AN EXTENSION OF A THEOREM OF LJUNGGREN
MICHAEL FILASETA and IKHALFANI SOLAN*
1. Introduction
E. S. Selmer [5] studied the irreducibility over the rationals of polynomials of the form xn"1xm"2 where nm and each "j2 fÿ1;1g. He obtained complete solutions in the casem1 and partial results form>1. Ljunggren [1] later extended the problem to polynomials of the form xn"1xm"2xp"3 where again each"j2 fÿ1;1g. He established**
Theorem(Ljunggren).For any distinct positive integers n, m, and p, and for any choice of"j2 fÿ1;1g, the polynomial
xn"1xm"2xp"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 withnm0. Similar studies and related problems can be found in [2], [3] and [4]. For example, in [2], Mikusinski and Schinzel proved that ifp is an odd prime then there is only a finite number of ratios n=m for which f x xnpxm1 is reducible; and in [3], Schinzel proved that for n>m the polynomial
g x xnÿ2xm1 x n;mÿ1
is irreducible unless n;mis 7k;2kor 7k;5kin which case
g x x3kx2kÿ1 x3kxk1 and x3kx2k1 x3kÿxkÿ1;
respectively.
* Both authors were supported by NSF Grant DMS-9400937. Research of the second author was done as partial fulfillment of the Ph.D. requirement at the University of South Carolina.
** Received September 2, 1996.
Consider now
f x xn"1xm"2xp"3xq"4; with each"j2 fÿ1;1g:
1
If we remove the cyclotomic factors of f x, must the resulting polynomial be 1 or irreducible? This is in fact not the case. A simple example is given by
x7x5x3x2ÿ1ÿx3xÿ1
x4x1
ÿ
: An infinite set of examples is given by
f x xnÿx21 xnÿ2x21 x2nÿ2xn2xnÿ2ÿx41 where n represents any integer 5 with n60;3;4;6; or 9 mod 12; here, Ljunggren's result for trinomials can be used to establish xnÿx21 and xnÿ2x21 are both irreducible. In fact, these examples show that if f x
has five terms as above, it may be reducible and still not have ``reciprocal'' factors (a factorg x 2Zxsatisfyingg x xdeggg 1=x). Nevertheless, it is reasonable still to considerf xas in (1) with added restrictions. Our main result is the following five term version of Ljunggren's theorem.
Theorem 1. Let f x xnxmxpxq1 be a polynomial with n>m>p>q>0. Then f xwith its irreducible reciprocal factors removed either is the identity1or is irreducible over the integers.
We do not know if the same result holds with the role of irreducible re- ciprocal factors replaced by cyclotomic factors. We were able to show that such a replacement is possible in the special case thatnis exactly one of 2m, 2q, 2p,mp,pq, ormq. We also have the following examples:
x11x8x6x2x1 x5ÿx31 x6x4x3x2x1
and
x12x112x71 x5x4ÿx3ÿx21 x7x5x3x21:
The first of these shows that Theorem 1 cannot be extended to polynomials with six non-zero terms with coefficients 1. The second example illustrates that we need p6q (and reciprocal considerations would imply we need m6p).
A more general theorem than Theorem 1 exists, and its proof would fol- low easily from the arguments given below. We emphasize Theorem 1 mainly because of its simplicity. The more general result replaces the condi- tion that the coefficients of f x in (1) are positive with the condition that when the product of the polynomial and its reciprocal polynomial is ex- panded there are no cancellation in terms. More precisely, we can show
Theorem 2. Let f x xn1xm2xp3xq4 be a polynomial with n>m>p>q>0 and each j 1. Suppose that the sum of the absolute values of the coefficients in the product
xn1xm2xp3xq4
4xn3xnÿq2xnÿp1xnÿm1 is equal to25. If f x x xwhere xand xare polynomials with integer coefficients, then at least one of x and x is a reciprocal poly- nomial.
Schinzel in [4] gives a general result which shows that any theorem similar to those stated above can be effectively established. Using this result directly involves performing a tremendous number of computations; we estimated establishing Theorem 1 directly in this manner would require over 10200 steps. Nevertheless, Schinzel's result is quite general giving a method of de- termining how all polynomials factor with Euclidean norm less than a pre- scribed 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 exponentsn,m, andp.
In the case of Theorem 1 (or Theorem 2), we were able to bypass considering as many cases, mainly because the coefficients are more restrictive. We make no pretense here, however, of developing new approaches; this paper is merely a note that a five 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
Supposef x x xwhere xand xare polynomials with integer coefficients. We show that at least one of x and x is a reciprocal polynomial. We explain first why this will imply Theorem 1. Suppose this has been established andf xhas more than two non-reciprocal irreducible factors (not necessarily distinct). Let u x denote one of these. The poly- nomial w x xdeguu 1=x will also be a non-reciprocal irreducible poly- nomial. If w x jf x, then we consider x and x such that f x x x, u x- x, and w x- x. If is a root of u x, then 0 and 1= 60 so that x is a non-reciprocal polynomial. Si- milarly, xis non-reciprocal, and we arrive at a contradiction to what we are about to show. Ifw x- f x, we consider a second non-reciprocal irre- ducible factor of f x, say v x, where possiblyv x u xif u x2jf x. If
w x xdegvv 1=xdividesf x, then we can repeat the above argument re- placing the role ofu xwithv x. So suppose now that bothxdeguu 1=xand xdegvv 1=xare not factors of f x. In this case, we consider xand x
such thatf x x x, u xj x, and v xj x. As before, we deduce that each of xand xis non-reciprocal, leading to a contradiction.
