If we remove the cyclotomic factors of fèxè, must the resulting polynomial be 1 or irreducible? This is in fact not the case. A simple example is give

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1 AN EXTENSION OF A THEOREM OF LJUNGGREN Michael Filaseta and Junior Solan* 1. Introduction E.S. Selmer ë5ë studied the irreducibility over the rationals of polynomials of the form x n + " 1 x m + " 2 where n ç m and each " j 2f,1;1g. He obtained complete solutions in the case m = 1 and partial results for mé1. Ljunggren ë1ë later extended the problem to polynomials of the form x n + " 1 x m + " 2 x p + " 3 where again each " j 2f,1;1g. He established Theorem èljunggrenè. For any distinct positive integers n, m, and p, and for any choice of " j 2f,1;1g, the polynomial x n + " 1 x m + " 2 x p + " 3 ; with its cyclotomic factors removed, either is the identity 1 or is irreducible over the integers. The analogous theorem also holds for the case of the trinomials studied by Selmer with n ç m ç 0. Similar studies and related problems can be found in ë2ë, ë3ë and ë4ë. For example, in ë2ë, Mikusinski and Schinzel proved that if p is an odd prime then there is only a ænite number of ratios n=m for which fèxè =x n æpx m æ 1 is reducible; and in ë3ë, Schinzel proved that for némthe polynomial gèxè = xn,2x m +1 x èn;mè, 1 is irreducible unless èn; mè isè7k; 2kè orè7k; 5kè in which case gèxè =èx 3k +x 2k,1èèx 3k + x k +1è and èx 3k + x 2k + 1èèx 3k, x k, 1è; respectively. Consider now è1è fèxè =x n +" 1 x m +" 2 x p +" 3 x q +" 4 ; with each " j 2f,1;1g: *Both authors were supported by NSF Grant DMS Research of the second author was done as partial fulællment of the Ph.D. requirement at the University of South Carolina. Typeset by AMS-TEX 1

2 If we remove the cyclotomic factors of fèxè, must the resulting polynomial be 1 or irreducible? This is in fact not the case. A simple example is given by An inænite set of examples is given by x 7 + x 5 + x 3 + x 2, 1=, x 3 +x,1 æ, x 4 +x+1 æ : fèxè =èx n,x 2 + 1èèx n,2 + x 2 +1è=x 2n,2 +x n+2 + x n,2, x 4 +1 where n represents any integer ç 5 with n 6ç 0; 3; 4; 6; or 9 èmod 12è; here, Ljunggren's result for trinomials can be used to establish x n, x and x n,2 + x are both irreducible. In fact, these examples show that if fèxè has æve terms as above, it may be reducible and still not have ëreciprocal" factors èa factor gèxè 2 Zëxë satisfying gèxè = æx deg g gè1=xèè. Nevertheless, it is reasonable still to consider fèxè as in è1è with added restrictions. Our main result is the following æve term version of Ljunggren's theorem. Theorem 1. Let fèxè =x n +x m +x p +x q +1 be a polynomial with némépéqé0. Then f èxè with its irreducible reciprocal factors removed either is the identity 1 or is irreducible over the integers. We do not know if the same result holds with the role of irreducible reciprocal factors replaced by cyclotomic factors. We were able to show that such a replacement is possible in the special case that n is exactly one of 2m, 2q,2p,m+p,p+q,orm+q. We also have the following examples: and x 11 + x 8 + x 6 + x 2 + x +1=èx 5,x 3 + 1èèx 6 + x 4 + x 3 + x 2 + x +1è x 12 + x 11 +2x 7 +1=èx 5 +x 4,x 3,x 2 + 1èèx 7 + x 5 + x 3 + x 2 +1è: The ærst of these shows that Theorem 1 cannot be extended to polynomials with six nonzero terms with coeæcients 1. The second example illustrates that we need p 6= q èand reciprocal considerations would imply we need m 6= pè. A more general theorem than Theorem 1 exists, and its proof would follow easily from the arguments given below. We emphasize Theorem 1 mainly because of its simplicity. The more general result replaces the condition that the coeæcients of f èxè in è1è are positive with the condition that when the product of the polynomial and its reciprocal polynomial is expanded there are no cancellation in terms. More precisely, we can show Theorem 2. Let fèxè =x n +æ 1 x m +æ 2 x p +æ 3 x q +æ 4 be a polynomial with némépé qé0and each æ j = æ1. Suppose that the sum of the absolute values of the coeæcients in the product èx n + æ 1 x m + æ 2 x p + æ 3 x q + æ 4 è, æ 4 x n + æ 3 x n,q + æ 2 x n,p + æ 1 x n,m +1 æ If f èxè = æèxèæèxè where æèxè and æèxè are polynomials with integer coeæcients, then at least one of æèxè and æèxè is a reciprocal polynomial. is equal to 25. Schinzel in ë4ë gives a general result which shows that any theorem similar to those stated above can be eæectively established. Using this result directly involves performing 2

