ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS

SHUICHI OTAKE AND TONY SHASKA

Abstract. In this paper, we compute the discriminant of a quartinomial n 2 of the form f(b,a,1)(t; x) = x + t(x + ax + b) by using the Bezoutian. Then, by using this result and another theorem of our previous paper, we construct a family of totally complex polynomials of the form f(b,a,1)(ξ, x) (ξ ∈ R, (a, b) 6= (0, 0)).

1. Introduction The discriminant of a polynomial has been a major objective of research in algebra because of its importance. For example, the discriminant of a polynomial allows us to know whether the polynomial has multiple roots or not and it also plays an important role when we compute the discriminant of a number field. Moreover the discriminant of a polynomial also tells us whether the Galois group of the polynomial is contained in the alternating group or not. This is why there are so many papers focusing on studying the discriminant of a polynomial ([B-B-G], [D-S], [Ked], [G-D], [Swa]). In the last two papers, the authors concern the computation of the discriminant of a trinomial and it has been carried out in different ways. n In this paper, we compute the discriminant of a quartinomial f(b,a,1)(t; x) = x + t(x2 + ax + b) by using the Bezoutian (Theorem2). Let F be a field of characteristic zero and f1(x), f2(x) be polynomials over F . Then, for any integer n such that n ≥ max{degf1, degf2}, we put n f1(x)f2(y) − f1(y)f2(x) X B (f , f ) : = = α xn−iyn−j ∈ F [x, y], n 1 2 x − y ij i,j=1

Mn(f1, f2) : = (αij)1≤i,j≤n.

0 The n × n matrix Mn(f1, f2) is called the Bezoutian of f1 and f2. When f2 = f1, the formal derivative of f1 with respect to the indeterminate x, we often write 0 0 Bn(f1) := Bn(f1, f1), Mn(f1) := Mn(f1, f1) and the matrix Mn(f1) is called the

Bezoutian of f1. We denote by Nf1 the number of distinct real roots of f1(x) and ∆(f1) the discriminant of f1(x). Moreover, for any real symmetric matrix M, we denote by σ(M) the index of inertia of M. The following are the list of important properties of Bezoutians. Proposition 1. Notations as above, we have

(i) Mn(f1, f2) is an n × n symmetric matrix over F . (Mn(f1, f2) ∈ Symn(F )). (ii) Bn(f1, f2)(Mn(f1, f2)) is linear in f1 and f2, separately. (iii) Bn(f1, f2) = −Bn(f2, f1)(Mn(f1, f2) = −Mn(f2, f1)).

(iv) Nf1 = σ(Mn(f1)). 1 2 SHUICHI OTAKE AND TONY SHASKA

1 (v) ∆(f1) = 2 det Mn(f1), where led(f1) is the leading coefficient of f1. led(f1) In particular, if f1 is monic, we have ∆(f1) = det Mn(f1). µ ν (vi) Let λ, µ, ν be integers such that λ ≥ µ > ν ≥ 0. Then Mλ(x , x ) = (mij)1≤i,j≤λ, where ( 1 i + j = 2λ − (µ + ν) + 1 (λ − µ + 1 ≤ i, j ≤ λ − ν), mij = 0 otherwise.

Proof. For (i)–(iv), see [Fuh, Theorem 8.25, 9.2]. For (v), see equation (4) of [Fro, p.217]. For (vi), see [O-S, Lemma 2] 

s+1 For any vector r = (r0, ··· , rs) ∈ R , let us put s X s−k gr(x) = g(r0, ··· , rs; x) = rs−kx ∈ R[x], k=0 (n) n fr(t; x) = f (r0, ··· , rs, t; x) = x + t · gr(x) ∈ R(t)[x]. In the previous paper [O-S], by using the above properties of Bezoutians, we ob- tained the next theorem

s+1 Theorem 1. [O-S, Theorem 2] Let r = (r0, ··· , rs) ∈ R be a vector such that gr(x) is a degree s separable polynomial satisfying Ngr (x) = γ (0 ≤ γ ≤ s). Let us (n) consider fr(t; x) = f (r0, ··· , rs, t; x) as a polynomial over R(t) in x and put 0 Pr(t) = det Mn(fr(t; x)) = det Mn(fr(t; x), fr(t; x)),

0 where fr(t; x) is a derivative of fr(t; x) with respect to x. Then, for any real number ξ > αr = max{α ∈ R | Pr(α) = 0}, we have  γ + 1 n − s : odd  Nfr (ξ;x) = γ n − s : even, rs > 0  γ + 2 n − s : even, rs < 0.

