Survey on Irreducibility of Trinomials

Home , Finite field, Irreducible polynomial, Reciprocal polynomial, Trinomial

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Theorem 1 ([22], [91], [115], [63]). Let n 2, a Q. The polynomial xn a is reducible over Q if and only if either a = bt for some≥ t n,t∈ > 1 or 4 n, a = 4b4 for− some b Q. | | − ∈ Naturally, the question arises about the irreducibility criterion for trinomials f(x)= axn + bxm + c, where a N, b,c Z 0 and (a, b, c) = 1. Starting from Eisenstein, many irreducibility criteria∈ appeared∈ for\{ different} types of trinomials. Yet the question is not fully solved. There is no single criterion which fits to every trinomial. Although several results appeared over time, to the best of our knowledge there is no survey article on this particular topic. Therefore, in this article, we tried to collect an exhaustive list of the irreducibility criteria for trinomials available in the literature. We compiled the existing irreducibility criteria of trinomials in section 2. In section 3, we studied the location of zeros of trinomials and their applications in the factorization of trinomials. The discriminant of trinomials was discussed in section 4. At the end of this article, we ask a few questions, answers to those questions will enhance the list of families of irreducible trinomials. Due to the diversity and intricacies of the theory applied in the polynomial factorization, we will concentrate on irreducibility criteria over rational numbers only. Unless specified, throughout the article, we will assume that all the polynomials have content 1. By this assumption, the irreducibility over Q and irreducibility over Z become the same. Further, by ǫ or ǫi, we will always mean that they belong to the set 1, +1 . For in depth study related to polynomial factorizations and for general references{ − on polynomials,} the reader should refer to [12], [90], and [97].

2 Reducibility of trinomials

This section is divided into various subsections depending upon the irreducibility criteria, which apply to a particular type of trinomials. We will ﬁrst discuss those families, whose reducibility nature has been resolved completely. Later we proceed to compile the results providing a partial answer to the question of the irreducibility of trinomials.

n m 2.1 Reducibility of x + bǫ1x + ǫ2

n m Let n, m, b N and let f(x) = x + bǫ1x + ǫ2 be a trinomial of degree n, ǫ1, ǫ2 ∈ n n−m ∈ 1, +1 . The polynomial f(x) is reducible if and only if ǫ2f˜(x)= x + bǫ1ǫ2x + ǫ2 is{ − reducible.} Thus, it is sufficient to check the reducibility of xn+bǫ xm+ǫ for 1 m n . 1 2 ≤ ≤ 2 n 2.1.1 Reducibility of x + bǫ1x + ǫ2 Perron [87] is the first mathematician who gave an irreducibility criterion of a polynomial depending upon the magnitude of the coefficients. He proved that,

2 n n−1 Perron’s criterion [87]: Let f(x) = x + an−1x + + a1x + a0 be a monic polynomial with integer coeﬃcients and a =0. If ··· 0 6 a > 1+ a + + a + a , | n−1| | n−2| ··· | 1| | 0| then f(x) is irreducible over Z. ˜ n n−1 By using Perron’s criterion, the polynomial ǫ2f(x)= x + bǫ1ǫ2x + ǫ2, and hence n f(x) = x + bǫ1x + ǫ2 is irreducible for every b 3. If b is either 1 or 2, then there are polynomials which are reducible. For example, ≥

x3 2x +1=(x 1)(x2 + x 1); − − − x8 + ǫx +1=(x2 + ǫx + 1)(x6 ǫx5 + ǫx3 x2 + 1). − − Selmer [104] came across the question of reducibility of trinomials of the form xn x 1 while studying continued fractions. With a change of sign in x, it is suﬃcient to consider± ± xn + x + 1 and xn x 1 only. For example, if n is odd and f(x) = xn x + 1, then − − − f( x) = xn x 1. Hence, if n is odd, then f(x) is reducible if and only if f( x) −is reducible− and− so− on. − − n If αi is a root of f(x)= x + ǫ(x + 1), then Selmer considered the following sum

n 1 S = α . f i − α i=1 i X Since S is a symmetric function of roots of f, S Q, and S Z for f(x)= xn+ǫ(x+1). f f ∈ f ∈ As any factor of f(x) over Z has a constant term 1, reducibility of f(x) over Z yields ± an integer partition of Sf . Using this fact, Selmer proved that Theorem 2 (Selmer [104]). The polynomial xn + ǫ(x +1) is irreducible except for ǫ =1 n and n 2 (mod 3). If ǫ = 1 and n 2 (mod 3), then x + x +1=Φ3(x)g(x), where g(x) is≡ either identically 1 or an irreducible≡ polynomial.

n m Selmer attempted to answer the reducibility problem of trinomials x + ǫ1x + ǫ2 for arbitrary values of m. By considering the sum

n 1 S = (αm ), f i − αm i=1 i X n m where αi is a root of x +ǫ1x +ǫ2, he gave an upper bound for the number of irreducible factors of it.

n m Theorem 3 (Selmer [104]). Let n = dn1, m = dm1, (n1, m1)=1 and f(x)= x +ǫ1x + ǫ2 be a polynomial of degree n 2m. Then all the roots of f(x) which lie on the unit circle are roots of unity and Φ (≥xd) divides f(x) only when n + m 0 (mod 3). Apart 3 1 1 ≡ from cyclotomic factors, f(x) has at most m irreducible factors, each is of degree greater than 5 for n> 7.

3 If b = 2, then a partial answer can be drawn from the work of Schinzel [93] on the trinomial xn 2xm + 1. However for complete answer one can refer either the works of Harrington [48]− (see Theorem 14 below) or by that of Filaseta et al. [40] (see Theorem n 9 below). The trinomial x + 2ǫ1x + ǫ2 factors as (x 1)g(x) for some irreducible polynomial g(x) Z[x] and n 3. If n =2, then x2 +2ǫ±x + ǫ is either irreducible or ∈ ≥ 1 2 has a cyclotomic factor. Combiningly we have

n Theorem 4. Let b N and f(x)= x + bǫ1x + ǫ2 be a trinomial of degree n 2. Then the followings are true.∈ ≥

(a) If b 3, then f(x) is irreducible. ≥

(b) If b =1, 2, then f(x) is reducible if and only if it factor as fc(x)fn(x), where fc(x) is a product of cyclotomic polynomials, and f (x) Z[x] is either identically 1 or n ∈ an irreducible polynomial. Further in reducible cases, if b =1, fc(x) is either Φ3(x) or Φ (x), and if b =2, then f (x) is either x +1, x 1, (x + 1)2 or (x 1)2. 6 c − − n m 2.1.2 Reducibility of x + bǫ1x + ǫ2, m ≥ 2 (A) b ≤ 2 Let f(x) = xn + bǫ xm + ǫ be a trinomial of degree n 2m and b is either 1 or 1 2 ≥ 2, ǫ1, ǫ2 1, +1 . One can get an upper bound on the number of irreducible ∈ { − n} m factors of f(x)= x +ǫ1x +ǫ2 from Theorem 3. In 1960, Tverberg [112] considered Newton’s formula for power sum of roots and proved a necessary and suﬃcient condition for the reducibility of f.

n m Theorem 5 (Tverberg [112]). Suppose f(x) = x + ǫ1x + ǫ2 is a polynomial of degree n 2m. Then f(x) is irreducible if and only if f(x) has no root on the unit ≥ circle. If f(x) has a root on the unit circle, then it is the roots of unity and the other factor of f(x) is irreducible.