3 a tremendous number of computations; we estimated establishing Theorem 1 directly in this manner would require over steps. Nevertheless, Schinzel's result is quite general giving a method of determining how all polynomials factor with Euclidean norm less than a prescribed amount. The methods used in this paper are essentially the same as those of Ljunggren. He presented some key ideas introducing reciprocal polynomials into the problem of determining how polynomials with small Euclidean norm factor. The proof he gave of his theorem above involved consideration of several cases depending on the relative sizes of the exponents n, m, and p. In the case of Theorem 1 èor Theorem 2è, we were able to bypass considering as many cases, mainly because the coeæcients are more restrictive. We make no pretense here, however, of developing new approaches; this paper is merely a note that a æve term version of Ljunggren's theorem does in fact exist. We give a proof of Theorem 1 below; a proof of Theorem 2 can be made with very few changes. 2. Proof of Theorem 1 Suppose f èxè = æèxèæèxè where æèxè and æèxè are polynomials with integer coeæcients. We show that at least one of æèxè and æèxè is a reciprocal polynomial. We explain ærst why this will imply Theorem 1. Suppose this has been established and fèxè has more than two non-reciprocal irreducible factors ènot necessarily distinctè. Let uèxè denote one of these. The polynomial wèxè =x deg u uè1=xè will also be a non-reciprocal irreducible polynomial. If wèxèjf èxè, then we consider æèxè and æèxè such that f èxè = æèxèæèxè, uèxè - æèxè, and wèxè - æèxè. If æ isarootofuèxè, then æèæè = 0 and æè1=æè 6= 0so that æèxè is a non-reciprocal polynomial. Similarly, æèxè is non-reciprocal, and we arrive at a contradiction to what we are about to show. If wèxè - fèxè, we consider a second nonreciprocal irreducible factor of fèxè, say vèxè, where possibly vèxè =uèxèifuèxè 2 jfèxè. If wèxè =x deg v vè1=xè divides fèxè, then we can repeat the above argument replacing the role of uèxè with vèxè. So suppose now that both x deg u uè1=xè and x deg v vè1=xè are not factors of fèxè. In this case, we consider æèxè and æèxè such that fèxè =æèxèæèxè, uèxèjæèxè, and vèxèjæèxè. As before, we deduce that each of æèxè and æèxè is non-reciprocal, leading to a contradiction. Now, let r = deg æ and s = deg æ. Write We have and è2è f 1 èxè =x r æèx,1 èæèxè = nx f 2 èxè=x n f 1 èx,1 è= f 1 èxèf 2 èxè =æèxèæèxè, x n æèx,1 èæèx,1 è æ c i x i and f 2 èxè =x s æèx,1 èæèxè: nx c i x n,i =èx n +x m +x p +x q +1è, x n +x n,q +x n,p +x n,m +1 æ : 3