Then, by using Theorem1 and Theorem2, we construct a certain family of totally complex polynomials of the form f(b,a,1)(ξ; x) (Theorem3). Here, note that a real polynomial f(x) to be called totally complex if it has no real roots, that is, Nf = 0.

2. Main results Let n ≥ 3 be an integer and a, b be real numbers. Let us put c = (b, a, 1) ∈ R3 and 2 n gc(x) = x + ax + b ∈ R[x], fc(t; x) = x + tgc(x) ∈ R(t)[x]. n
n
We denote by m (n, m ∈ Z≥0) the binomial coefficient. Note that 0 = 1 for any n
n
integer n ≥ 0 and m = 0 if n < m. Moreover, we define m = 0 for any n ∈ Z≥0 and m ∈ Z<0. In the following, we prove the next two theorems. ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS3

Theorem 2. Put m0 = b(n − 3)/2c, m1 = d(n − 3)/2e. Moreover, for any integers n ≥ 4 and k (0 ≤ k ≤ m0), put !  2 ! n − k − 3 n(n − 1) 5n − (6k + 23)n + 10k + 24 n − k − 3 S =(n − 1)3 a4 − a2b k k n − k − 3 k ! n − k − 4 + 4n2(n − 2) b2. k Here, b·c is the floor function and d·e is the ceiling function. (1) Suppose n = 3. Then, we have ( ) 2 2 2 3 2 ∆ (fc(t; x)) = t (a − 4b)t − (4a − 18ab)t − 27b .

(2) Suppose n ≥ 4. Then, we have ( ) m1 n−1 n−2 2 2 n n−1 ∆ (fc(t; x)) = (−1) t (n − 2) (a − 4b)t + γct − n b , where  m0 X n+k k n−2k−4 k n−2k−4 k  (−1) n (n − 1) (n − 2) a b Sk (a 6= 0 or a = 0, n : even), γc = k=0  0 (a = 0, n : odd). Theorem 3. Suppose n ≥ 4 is an even integer. Moreover, let us suppose b 6= 0 and 2 2 (n − 1) a /4n(n − 2) ≤ b. Then we have Nfc(ξ;x) = 0 for any positive real number ξ. By a direct computation, we have ∆ x3 + t(x2 + ax + b)
= t2 (a2 − 4b)t2 − (4a3 − 18ab)t − 27b2 , ∆ x4 + t(x2 + ax + b)
= −t3 (4a2 − 16b)t2 + (27a4 − 144a2b + 128b2)t − 256b3 and we get Theorem2 for n = 3, 4. Moreover, if n = 4 and (n−1)2a2/4n(n−2) = b, we have 1  27 2 P (t) = ∆ x4 + t(x2 + ax + b)
= a2 t + a2 t3 c 2 8 and hence αc = 0, which implies Nfc(ξ;x) = 0 for any ξ > 0 by Theorem1 since x2 + ax + b has no real root in this case. Then, by considering the graph of the 4 2 2 2 function y = x +ξ(x +ax+b), we have Nfc(ξ;x) = 0 whenever (n−1) a /4n(n−2) ≤ b and ξ > 0, which is the claim of Theorem3 for n = 4. Therefore, let us assume n ≥ 5 hereafter. In the following, we put (c) Ac(t) = (aij (t))1≤i,j≤n = Mn(fc(t; x)) ∈ Symn(R(t)), (c) Bc = (bij )1≤i,j≤2 = M2(gc(x)) ∈ Sym2(R),

Pc(t) = det Ac(t) = ∆(fc(t; x)). Then, by Theorem1, we have the next lemma. 2 Lemma 1. Suppose n is even and a /4 < b. Put αc = max{α ∈ R | Pc(α) = 0}.