In the same year, Ljunggren [72] developed an ingenious tool to study the behavior of the factors of a polynomial f(x) having a small Euclidean norm. If f(x) = n n−1 anx + an−1x + + a1x + a0, then the Euclidean norm of f, denoted by f 2, is deﬁned as ··· k k f = a2 + a2 + + a2 + a2 1/2 . k k2 n n−1 ··· 1 0

Suppose f(x) = f1(x)f2(x) is a non-trivial factorization of f(x). Ljunggren con- ˜ ˜ sidered the polynomials g(x) = f1(x)f2(x) andg ˜(x) so that g(x)˜g(x) = f(x)f(x). 2 If the value of f 2 is small, then one can eﬀectively compute the nature of the factors of f. Itk is interestingk to note that f 2 is the coeﬃcient of xn in f(x)f˜(x). k k2 Ljunggren applied this method to study trinomials and quadrinomials of the form n m n m r x +ǫ1x +ǫ2 and x +ǫ1x +ǫ2x +ǫ3, respectively. For several other applications of this method, readers can refer [13], [32], [33], [39], [41], [45]. In the case of trinomials, Ljunggren proved that

4 Theorem 6 (Ljunggren [72]). If n = n1d, m = m1d, (n1, m1)=1, n 2m, then the polynomial ≥ n m f(x)= x + ǫ1x + ǫ2, is irreducible, apart from the following three cases, where n + m 0 (mod 3): 1 1 ≡

n1, m1 both odd, ǫ1 =1; n1 even, ǫ2 =1; m1 even, ǫ1 = ǫ2.

2d m n d In these three cases, f(x) is the product of the polynomial x + ǫ1 ǫ2 x +1 and a second irreducible polynomial.

However, the above factor is not correct in all cases. For example, by Theorem 6, the polynomial x50 x4 1 has factor x4 + x2 + 1 but they are divisible by 4 2 − − x x + 1. More generally, if d is even, d1 5 (mod 6), d2 1 (mod 6) and d3 − ≡ ≡ d3 is odd, then the polynomials xdd1 + xd 1, xdd2 x2d 1 and x2 d xd +1 are divisible by x2d + xd + 1 by Theorem 6.− But they− are− divisible by x−2d xd + 1. − One can check that Ljunggren’s result is true only when ǫ1 = ǫ2 = 1. We corrected these errors in [65] and proved that

a b Theorem 7 (Koley & Reddy [65]). Let n, m N, n 2m and m = 2 3 m1, p q ∈ ≥ · · n 2m = 2 3 n1, where a, b, p, q 0, (m1n1, 6) = 1. If ǫ1 + ǫ2 = 2 and − n · m · ≥ 6 f(x)= x + ǫ1x + ǫ2 is reducible, then q > b and one of the following holds: (a) ǫ = ǫ =1, a = p, 1 − 2 (b) ǫ = 1, aǫ < pǫ . 1 − 2 2 (n,m) In each of these cases, the cyclotomic part of f is fc(x) = Φ6(x ).

Schinzel [93] studied the polynomials of the form xn 2xm +1. If n = 2m, then xn 2xm +1=(xm 1)2 is a product of cyclotomic polynomials.− For n =2m, he − − 6 proved that

Theorem 8 (Schinzel [93]). Let n = 2m. The polynomial xn 2xm + 1 can 6 − be factored as (x(n,m) 1)g(x) where g(x) Z[x] is an irreducible non-reciprocal polynomial except in the− following cases: ∈

x7k 2x2k +1=(xk 1)(x3k + x2k + 1)(x3k + xk + 1); − − x7k 2x5k +1=(xk 1)(x3k + x2k + 1)(x3k xk 1), − − − − for every k 1. ≥ If n =2m, then xn +2xm +1=(xm +1)2 is a product of cyclotomic polynomials. On the other hand, if we replace x by x in Theorem 8, then one can conclude the following. −

5 If n is even, m is odd, then xn +2xm +1 can be factored as (x(n,m) +1)g(x) where g(x) is an irreducible non-reciprocal polynomial. Similarly, partial results can be drawn for xn +2xm 1 and xn 2xm 1 by − − − replacing x with x in Theorem 8. Recently Filaseta et al. [40] gave a complete − n m characterization of the irreducibility of x +2ǫ1x + ǫ2.

n m Theorem 9 (Filaseta et al. [40]). Let n>m> 0 and f(x) = x +2ǫ1x + ǫ2. Then the non-cyclotomic part of f(x) is reducible in the following cases:

x7k 2x2k +1=(xk 1)(x3k + x2k 1)(x3k + xk + 1); − − − x7k 2x5k +1=(xk 1)(x3k + x2k + 1)(x3k xk 1); − − − − x7k +2x2k 1=(xk + 1)(x3k x2k + 1)(x3k + xk 1); − − − x7k 2x5k 1=(xk + 1)(x3k xk + 1)(x3k xk 1); − − − − − x7k +2x3k +1=(x3k x2k + 1)(x4k + x3k + x2k + 1); − x7k +2x4k +1=(x3k xk + 1)(x4k + x2k + xk + 1); − x7k +2x3k 1=(x3k + x2k 1)(x4k x3k + x2k + 1); − − − x7k 2x4k 1=(x3k xk 1)(x4k + x2k xk + 1), − − − − − for every k 1. ≥ (B) b ≥ 3 n m Let f(x)= x + bǫ1x + ǫ2 be a polynomial of degree n 2m and let b 3, ǫ1, ǫ2 1, +1 . Recall from the subsection 2.1.1 that these≥ polynomials are≥ irreducible∈ when{ − m }= 1. Minkusinski and Schinzel [76] considered trinomials of the form n m x + pǫ1x + ǫ2, where p is an odd prime. They proved that there are only ﬁnitely many such trinomials that are reducible. Theorem 10 (Minkusinski & Schinzel [76]). If p is an odd prime, then there are n n m only a ﬁnite number of ratios m for which f(x)= x + pǫ1x + ǫ2 is reducible.

In 1971, Schinzel [96] asked the following question: Is it possible to ﬁnd a reducible trinomial of the form xn + Axm +1 with A > 2, n =2m and n>m> 0? | | 6 Coray answered this question in a letter to Schinzel furnishing the following examples:

x13 3x4 +1=(x3 x2 + 1)(x10 + x9 + x8 x6 + x2 + 1); − − − −··· x13 +3x7 +1=(x4 x3 + 1)(x9 + x8 + x7 + x6 x4 + x3 + 1). − − A. Bremner [16] considered the trinomials xn + Axm + 1 with A > 2 and proved more than what Schinzel asked. He showed that there are| only| ﬁnitely many trinomials with an irreducible cubic factor. More precisely,

6 Theorem 11 (Bremner [16]). Let n 2m> 0 and the trinomial xn + Axm +1 has an irreducible factor of degree three, 0≥= A Z. If n> 3, then the only possibilities are the following: 6 ∈ x4 +2ǫx +1=(x + ǫ)(x3 ǫx2 + x + ǫ); − x5 + x +1=(x2 + x + 1)(x3 x2 + 1); − x7 2x2 +1=(x 1)(x3 + x2 1)(x3 + x + 1); − − − x7 +2x3 +1=(x3 x2 + 1)(x4 + x3 + x2 + 1); − x13 3x4 +1=(x3 x2 + 1)(x10 + x9 + x8 x6 2x5 2x4 x3 + x2 + 1); − − − − − − x33 + 67x11 +1=(x3 + x + 1)(x30 x28 x27 + x26 +2x25 2x3 + x2 x + 1). − − −···− − If n is odd, then by changing the variable x by x, the polynomial xn + Axm 1 − − has irreducible cubic factors in the above cases only. If n is even, Lutczyk [96] provided the following example, x8 +3x3 1=(x3 + x 1)(x5 x3 + x2 + x + 1). − − − Bremner [16] also gave several examples when n is even. For example, x6 + (4b4 4b)x2 1=(x3 +2bx2 +2b2x + 1)(x3 2bx2 +2b2x 1), − − − − where b =0, 1, b Z is an infinite family of polynomials having a cubic factor. By 6 ∈ 3 2 2 Perron’s criterion, x +2b ǫ1x +2bx + ǫ2 is irreducible when b = 0, 1 and hence so is its reciprocal. However, it is not known whether there are only6 finitely many such trinomials of the form xn + Axm 1 having a cubic factor when n is even. We, therefore, ask the following question,− Question 12. Let n, m, A N, n 2m, A> 2 and let n be an even integer. Are there finitely many trinomials∈ of the≥ form xn Axm 1 which have an irreducible ± − cubic factor?

