are not divisible by e .Proof . Let α be any root of f in a certain extension of F . If am is divided by e ,then α am = 1 , so f divides a polynomial x am + 1 . Since e > , f (0) = 0 , andthus f does not divide x bs . Therefore f cannot divide the trinomial x am + x bs + 1 .The case where bs is divided by e is very similar. Suppose am − bs is divided by e .Then in the same way as above we see easily that x am − bs + 1 is divided by f and thus x am + x bs + 1 = x bs (cid:0) x am − bs + 1 (cid:1) + 1 is not divisible by f . ✷ If f is an irreducible polynomial of order e and degree n over F , then e is a divisorof n − . Thus the above theorem derives directly the result in [3]. And if a = b = 1 and f = x + x + 1 then the converse of Theorem 5 is also true. Corollary 4
The trinomial x n + x k + 1( n > k ) is divided by x + x + 1 if and only if n, k and n − k are not divided by 3. Proof . Since the order of x + x + 1 is 3, the necessity is clear from above theorem.Suppose that n, k and n − k are not divided by 3. Then we get two cases: n ≡ , k ≡ , n − k ≡ or n ≡ , k ≡ , n − k ≡ . Let α be any root of x + x + 1 then in the first case we have α n + α k + 1 = α n +2 + α k +1 + 1 = α + α + 1 = 0 . Hence x + x + 1 divides x n + x k + 1 . The second case is similar. ✷ Finally we consider the criterion for testing if an irreducible polynomial divides tri-nomials of type x am + x bs + 1 over F . Theorem 6
Let f be an irreducible polynomial of order e and degree n over F and a and b be positive integers. Then f divides trinomials x am + x bs + 1 if and only if gcd (cid:0) x e , x ) e (cid:1) has degree greater than 1, where e = e gcd ( a, e ) , e = e gcd ( b, e ) . Proof . Let α be any root of f . Then the order of α in the multiplicative group F ∗ n is e and , α, α , · · · , α e − are distinct roots of x e − . Since x e − Y d | n Q d for every i (0 ≤ i ≤ e − , α i is a root of an irreducible polynomial whose order is a di-visor of e . In particular, α a has order e = e gcd ( a,e ) and α a , α a , · · · , α ( e − a are all rootsof C e ( x ) := x e − x − . Similarly α b , α b , · · · , α ( e − b are all roots of C e ( x ) := x e − x − and thus α b , α b , · · · , α ( e − b are all roots of C e ( x + 1) . Hence α is a rootof trinomial x am + x bs + 1 if and only if C e ( x ) and C e ( x + 1) have common root. Thisis equivalent to the fact that gcd (cid:0) x e , x ) e (cid:1) has degree greater than 1. ✷ Put a = b = 1 in Theorem 6. Then we have Welch’s criterion. Corollary 5 ([5])
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