4 On the other hand, n è3è f 1 èxèf 2 èxè =èx!è c i X n x i c i x n,i! Equating the coeæcients of x 2n and x n in the two expressions for f 1 èxèf 2 èxè we ænd Thus, c 0 c n =1 and c c æææ+c 2 n =5: c 0 c n =1 and c c æææ+c 2 n,1 =3: We deduce that three of the c i 's with i 2 f1; 2;:::;n, 1g, say c k1 ;c k2 and c k3 with k 1 ék 2 ék 3,must be æ1 and the other c i 's are equal to 0. Furthermore, so that èc n + c k3 + c k2 + c k1 + c 0 è 2 = f 1 è1èf 2 è1è = 25 c 0 = c k1 = c k2 = c k3 = c n =1 or c 0 = c k1 = c k2 = c k3 = c n =,1: We may suppose the former occurs and do so. Thus, f 1 èxè =x n +x k 3 +x k 2 +x k 1 +1 and f 2 èxè =x n +x n,k 1 +x n,k 2 +x n,k 3 +1: We suppose as we may that n ç m+q, since otherwise we may replace fèxè with x n fè1=xè èso that the role of m gets replaced by n, q, the role of p gets replaced by n, p, and the role of q gets replaced by n, mè. It suæces also to take n ç k 1 + k 3, since otherwise we can interchange the role of f 1 èxè and f 2 èxè èreplacing k 3 with n, k 1, k 2 with n, k 2, and k 1 with n, k 3 è. From è2è, we deduce that : è4è f 1 èxèf 2 èxè =x 2n +x 2n,q +x 2n,p +x 2n,m +x n+m +x n+p + x n+q + x n+m,q + x n+m,p + x n+p,q +5x n +æææ : From è3è, we obtain è5è f 1 èxèf 2 èxè =x 2n +x 2n,k 1 +x 2n,k 2 +x 2n,k 3 +x n+k 3 +x n+k 2 + x n+k 1 + x n+k 3,k 1 + x n+k 3,k 2 + x n+k 2,k 1 +5x n +æææ : In è4è and è5è, the terms shown are those having an exponent ofxbeing at least n. The condition n ç m + q implies that the second largest exponent in è4è is 2n, q. The condition n ç k 1 + k 3 implies that the second largest exponent in è5è is 2n, k 1. It follows that k 1 = q. The sum of the exponents greater than n in the expanded product of f 1 èxèf 2 èxè given in è4è is 14n +2m,2q. The sum of those exponents greater than n in the expanded 4

5 product of f 1 èxèf 2 èxè given in è5è is 14n +2k 3,2k 1. We deduce that k 3, k 1 = m, q. Since k 1 = q, we obtain k 3 = m. Making the substitutions k 1 = q and k 3 = m in è5è and comparing the resulting exponents with è4è, we see that f2n, p; n + p; n + m, p; n + p, qg = f2n, k 2 ;n+k 2 ;n+k 3,k 2 ;n+k 2,k 1 g: The largest element in the representation of the set given on the left is either 2n, p or n + p, and similarly the largest element on the right is either 2n, k 2 or n + k 2. So one of 2n, p and n + p must equal one of 2n, k 2 and n + k 2. If 2n, p =2n,k 2 or n + p = n + k 2, then k 2 = p. In this case, we obtain hk 1 ;k 2 ;k 3 i= hq; p; mi. Thus, f 1 èxè =x n +x m +x p +x q +1=fèxè so that è6è hk 1 ;k 2 ;k 3 i=hq; p; mi =è æèxè =x r æèx,1 è: If 2n, p = n + k 2 or n + p =2n,k 2, then k 2 = n, p. Comparing exponents in è4è and è5è with this additional substitution, we deduce that fn + m, p; n + p, qg = fn + k 3, k 2 ;n+k 2,k 1 g=fm+p; 2n, p, qg: If n + m, p = m + p, then n =2pso that k 2 = n, p = p, and we can apply è6è. If n + m, p =2n,p,q, then n = m + q so that k 3 = m = n, q and k 1 = q = n, m. Thus, hk 1 ;k 2 ;k 3 i=hn,m; n, p; n, qi. An argument analogous to the argument for è6è gives in this case that æèxè =x s æèx,1 è. This completes the proof of the theorem. References 1. W. Ljunggren, On the irreducibility of certain trinomials and quadrinomials, Math. Scand. 8 è1960è, 65í J. Mikusinski and A. Schinzel, Sur la rçeductibilitçe de certains trin^omes, Acta Arith. 9 è1964è, 91í A. Schinzel, Solution d'un problçeme de K. Zarankiewicz sur les suites de puissances consçecutives de nombres irrationels, Colloq. Math. 9 è1962è, 291í A. Schinzel, Reducibility of lacunary polynomials I, Acta Arith. 16 è1969è70è, 123í E. Selmer, On the irreducibility of certain trinomials, Math. Scand. 4 è1956è, 287í302. 5