n m 2.2 Reducibility of x + bǫ1x + cǫ2,c> 1

n m Suppose f(x)= x + bǫ1x + cǫ2 is a trinomial of degree n, where b, c N,c> 1, ǫ1, ǫ2 n m ˜ ∈ ∈ 1, +1 . Unlike in the case of x + bǫ1x + ǫ2, here ǫ2f(x) is not of the form of f(x). So,{ − we have} to consider the reducibility of f(x) for 1 m n 1. Nagell [79] gave the ≤ ≤ − following irreducibility criterion for trinomials with a large middle coeﬃcient. Theorem 13 (Nagell [79]). Let b, c N and let f(x)= xn + bǫ xm + cǫ be a polynomial ∈ 1 2 of degree n. Then f(x) is irreducible if (a) b> 1+ cn−1 and (b) if d n,d > 1, then c is not a dth power. In particular, c> 1. | Nagell’s criterion is weaker than that of Perron’s criterion in the case of m = n 1. If m = n 1, then the second condition of Nagell’s theorem becomes redundant. − − 7 n n−1 2.2.1 Reducibility of x + bǫ1x + cǫ2, c > 1 J. Harrington [48] asked the values of n and b for which f(x)= xn + b(xn−1 + xn−2 + + x + 1) ··· is irreducible. If b = 1, then f(x) is a product of cyclotomic polynomials. For b = 1, one can see from Theorem 8 and −

xn 2xn−1 +1=(x 1)(xn−1 xn−2 x 1), − − − −···− − that f(x) is irreducible. However, the answer is not known for b 2. Harrington n n−1 | | ≥ studied the trinomials of the form x + bǫ1x + cǫ2 to answer these questions and proved that Theorem 14 (Harrington [48]). Let n 3,b,c N, nb =9, b = c, b 2 and c 2(b 1). n n−1 ≥ ∈ 6 6 ≥ ≤ − If the trinomial f(x)= x + bǫ1x + cǫ2 is reducible, then f(x)=(x 1)fn(x), where f (x) Z[x] is a non-reciprocal irreducible polynomial. ± n ∈ Theorem 14 is no longer true if c> 2(b 1). For example, − x6 +2ǫx5 +4=(x3 2x +2ǫ)(x3 +2ǫx2 +2x +2ǫ); − x7 +2x6 +9=(x3 3x + 3)(x4 +2x3 +3x2 +3x + 3). − By Perron’s criterion, if c < b 1, then xn + bǫ xn−1 + cǫ is irreducible. Hence, − 1 2 n n−1 Theorem 15. Let n 4, b 2,c N, and let f(x)= x + bǫ1x + cǫ2 be a polynomial of degree n. ≥ ≥ ∈ (a) If c < b 1, then f(x) is irreducible. − (b) If b 1 c 2(b 1), b = c and f(x) is reducible, then f(x)=(x 1)fn(x), where f (x−) ≤Z[x≤] is a− non-reciprocal6 irreducible polynomial. ± n ∈ If n is even, then xn + bxn−1 + b 1=(x +1)f (x) shows that the lower bound of c − n can not be improved further. For b = c, Harrington conjectured that Conjecture 16 (Harrington [48]). Let n, b N, b 2 and f(x) = xn + bǫ xn−1 + bǫ . ∈ ≥ 1 2 If f(x) = x2 +4x +4 and f(x) = x2 4x +4, then f(x) is irreducible. 6 6 − 2m m 2.2.2 Reducibility of x + bǫ1x + cǫ2, c > 1

2m m Let b, c N,c > 1 and f(x) = x + bǫ1x + cǫ2 be a polynomial of degree 2m. From Nagell’s∈ criterion, f(x) is irreducible whenever b > 1+ c2m−1 and c = ud, where 2 2 6 d 2m,d > 1,u N. Also note that if b 4cǫ2 is a square, then x +bǫ1x+cǫ2 is reducible | 2m∈ m − and hence x + bǫ1x + cǫ2 is a product of two m degree polynomials. A. Schinzel studied the problem of reducibility of trinomials extensively in [98], [99], [100]. He even considered the factorization problem of trinomials over an arbitrary ﬁeld and provided many necessary and suﬃcient conditions for several families of trinomials. One such criterion is

8 Theorem 17 (Schinzel [98]). Let K be any field of characteristic different from 2 and 2m m b, c N. Then the polynomial x + bǫ1x + cǫ2 is reducible over K if and only if either b2 ∈4cǫ is a square in K or there is a prime p dividing m such that x2p + bǫ xp + cǫ is − 2 1 2 reducible over K or 4 m and x8 + bǫ x4 + cǫ is reducible over K. | 1 2 2p p It is, therefore, sufficient to find the reducibility condition of x + bǫ1x + cǫ2 to 2m m completely determine the reducibility of x + bǫ1x + cǫ2, where p is a prime dividing m. In the course of determining a Galois group of a family of polynomials, Filaseta et al. [38] proved that

2p p p Theorem 18 (Filaseta et al. [38]). Let p be an odd prime and u 2. Then x +ǫ1x +u is irreducible over Q. ≥

2p p Later they separately considered the reducibility of trinomials of the form x +bǫ1x + 2 2 cǫ2, where b, c N and p is a prime number in [37]. They noticed that if b 4cǫ2 = d 2 2p∈ b2 p 2p b2 p 2p − or c = b = d ,p 5 or c = 2 =2 d ,p 3 or c = 3 =3 d ,p 5, for some integer 2p ≥ p ≥ ≥ d Z, then x + bǫ1x + cǫ2 is reducible over Z. It has been proved in [37] that if b, c do ∈ 2p p not satisfy any of these conditions, then x + bǫ1x + cǫ2 is irreducible over Z provided

p ∤ (1812 1) (q2 1). − − q prime Yq|b

However, there are irreducible polynomials even when p divides the above product. For example, x4 +2x2 + 3 is irreducible. Here, p = b =2,c = 3; b, c do not satisfy any of the above conditions and the product is (1812 1)(22 1) divisible by 2. − − n 2.2.3 Reducibility of x + bǫ1x + cǫ2, c > 1 Suppose f(x) = xn + bǫ x + cǫ is a polynomial of degree n 2, where b N and 1 2 ≥ ∈ c 2, ǫ1, ǫ2 1, +1 . Serret [105] and Ore [81] studied the irreducibility of trinomials of≥ the form ∈xn { −x + cǫ }independently. By the use of reducibility of polynomials modulo − 2 a prime, they gave an irreducibility criterion for them.

p Theorem 19 (Serret [105], Ore [81]). Let c N and f(x)= x x + cǫ2 be a polynomial of prime degree p. Then f(x) is irreducible if∈ p ∤ c. − There are irreducible polynomials with p c. For example, x5 x + 10 and x2 x +2 both are irreducible. Irreducible polynomials| can be found for compo− site values− of n as well. For example,

x4 x + 2; x15 x + 3; − − x10 x + 6; x6 x + 18, − − are all irreducible. A.I. Bonciocat and N.C. Bonciocat [11] gave an irreducibility criterion for arbitrary polynomials with constant term divisible by large prime power. A version of their result in case of trinomials is