Now, letrdegandsdeg . Write f1 x xr xÿ1 x Xn
i0
cixi and f2 x xs xÿ1 x:
We have
f2 x xnf1 xÿ1 Xn
i0
cixnÿi
and
f1 xf2 x x xÿxn xÿ1 xÿ1 2
xnxmxpxq1 xnxnÿqxnÿpxnÿm1:
On the other hand,
f1 xf2 x Xn
i0
cixi
! Xn
i0
cixnÿi
! : 3
Equating the coefficients ofx2n andxn in the two expressions for f1 xf2 x
we find
c0cn1 and c20c21 c2n5:
Thus,
c0cn1 and c21c22 c2nÿ13:
We deduce that three of theci's withi2 f1;2;. . .;nÿ1g, sayck1;ck2 andck3
with k1<k2<k3, must be 1 and the other ci's are equal to 0. Further- more,
cnck3ck2ck1c0
2f1 1f2 1 25 so that
c0ck1 ck2 ck3cn1 or c0ck1ck2 ck3 cn ÿ1:
We may suppose the former occurs and do so. Thus,
f1 x xnxk3xk2xk11 and f2 x xnxnÿk1xnÿk2xnÿk31:
We suppose as we may thatnmq, since otherwise we may replace f x
withxnf 1=x(so that the role ofmgets replaced bynÿq, the role ofpgets replaced bynÿp, and the role ofqgets replaced bynÿm). It suffices also to take nk1k3, since otherwise we can interchange the role of f1 x and f2 x(replacingk3withnÿk1,k2 withnÿk2, andk1 withnÿk3). From (2), we deduce that
f1 xf2 x x2nx2nÿqx2nÿpx2nÿmxnmxnp 4
xnqxnmÿqxnmÿpxnpÿq5xn : From (3), we obtain
f1 xf2 x x2nx2nÿk1x2nÿk2x2nÿk3xnk3xnk2 5
xnk1xnk3ÿk1xnk3ÿk2xnk2ÿk15xn : In (4) and (5), the terms shown are those having an exponent ofx being at leastn.
The conditionnmq implies that the second largest exponent in (4) is 2nÿq. The conditionnk1k3implies that the second largest exponent in (5) is 2nÿk1. It follows that k1q.
The sum of the exponents greater than n in the expanded product of f1 xf2 xgiven in (4) is 14n2mÿ2q. The sum of those exponents greater thann in the expanded product off1 xf2 xgiven in (5) is 14n2k3ÿ2k1. We deduce thatk3ÿk1mÿq. Sincek1q, we obtaink3m.
Making the substitutions k1q andk3 min (5) and comparing the re- sulting exponents with (4), we see that
f2nÿp;np;nmÿp;npÿqg f2nÿk2;nk2;nk3ÿk2;nk2ÿk1g:
The largest element in the representation of the set given on the left is either 2nÿp or np, and similarly the largest element on the right is either 2nÿk2 ornk2. So one of 2nÿpandnpmust equal one of 2nÿk2 and nk2.
If 2nÿp2nÿk2 ornpnk2, thenk2p. In this case, we obtain hk1;k2;k3i hq;p;mi. Thus,
f1 x xnxmxpxq1f x
so that
hk1;k2;k3i hq;p;mi ) x xr xÿ1:
6
If 2nÿpnk2 or np2nÿk2, then k2nÿp. Comparing ex- ponents in (4) and (5) with this additional substitution, we deduce that
fnmÿp;npÿqg fnk3ÿk2;nk2ÿk1g fmp;2nÿpÿqg:
Ifnmÿpmp, then n2p so that k2 nÿpp, and we can apply (6). If nmÿp2nÿpÿq, then nmq so that k3mnÿq and k1qnÿm. Thus, hk1;k2;k3i hnÿm;nÿp;nÿqi. An argument ana- logous to the argument for (6) gives in this case that x xs xÿ1.
This completes the proof of the theorem.
Added in proof. The example
x8x7xÿ1 x21 x3x2ÿ1 x3ÿx1
shows that the theorem of Ljunggren stated in the introduction is not cor- rect. This was first noted by W. H. Mills (The factorization of certain quad- rinomials, Math. Scand. 57 (1985), 44^50); he further supplied a corrected version of the theorem of Ljunggren in which the cases where the non-cy- clotomic part of xn1xm2xp3 is reducible are classified. Mills' ar- guments are based on carrying out, with more careful analysis, the ideas in Ljunggren's paper. The authors appreciate Robert Murphy and Andrzej Schinzel for bringing the paper of Mills to their attention.
REFERENCES
1. W. Ljunggren,On the irreducibility of certain trinomials and quadrinomials,Math. Scand. 8 (1960), 65^70.
2. J. Mikusinski and A. Schinzel,Sur la re¨ductibilite¨ de certains trinoªmes,Acta Arith. 9 (1964), 91^-95.
3. A. Schinzel,Solution d'un proble©me de K. Zarankiewicz sur les suites de puissances conse¨cutives de nombres irrationels,Colloq. Math. 9 (1962), 291^296.
4. A. Schinzel,Reducibility of lacunary polynomials I,Acta Arith. 16 (1969/70), 123^159.
5. E. Selmer,On the irreducibility of certain trinomials,Math. Scand. 4 (1956), 287^302.
MATHEMATICS DEPARTMENT UNIVERSITY OF SOUTH CAROLINA COLUMBIA, SC 29208
USAfilaseta@math.sc.edu
MATHEMATICS DEPARTMENT U.W.I. MONA
JAMAICA, W.I.
solan@uwimona.jm.edu