9 Theorem 20 (Bonciocat & Bonciocat [11]). Let a, b, c N, p be a prime number, and p ∤ bc, r 0,k 1. The polynomial axn + bprǫ x + pkcǫ∈is irreducible if ≥ ≥ 1 2 pk > bp2r + cn−1apnr. It is not possible to improve Theorem 20 for arbitrary values of a, b, c. For example, let f(x)= x2 +2x 8. Then a = b = c =1,p = n =2,r =1,k =3, 2k =22r +2nr and − f(x)=(x 2)(x + 4). − n r k Reducibility of x + p ǫ1x + p ǫ2 Let p be a prime number and f(x) = xn + prǫ x + pkǫ be a trinomial of degree n 2, 1 2 ≥ r, k 0, ǫ1, ǫ2 1, +1 . If a = b = c = 1 in Theorem 20, then ≥ ∈ { − n }r k The polynomial x + p ǫ1x + p ǫ2 is irreducible if

pk >p2r + pnr, where k 1,r 0. ≥ ≥ There are irreducible polynomials that do not satisfy the condition of Theorem 20. 4 4 4 2 4 For example, x +2ǫ1x +2 ǫ2 is irreducible while 2 < 2 +2 . n r It is possible to ﬁnd more irreducible families of polynomials of the form x + p ǫ1x + k p ǫ2, where p is a prime number and r, k 0. The case k = 0 has already been solved n ≥ in x + bǫ1x + ǫ2 and k =1,r 1 case follows from Eisenstein’s criterion. ≥ n Eisenstein’s criterion [34]: Let p be a prime number and let f(x)= anx + + 2··· a1x + a0 Z[x] be a polynomial of degree n. If p ai for 0 i n 1, and p ∤ an,p ∤ a0, then f(x)∈ is irreducible over Z. | ≤ ≤ − The case k = 1,r = 0 and p is an odd prime follows from Theorem 20. In [66] (see n Theorem 25 below), we have shown that if x +ǫ1x+2ǫ2 is reducible, then it is a product of cyclotomic factors and an irreducible non-reciprocal polynomial. If k 2 and r = 0, then it is irreducible by Theorem 20. An alternate proof of this particular≥ case can also be found in [65]. For k 2,r = 1, we will show that ≥ n k Theorem 21. Let p be a prime number and k 1, n 2. Then x + pǫ1x + p ǫ2 is irreducible. ≥ ≥

Proof. Suppose f(x) is reducible. Let f(x) = f1(x)f2(x) be a proper factorization of f(x), where f (x)= xs + c xs−1 + + c x + c ; 1 1 ··· s−1 s f (x)= xt + d xt−1 + + d x + d , t + s = n. 2 1 ··· t−1 t Comparing the coeﬃcient of x in f(x)= f1(x)f2(x), we have

cs−1dt + csdt−1 = pǫ1, (1)

n Reducing the coeﬃcient modulo p, we get fp(x)= x =(f1)p(x)(f2)p(x). This is possible 2 only when p ci,dj for every i and j. In other words, p divides the left hand side of (1), | n k which is not the case for the right hand side of (1). Hence, x + pǫ1x + p ǫ2 is irreducible for every k 1, n 2. ≥ ≥ 10 If k 2 and r 2, then there are polynomials which are reducible. For example, ≥ ≥ x5 9x 27=(x2 +3x + 3)(x3 3x2 +6x 9); − − − − x5 + 81x +243=(x2 +3x + 9)(x3 3x2 + 27); − x8 + 81x 81=(x2 3x + 3)(x6 +3x5 +6x4 +9x3 +9x2 27). − − − The complete characterization of the case k 2 and r 2 is not known. The condition ≥ ≥ of Theorem 20, pk >p2r + pnr is equivalent to k >nr except for k = nr +1,p =2, n = 2. 2 r 2r+1 If p =2, n = 2 and k = nr + 1, then f(x)= x +2 ǫ1x +2 ǫ2 is reducible only when r−1 r−1 ǫ2 = 1, and in that case f(x)=(x +2 (3 + ǫ1))(x 2 (3 ǫ1)). − n − r − k On the other hand, the polynomial f(x) = x + p ǫ1x + p ǫ2 is irreducible over k k k(n−1) n r Z if and only if f(p x) = p ǫ2(p ǫ2x + p ǫ1ǫ2x + 1) is irreducible over Q, that k(n−1) n r is, if and only if p ǫ2x + p ǫ1ǫ2x + 1 is irreducible over Z. The polynomialg ˜(x) = k(n−1) n r n r n−1 k(n−1) p ǫ2x +p ǫ1ǫ2x+1 is irreducible over Z if and only if g(x)= x +p ǫ1ǫ2x +p ǫ2 is irreducible over Z. From Perron’s criterion, the polynomial g(x) is irreducible if pr >pk(n−1) + 1. Since k 2, this is same as r>k(n 1). ≥ n − r k Therefore, if we summarize all these results for x + p ǫ1x + p ǫ2, then it can be put in the following theorem and table below.

n r k Theorem 22. Let p be a prime number, k,r 0 and let f(x)= x + p ǫ1x + p ǫ2 be a polynomial of degree n 2. ≥ ≥ (a) If f(x) is reducible and either r 0, 1 or k 0, 1 , then f(x) = f (x)f (x), ∈ { } ∈ { } c n where fc(x) is a product of cyclotomic polynomials and fn(x) is either 1 or a non- reciprocal irreducible polynomial.

2 r 2r+1 (b) If k 2,r 2 and f(x) = x +2 ǫ1x 2 , then f(x) is irreducible except possibly≥ for k≥ r k(n 1)6 . − n ≤ ≤ −

11 k r f(x) Irreducible Reducible 0 0 xn x 1 n 2 − − ≥ xn + x 1 n 5 (mod 6) n 5 (mod 6), − 6≡ ≡ f(x) = Φ6(x)fn(x) xn + ǫx +1 n 2 (mod 6) n 2 (mod 6), 6≡ ≡ f(x) = Φ3(ǫx)fn(x) 0 1 xn + ǫ prx + ǫ pr > 2 ≥ 1 2 xn +2x 1 n 2 − ≥ xn 2x 1 n 0 (mod 2) n 1 (mod 2), − − ≡ ≡ f(x)=(x + 1)fn(x) xn 2x +1 n 3,f(x) = (x 1)f (x) − ≥ − n n = 2,f(x) = (x 1)2 − xn +2x +1 n 1 (mod 2) n 0 (mod 2), ≡ ≡ n 3,f(x) = (x + 1)f (x) ≥ n n = 2,f(x) = (x + 1)2 1 0 xn + ǫ x + pǫ p 3 1 2 ≥ xn + x 2 f(x)=(x 1)f (x) − − n xn x +2 n 2 − ≥ xn + x +2 n 0 (mod 2) n 1 (mod 2), ≡ ≡ f(x)=(x + 1)fn(x) xn x 2 n 1 (mod 2) n 0 (mod 2), − − ≡ ≡ f(x)=(x + 1)fn(x) 1 1 xn + prǫ x + pkǫ n 2 ≥ 1 2 ≥ 2 1 xn + prǫ x + pkǫ n 2 ≥ ≤ 1 2 ≥ 2 xn + prǫ x + pkǫ According to Theorem 22 ≥ 1 2 Question 23. Let p be a prime number, r, k, n 2. For what values of r, where k ≥ n ≤ r k(n 1), the polynomial xn + prǫ x + pkǫ , ǫ , ǫ 1, +1 is irreducible over Z? ≤ − 1 2 1 2 ∈ { − } n m k 2.2.4 Reducibility of x + bǫ1x + p ǫ2, k ≥ 1

n m k Instead of restricting the value of m and the middle coefficient, let f(x)= x +bǫ1x +p ǫ2 be an n degree trinomial, where p is a prime number, b, k N, ǫ1, ǫ2 1, +1 . If k = 1 and p b, then f(x) is irreducible due to Eisenstein’s criterion.∈ If ∈k = { − 1 and p} ∤ b, there are only| finitely many b values for which f(x) is reducible. On the other hand, for k 2, the polynomial is irreducible provided pk is sufficiently large. ≥ (A) k = 1

12 n m n−1 By Nagell’s criterion, x + bǫ1x + pǫ2 is irreducible whenever b>p + 1. If the constant term is a large prime number, then Panitopol and Stef¨anescu [86] gave an eﬀective criterion for the irreducibility of polynomials.

n Theorem 24 (Panitopal & Stef¨anescu [86]). Let f(x)= anx + +a1x+a0 Z[x] be a polynomial with a > a + + a . If a is prime or ···a a ∈ < 1, | 0| | 1| ··· | n| | 0| | 0| − | n| then f(x) is irreducible in Z[x]. p p

n m Thus, x + bǫ1x + pǫ2 is irreducible for every prime p > b + 1. Theorem 24 fails n m to answer the irreducibility of x + ǫ1x +2ǫ2. We have considered the equality condition in Theorem 24 and proved that

nr nr−1 n1 Theorem 25 (Koley & Reddy [66]). Let f(x)= a x +a − x + +a x + nr nr 1 ··· n1 a Z[x],r 2, where a =0 for every i and let a + + a − + a = a 0 ∈ ≥ i 6 | n1 | ··· | nr 1 | | nr | | 0| be a prime number. If f(x) is reducible, then it is the product of an irreducible non-reciprocal polynomial and distinct cyclotomic polynomials. Furthermore, the nr nr−1 product of all of its cyclotomic factors is given by fc(x)=(x +sgn(a0anr ), x + n1 sgn(a a − ),...,x + sgn(a a )), where sgn(x) denotes the sign of x R. 0 nr 1 0 n1 ∈ Theorem 25 is not true if the constant term is not a prime number. For example,

x4 +3x2 +4=(x2 x + 2)(x2 + x +2); (2) − 2x3 + 11x2 +8x 21=(x 1)(x + 3)(2x + 7). − − From Theorem 25, the polynomial xn +(p 1)ǫ xm +pǫ has only one non-reciprocal − 1 2 irreducible factor apart from its cyclotomic factors. By Eisenstein’s criterion, if b 0 (mod p), then xn + bǫ xm + pǫ is irreducible. Therefore, combining all of ≡ 1 2 them, one can see that

n m Theorem 26. Let p be a prime, b N and let f(x) = x + bǫ1x + pǫ2 be a polynomial of degree n. Then f(x) is∈ irreducible except possibly for

p 1 b pn−1 +1, b 0 (mod p). − ≤ ≤ 6≡ n If b = p 1 and f(x) is reducible, then f(x) = fc(x)fn(x), where fc(x)=(x + m − ǫ2, x + ǫ1ǫ2) is the cyclotomic part of f(x) and fn(x) Z[x] is a non-reciprocal irreducible polynomial. ∈

(B) k ≥ 2

For m = 1, we have the irreducibility criterion given in Theorem 20. If m = 2, then A.I. Bonciocat and N.C. Bonciocat provided a criterion similar to Theorem 20.

13 Theorem 27 (Bonciocat & Bonciocat [11]). Let a, b, c N and p be a prime ∈n r 2 k number, p ∤ bc,r 0,k 1. The polynomial f(x) = ax + bp ǫ1x + p cǫ2 is irreducible if k is odd≥ and≥

pk > bcp3r + cn−1apnr.

The theorem is not true if k is an even integer. For example, let f(x)= x4+7x2+24. Then a =1= c, b = 7, n = k = 4,p = 2,r = 0 and 24 > 7 + 1. However, f(x)=(x2 x + 4)(x2 + x + 4). − n m When k is an even integer, Jonassen [56] studied trinomials of the form x +ǫ1x + 4ǫ2. By applying a reﬁnement of Ljunggren’s method, he proved that Theorem 28 (Jonassen [56]). Suppose n > m are positive integers. Then the n m polynomial x + ǫ1x +4ǫ2 is irreducible except in the following cases: x3k + ǫ x2k +4ǫ =(xk +2ǫ )(x2k ǫ xk + 2); 1 1 1 − 1 x5k + ǫ x2k 4ǫ =(x3k + ǫ x2k xk 2ǫ )(x2k ǫ xk + 2); 1 − 1 1 − − 1 − 1 x11k + ǫ x4k +4ǫ =(x5k x3k ǫ x2k +2ǫ )(x6k + x4k + ǫ x3k + x2k + 2), 1 1 − − 1 1 1 for any k 1 and the factors on the right hand side are irreducible in each cases. ≥ If c = 1,r = 0 in both Theorem 20 and Theorem 27, then the conditions read pk > a + b, which is similar to that of Theorem 24. Because of this reason, we n 3 k considered polynomials of the form f(x) = anx + + a3x + p ǫ, where p is a prime number, p ∤ a a and k 2 in [64]. To frame the··· results of [64] conveniently, 3 n ≥ we introduce the following sets: Let n>m> 0, a, b N and p be a prime number. We deﬁne ∈

= axn + bǫ xm + pkǫ p ∤ ab, k 2, ǫ , ǫ 1, +1 Sm { 1 2 | ≥ 1 2 ∈ { − }} and for m 2, ′ = f k 0 (mod m) . Then it has been shown that ≥ Sm { ∈ Sm | 6≡ } Theorem 29 (Koley & Reddy [64]). Let f(x) = axn + bǫ x3 + pkǫ ′ be a 1 2 ∈ S3 trinomial of degree n 4. If pk > a + b, then f(x) is irreducible. ≥ The condition p ∤ ab in the deﬁnition of cannot be dropped. For example, Sm 2x4 3x3 81=(x 3)(2x3 +3x2 +9x + 27). − − − Further, the necessity of the condition 3 ∤ k can be seen from the following examples:

x4 +4ǫx3 +27=(x +3ǫ)2(x2 2ǫx +3); (3) − 2x5 3x3 +27=(x2 3x + 3)(2x3 +6x2 +9x + 6). − − k n m k Later we consider the case p = a + b for f(x) = ax + bǫ1x + p ǫ2, m 3 and proved that ≤

14 n m k ′ ′ Theorem 30 (Koley & Reddy [64]). Let f(x)= ax +bǫ1x +p ǫ2 1 2 3 be a k ∈ S ∪S ∪S polynomial of degree n. If p = a+b and f(x) is reducible, then f(x)= fc(x)fn(x), n m where fc(x)=(x + ǫ2, x + ǫ1ǫ2) is the product of all cyclotomic factors of f and fn(x) is an irreducible non-reciprocal polynomial. In particular, f(x) is irreducible except for the following families of polynomials:

f(x) fc(x) f(x) fc(x) t ax2u+1 + bx + pkǫ x + ǫ ax22 (2u+1) bx2 pk x2 +1 2 2 − − ax2u bx pk x +1 ax2(2u+1) + bx2 + pkǫ x2 + ǫ − − 2 2 2(2u+1) 2 k 2 2u+1 3 k (2u+1,3) ax + bx + p ǫ2 x + ǫ2 ax + bx + p ǫ2 x + ǫ2 t ax2u+1 bx2 + pk x +1 ax2 (2u+1) bx3 pk x(2u+1,3) +1 − − − for every u, t 1. ≥ Example (2) shows that Theorem 30 is not true for f(x) ′ . The following ∈ S2 \ S2 example illustrate that it is neither true for f(x) ′ as well ∈ S3 \ S3 3x4 5x3 8=(x + 1)(x 2)(3x2 2x + 4). − − − − n m 2.3 Reducibility of ax + bǫ1x + cǫ2

n m Suppose f(x)= ax +bǫ1x +cǫ2 is a trinomial of degree n, where a, b, c N. If a,c > 1, then there are very few results available related to the reducibility criterion.∈ Theorem 20 and Theorem 27 gives an irreducibility criterion when the constant term is divisible by a large prime power. A. Schinzel [94] found the conditions on coeﬃcients so that an arbitrary trinomial is divisible by a cyclotomic polynomial. Theorem 31 (Schinzel [94]). Let a, b, c are non-zero integers, 0

2d m1+n1 m1 d m1 n1 x + ǫ1 ǫ2 x +1, if c = ǫ1a = ǫ2b,n1 + m1 0 (mod 3),ǫ1 = ǫ2 ; d m1 2 m1 ≡ n1 (x ( ǫ1) ǫ1ǫ2) , if c = ǫ1a + ǫ2b, ( ǫ1) = ( ǫ2) ,anǫ1 + bmǫ2 =0; fc x  ( )= d − − m1 − m1 − n1 x ( ǫ1) ǫ1ǫ2, if c = ǫ1a + ǫ2b, ( ǫ1) = ( ǫ2) ,anǫ1 + bmǫ2 =0;  − − − − 6 1 otherwise.  Schinzel also proved that there are only ﬁnitely many trinomials axn + bxm + c whose non-cyclotomic part is reducible. More precisely, Theorem 32 (Schinzel [94]). For any non-zero integers a, b, c there exist two eﬀectively computable constants A(a, b, c) and B(a, b, c) such that if n>m> 0 and the non- cyclotomic part of axn + bm + c is reducible, then (a) n A(a, b, c), (n,m) ≤ (b) there exists integers α and β such that m = n is integral, 0 <β<α B(a, b, c) β α ≤ and if g(x) is an irreducible non-reciprocal factor of axα + bxβ + c then g(xn/α) is an irreducible non-reciprocal factor of axn + bxm + c.

15 All the trinomials that we came across so far have at most two non-reciprocal irre- ′ ducible factors. We recall the two sets m and m introduced earlier. From Theorem ′ k S S 27, if f(x) 2 and p > a + b, then f(x) is irreducible. We have shown in [64] that ∈ S ′ k if f(x) 2 2 and p > a + b, then f(x) can have at most two non-reciprocal irreducible∈ factors. S \ S Apart from the polynomials given in Theorem 28, many polynomials having two irreducible non-reciprocal factors and k even. For example, if p is a prime and k 1, a 1, then ≥ ≥ x4 + (2 pk a2)x2 + p2k =(x2 + ax + pk)(x2 ax + pk) ′ · − − ∈ S2 \ S2 is a product of two irreducible non-reciprocal factors. Later we extend it and proved that if f(x) ′ and pk > a + b, then f(x) can have at most three non-reciprocal ∈ S3 \ S3 irreducible factor. The sharpness of this result follows from example (3). However, the number of non-reciprocal irreducible factors of axn +bxm +c is unbounded. For example,

2 x x2k 2k+1xk +22k =(xk 2k)2 = 2ϕ(d)Φ , (4) − −  d 2  Yd|k   ϕ(d) x where 2 Φd 2 Z[x] is an irreducible non-reciprocal polynomial for every d N. Over time, several∈ other questions were asked related to the irreducibility of trinomials.∈ For example, Schinzel [103] asked:

Question 33. Does there exist an integer K, independent of the polynomial, such that every trinomial in Q[x] has an irreducible factor in Q[x] with at most K terms?

Mrs. H. Smyczek [103] showed that if K exists, then K 6. Bremner [15] improved ≥ to K 8 by furnishing an exact trinomial of the form x14 + bx2 + c, having precisely two factors≥ each of length 8. However, the question remains open as of today. Inspired by the work of Bremner [15], several authors tried to characterize trinomials as having a factor of ﬁxed length. For further studies in this direction, one can look into [24], [49], [68], [70],[71], [89], [92], [95],[98], [99], [125]. Let 0 < n n n . A trinomial xn + bxm + c Q[x] is said to have 1 ≤ 2 ≤ ··· ≤ k ∈ reducibility type (n1, n2,...,nk), if it is a product of irreducible polynomials in Q[x] of degree n1, n2,...,nk. For example, if we take k = 15 in (4), then x30 216x15+230 = (x 2)2(x2+2x+4)2(x4+2x3+4x2+8x+16)2(x8 2x7+ +32x3 128x+256)2 − − − · · · − has a reducibility type (1, 1, 2, 2, 4, 4, 8, 8). A. Bremner and M. Ulas [17] studied possible reducibility type for n 10 and m 3. They showed that if m =1, then the reducibility ≤ ≤ type (1, 1, 1, 1) is impossible. It has been conjectured that

Conjecture 34 (Bremner & Ulas [17]). Let b, c N and f(x) = xn + bǫ x + cǫ be a ∈ 1 2 trinomial of degree n 4. Then the reducibility type (1, 1, 1, n 3) doesn’t exist. ≥ −

16 Suppose f(x) Q[x] is a polynomial which divides infinitely many trinomials. Can ∈ r f(x) divide a linear or quadratic polynomial in x for some r 1? For example, Φ3(x) divides x3k+2 + x + 1 for every k 1, by Theorem 2, and Φ (x)≥Φ (xr)= x2r + xr +1 for ≥ 3 | 3 every r 1, (r, 3) = 1. Is this true for arbitrary f(x) Q[x]? Posner and Rumsey [88] investigated≥ this question and subsequent development∈ can be found in [46], [102]. Till now we were confined to mainly questions regarding trinomials over field of rational numbers. Irreducibility of trinomials over arbitrary field is not that much common in the literature. Let K be an algebraic number field and f(x) = xn + bxm + c K[x] ∈ has a linear or quadratic factor g(x). Schinzel [101] studied reducibility of the part f(x) h(x) = g(x(n,m)) and gave several necessary and sufficient condition for reducibility of h(x) over K. For irreducibility of trinomials over the function field, we refer to [98], [99], [100]. The reducibility of trinomials over finite field has also been studied extensively. There is a classical result, known as modulo p-test (p being a prime number), which relates the irreducibility of polynomials over a finite field and the irreducibility of polynomials over Q. Modulo p-test([43], Theorem 17.3): Let p be a prime and suppose that f(x) Z[x] ∈ with deg(f(x)) 1. Let f˜(x) be the polynomial in Z [x] obtained from f(x) by reduc- ≥ p ing all the coefficients of f(x) modulo p. If f˜(x) is irreducible over Zp and deg(f˜(x)) = deg(f(x)), then f(x) is irreducible over Q.

For a general overview of polynomial factorization over finite field, the reader can consider [82], [118]. Swan [110] proved that if F is a field of odd characteristic and f(x) F[x] is a polynomial of degree n, having no repeated roots, then r n (mod 2) if and only∈ if discriminant (see section 4 for definition) of f(x) is a square in F≡, where r is the number of irreducible factors of f(x). Dickson [28] also proved this result independently. n n L. Carlitz [23] considered the trinomial g(x) = x2q +1 + xq −1 + 1 and proved that the degree of every irreducible factor of g(x) over Fq divides either 2n or 3n, where q is a prime power. The distribution of irreducible trinomials over F3 has been studied by O. Ahmadi [2]. It has been proved that if n is even, q is an odd prime power, then xn + bxm + c n m is irreducible over Fq if and only if x bx + c is irreducible over Fq. Later Ahmadi proved that if 4 n, then xn xm + 1 is irreducible− over F . A necessary condition for the | − 3 irreducibility of trinomials over a finite field is given by J. Gathen [121]: Theorem 35 (Gathen [121]). Let p be an odd prime, f(x) = xn + bxm + c,b,c r ∈ F r 0 ,r 1,n > m 1,d = (n, m), n = n/d, m = m/d,k = p(p 1),k = p \{ } ≥ ≥ 1 1 1 − 2 lcm(4,k1). Then the discriminant of f and the property that f is squarefree with odd number of irreducible factors depends only on the following residues:

n (mod k ); m (mod k ); n (mod pr 1) and m (mod pr 1). 2 1 1 − 1 − For further study on the factorization of trinomials over ﬁnite ﬁeld, reader should refer to [1], [6], [10], [35], [42], [120], [122].

17 3 Location of zeros of trinomials

The location of zeros of a polynomial sometimes helps us in determining the irreducibility of the polynomial. For example, if a monic polynomial f(x) Z[x], f(0) = 0 has ∈ 6 exactly one root in z > 1 and all the remaining roots are in z < 1, then f(x) is | | | | irreducible. For if f(x) = f1(x)f2(x) in Z[x], then either f1(x) or f2(x) has all its root in z < 1, contradicting the fact f (x), f (x) Z[x]. Similarly, if a monic polynomial | | 1 2 ∈ f(x) Z[x], f(0) = 0 has k roots in z > 1 and remaining roots are in z < 1, then f(x) can have∈ at most 6k irreducible factors.| | In this section, we will discuss about| | the location of zeros of trinomials. Unless specified otherwise, throughout the section. it is assumed that all the coefficients are non-zero real numbers. Kennedy [61] specified the region in which all the roots of the trinomial xn + bxm + c lie. Theorem 36 (Kennedy [61]). If α is the positive real root of the equation xn+ b xm c = | | −| | 0 and β is the positive real root of xn b xm c =0, then the roots of the trinomials xn + bxm + c lies in the annulus α −|z | β. −| | ≤| | ≤ However, finding a positive real root of xn +bxm c is not an easy task to accomplish. −| | Because of this reason, Kennedy gave a weaker but easily computable bound. Theorem 37 (Kennedy [61]). If α and β are the positive real roots of the equation xn−m = c n−m/n b and xn−m = c n−m/n + b respectively, then the roots of xn +bxm +c lie in the| annulus| −|α<| z < β. | | | | | | Suppose a, b, c are non-zero real numbers, a + c < b , and f(x) = axn + bxm + c. For z = 1, | | | | | | | | azn + bzm + c bzm = azn + c a + c < b = bzm . | − | | |≤| | | | | | | | From Rouché’s theorem, f(x) = axn + bxm + c, a + c < b has m number zeros in z < 1. If a + c = b , then the location of zeros| | has| | been| | calculated by Dilcher et al.| | [29]. They| | considered| | | | trinomials of the form bxn axm + a b with a>b> 0 and − − proved that Theorem 38 (Dilcher et al. [29]). Suppose a>b> 0 are real numbers. The number of zeros of f(x) = bxn axm + a b strictly inside the unit circle is m gcd(n, m) if − − − a n , and m if a < n . b ≥ m b m M. A. Brilleslyper and L.E. Schaubroeck [18] counted the number of roots in the interior of the unit circle for the trinomial xn + xm 1. They proved that if m = 1, then there are exactly 2 n + 1 roots in the interior of the− unit circle, where x denotes the ⌊ 6 ⌋ ⌊ ⌋ greatest integer less than or equal to x. They further conjectured that Conjecture 39 (Brilleslyper & Schaubroeck [18]). Let n, m N and (n, m)=1. Then the number of roots in the interior of the unit circle for the p∈olynomial xn + xm 1 is − n + m 1 2 − +1. 6 j k 18 The remaining results available in the literature focus on the zeros of trinomials of the form axn + bx + c. For example, Nicolas and Schinzel [80] studied the location of zeros of xn+1 ax +(a 1), and later the results have been improved further by Hernane − − and Nicolas [51]. The zeros of (a 2)xn +(a 1)x b, a > 2 has been studied in [3]. Fejér [36] and Szegô [111] independently− showed− that− the trinomial cxn x +1 has root − both in the regions z 1 1 and z 1 1, for any non-zero real number c. Joyal et.al. [58] proved that| if−n| ≥3, then the| − polynomial| ≤ cxn x + 1 has a root outside every circle that passes through the≥ origin. − P. Vassilev [119] considered trinomials with complex coefficients. He proved that the trinomial zn pz 1 has only one root in the interval (0, 1) when p < 0. Recently A. − − Melman [75] studied the problem of the location of zeros of trinomials extensively and improved many existing known results. The number of roots of a trinomial over a finite field has been estimated in [60]. For similar studies over finite fields, one may look into [27], [55], [116], [117].

4 Discriminant of trinomials

Suppose f(x)= a xn + a xn−1 + + a x + a Z[x] is a polynomial of degree n. Let n n−1 ··· 1 0 ∈ α1,α2,...,αn be the roots of f(x). The discriminant of f(x) is deﬁned as

2n−2 2 n 2n−2 D = a (α α ) =( 1)(2)a (α α ). f n i − j − n i − j iresultant of f(x). More precisely, the resultant of f(x), sometimes− denoted by Res(f, f ′), is deﬁned as

′ n 2n−1 2 2n−1 Res(f, f )=( 1)(2)a (α α ) = a (α α ), − n i − j n i − j iformal derivative of f(x). From the definition, it follows− that f(x)··· Z[x] has multiple roots if and only if ∈ Df = 0. If f(x) Z[x] has multiple roots, then f(x) is said to be a non-separable polynomial over Q.∈ Otherwise, it is said to be a separable polynomial over Q. There is another concept, called the discriminant of an algebraic number field. Sup- pose K is an algebraic number field and K is the ring of integers of K. Let b1, b2,...,bn be an integral basis of and let σ ,...,σO be the set of embeddings of K into C. OK { 1 n } Then the discriminant of K, denoted by ∆K, is defined as

2 ∆K = det(B) , where B =(σi(bj))i,j. Example. Let d be a square-free integer and let K = Q(√d) be a quadratic ﬁeld. Then the discriminant of K is (see [74] for details)

d, if d 1 (mod 4); ∆K = ≡ 4d, if d 2, 3 (mod4). ( ≡ 19 The discriminant of a polynomial and discriminant of an algebraic number ﬁeld are closely related. Let f(x) be a monic irreducible polynomial with a root α and K = Q(α). Then the discriminant of f(x) is

D = i(α)2∆ , where i(α) = #[ : Z[α]]. f K OK 2 If Df is square-free, then i(α) = 1 would imply that K = Z[α], and hence K has a basis in powers of α. In this case, K is said to be a monogenicO field. There areO several advantages of knowing a field to be monogenic. For example, let K be a monogenic field. Then the Galois closure L of K has Galois group Sn and L is everywhere unramified over its subfield Q( Df ). Therefore, the knowledge of the square-freeness of the discriminant of a polynomial is useful. However, it is very difficult to provide a formula for the p discriminant of a polynomial with more than three non-zero terms. To the best of our knowledge, only quadrinomials for which the discriminant is known as of today are of the form xn + t(x2 + bx + c) (see [85] for details). The reader can refer to Chapter 10 of [7] for a detailed literature review on the discriminants of trinomials up to 1980. P. Lefton [69] proved the discriminant formula of an arbitrary trinomial by using a polynomial and its derivatives along with the properties of roots of unity. He showed that the discriminant of axn + bxm + c is

n n−m−1 m−1 n/d m/d (n−m)/d n/d (n−m)/d m/d n/d d D =( 1)(2)a c (n a c ( 1) (n m) m b ) , − − − − where d =(n, m). D. Drucker and D. Goldschmidt [31] developed an ingenious method to compute the determinant of a matrix. By using this method, G.R.Greenﬁeld and D. Drucker [44] proved that

Theorem 40 (Greenﬁeld & Drucker [44]). The discriminant of the trinomial xn +bxm +c is n m−1 n/d n−m/d n/d n−m/d m/d n/d d D =( 1)(2)c n c ( 1) (n m) m b , − − − − where d =(n, m). Let (n, m) = 1 and f(x)= xn xm 1. From Theorem 40, the discriminant of f is ± ± D = nn (n m)n−mmm. f ± ± − Recently D. Boyd et al. [14] attempted to ﬁnd the square-free values of this discriminant. They found several families of polynomials whose discriminant has a square 2 2 factor. For example, if n 2 (mod 6), then (n −n+1) divides nn (n 1)n−1. That ≡ 9 − − is, the discriminant of xn x 1 has a square-factor when n 2 (mod 6). However, such a discriminant is sporadic.− − If n 1000, then there are only≡ 6 such values of n, n 130, 257, 487, 528, 815, 897 . It can≤ be seen that 832 130130 +129129 and 592 divide ∈{ } | nn +( 1)n(n 1)n−1 for the remaining values of n. They conjectured that − − Conjecture 41 (Boyd et al. [14]). The set of positive integers n such that nn +( 1)n(n 1)n−1 is square-free has density 0.9934466 ..., correct to that many decimal places.− −

20 They obtain 0.99344674 as an upper bound for this density. I.E. Shparlinski [107] gave a lower bound of the above density. If the abc conjecture is true, then Anirban Mukhopadhyay et al. [77] showed that for every odd n 5, there is a positive proportion of pairs (b, c) for which the trinomial xn + bx + c is irreducible≥ and has square-free discriminant. By using square-sieve and bounds of character sums, I.E. Shparlinski [106] proved a weaker but unconditional version of this result. K. Kedlaya[59] gave a method to construct a monic irreducible polynomial f(x) Z[x] of degree n 2 having exactly r real roots, 0 r n, with a ∈ ≥ ≤ ≤ square-free discriminant. On the other hand, discriminants of algebraic number fields have also been explored by several mathematicians. Suppose α is a root of the irreducible polynomial xn + bx + c and K = Q(α). K. Komatsu [67] found the discriminant of K as well as an integral basis for K. Later, P. Llorente et al. [73] improved the results of Komatsu. B. K. Spearman and K.S. Williams [109] restricted to trinomials f(x)= x5 + ax + b, whose Galois group is a dihedral group of order 10. If α is a root of f and K = Q(α), then they give an explicit formula of ∆K in terms of a and b. A. Alaca and S¸. Alaca [4] utilized a method described by S¸. Alaca [5] and found a p-integral base of K for every prime p, which in turn helps them to provide a formula for ∆K, where K = Q(α),α being a root of the irreducible polynomial x5 + bx + c. We have seen that the polynomial xn x 1 is irreducible with discriminant nn + ( 1)n(n 1)n−1 and has Galois group S .− Boyd− et al. [14] showed that there are certain − − n values of n (see above) for which the discriminant of xn x 1 has a square factor. J. Lagarias(see problem 99:10, [78]) asked − − Is there a monogenic field whose Galois group is Sn for every n 5? K. Kedlaya [59] answered this question affirmatively through a≥ different approach. Boyd et al. [14] answered this by using square freeness of discriminant. For similar studies in this direction, we refer to [113], [114], [123], [124]. Galois group of trinomials of different forms has been considered by several authors. One may consult [9], [19], [20], [21], [25], [26], [50], [57], [83], [84], [108] for further studies. n For example, H. Osada [83] proved that the Galois group of x x 1 over Q is Sn for all n 2. Later Osada [84] gave some number-theoretic conditions− − on the coefficients ≥ n m b, c so that x + bx + c has Galois group either Sn or An over Q. In contrast, a class of trinomials has been determined by A. Bishnoi and S.A. Khan- n duja [9] having Galois group Sn. In particular, they showed that if n 8, 2 permutation polynomial if c f(c) is a permutation from F to itself.∈ Permutation trinomials has been studied in [8],7→ [30], [52], [53], [54].

21 5 Open Problems

We have already mentioned several conjectures, open problems, and asked a few questions in respective places. In this section, we would like to ask few more questions related to the irreducibility of trinomials and locations of zeros of trinomials which may give impetus to the reader to work further in this direction. n m Recall that the reducibility nature of x +2ǫ1x + ǫ2 has been resolved by Theorem n 9. If p is an odd prime, from Theorem 10, there are only ﬁnitely many values of m for n m which x + pǫ1x + ǫ2 is reducible.

Question 42. Let p be an odd prime, ǫ1, ǫ2 1, +1 and n > m 1. What are the n m ∈ { − } ≥ values of n and m for which x + pǫ1x + ǫ2 is reducible?

n m From Theorem 32, there are only ﬁnitely many trinomials x + bǫ1x + cǫ2 whose non-cyclotomic part is reducible. However, it is not known whether there are only ﬁnitely n m many reducible trinomials of the form x + bǫ1x + cǫ2. In particular,

Question 43. Let n > m 1, b N, ǫ1, ǫ2 1, +1 . Are there ﬁnitely many ≥ n ∈ m ∈ { − } reducible trinomials of the form x + bǫ1x + ǫ2?

n n−1 From Theorem 6 and Theorem 7, the trinomial x + ǫ1x + ǫ2 is reducible only when n = 6t +2, ǫ1 = ǫ2 = 1 or n = 6t +5, ǫ1 = ǫ2 = 1, t N. If b 2, then the n −n−1 − − ∈ ≥ reducibility of x + bǫ1x + cǫ2, b = c has been determined under the certain condition on c in Theorem 15. For b = c 2,6 Harrington conjectured (recall Conjecture 16) that n≥ n−1 apart from two polynomials, x + bǫ1x + bǫ2 is irreducible.

Question 44. Let c N and c > 1, ǫ1, ǫ2 1, +1 . What are the values of c for n n−1 ∈ ∈ { − } which x + ǫ1x + cǫ2is irreducible? More generally, we ask the following question

Question 45. Let b, c N,b,c 2, and c > 2(b 1), ǫ1, ǫ2 1, +1 . For what ∈ ≥n n−1 − ∈ { − } values of b and c, the polynomial x + bǫ1x + cǫ2 is irreducible?

2p p Filaseta et al. [38] gave a suﬃcient condition for the irreducibility of x +bǫ1x +cǫ2. There are only ﬁnitely many values of p for which the irreducibility question remains open. Question 46. Let p be a prime number, b, c N, ǫ , ǫ 1, +1 and ∈ 1 2 ∈ { − } p (1812 1) (q2 1). | − − qprime Yq|b If b, c do not satisfy any of the following conditions b2 b2 b2 4cǫ = d2 or c = b2 = d2p,p 5 or c = =2pd2p,p 3 or c = =3pd2p,p 5 − 2 ≥ 2 ≥ 3 ≥ for some d Z, then which are the families among x2p + bǫ xp + cǫ irreducible? ∈ 1 2 22 n m In Theorem 26, the irreducibility of x + bǫ1x + pǫ2 has been discussed except for p +1 b pn−1 +1,p ∤ b. ≤ ≤ n m Question 47. Let p be a prime number, b N, and let f(x) = x + bǫ1x + pǫ2. For what values of b, p +1 b pn−1 +1,p ∤ b,∈ the polynomial f(x) is irreducible? ≤ ≤ The number of roots of xn + x 1 which lies inside the unit circle has been counted − in [18] and they conjectured (see Conjecture 39) the exact number of roots within the unit circle for arbitrary values of m.

n m Question 48. Let n > m 1 and ǫ1, ǫ2 1, +1 . How many roots of x + ǫ1x + ǫ2 lies in the interior of the unit≥ circle? ∈ { − }

More generally, one can look for polynomials of the form f(x)= axn +bxm +c, where a, b, c are non-zero real numbers. From Theorem 38 and related discussion, one can ﬁnd the number of roots of f(x) that are inside the unit circle provided b a + c . | |≥| | | | Question 49. Let a, b, c R 0 and n > m 1. How many roots of axn + bxm + c lies in the interior of the∈ unit\{ circle?} ≥

Acknowledgement: We thank the reviewer for providing valuable suggestions and comments on an earlier version of the manuscript.

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