Then, for any real number ξ > αc, we have Nfc(ξ,x) = 0. 4 SHUICHI OTAKE AND TONY SHASKA

To prove Theorem2, we will use the equality ∆( fc(t; x)) = det Ac(t). Also, to 2 2 prove Theorem3, it is enough to prove αc = 0 when (n − 1) a /4n(n − 2) ≤ b and (a, b) 6= (0, 0), which implies we need to compute the polynomial Pc(t) = det Ac(t) precisely. Here, let Qm(k; c) = (qij)1≤i,j≤m, Rm(k, l; c) = (rij)1≤i,j≤m be m × m elementary matrices such that    1  1 .  ..   ..   .         1 c   1       ..  Qm(k; c)=  c , Rm(k, l; c)=  .  ,      1   1   .     ..   ..     .  1 1 where qkk = c, rkl = c, respectively and, for any m × m matrices M1, M2, ··· , Ml, Ql put k=1 Mk = M1M2 ··· Ml. Moreover, put lk = n + k. Then, as is the case in [O-S], we inductively define the matrix Ac(t)k (1 ≤ k ≤ n − 2) as follows;

(c) t (1) Ac(t)1 = (aij (t)1)1≤i,j≤n = Sc(t)1Ac(t)Sc(t)1, where 1 √ Y (c) √ S (t) = Q (1; 1/ n) R (1, l − 1; −a (t)/ n). c 1 n n k 1,lk−1 k=0 (2) Put ( (n − 1)/2, n : odd, n0 = n/2, n : even.

(c) t Then, Ac(t)k = (aij (t)k)1≤i,j≤n = Sc(t)kAc(t)k−1Sc(t)k, where  n (c) !  Y akm(t)k−1  Rn l0 − k, m; − (2 ≤ k ≤ n0)  −(n − 2)t m=l0−k+1  (c) ! n (c) ! Sc(t)k = akk (t)k−1 Y akm(t)k−1 Rn l0 − k, k; − Rn l0 − k, m; −  −2(n − 2)t −(n − 2)t  m=k+1   (n0 < k ≤ n − 2).

By the definition of Ac(t)k (1 ≤ k ≤ n − 2), we have det Ac(t) = n det Ac(t)n−2. Therefore, to compute the polynomial Pc(t) = det Ac(t), let us first compute the matrix Ac(t)n−2 concretely.

3. Computation of the matrix Ac(t)n−2 By Proposition1, we have 2 Bc = M2(x + ax + b, 2x + a) 2 2 2 = 2M2(x , x) + aM2(x , 1) + (a − 2b)M2(x, 1)  2 a  = . a a2 − 2b

Here, let us give some examples of the matrices Ac(t) and Ac(t)1 for some small n. ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS5

Example 1. (1) Let n = 5. Then,

 5 0 0 2t at   0 0 −3t −4at −5bt    Ac(t) =  0 −3t −4at −5bt 0 ,    2t −4at −5bt 2t2 at2  at −5bt 0 at2 (a2 − 2b)t2  1 0 0 0 0   0 0 −3t −4at −5bt    Ac(t)1 =  0 −3t −4at −5bt 0 .    0 −4at −5bt (6/5)t2 (3/5)at2  0 −5bt 0 (3/5)at2 (4/5)a2 − 2b t2

(2) Let n = 6. Then,

 6 0 0 0 2t at   0 0 0 −4t −5at −6bt     0 0 −4t −5at −6bt 0  Ac(t) =  ,  0 −4t −5at −6bt 0 0     2t −5at −6bt 0 2t2 at2  at −6bt 0 0 at2 (a2 − 2b)t2  1 0 0 0 0 0   0 0 0 −4t −5at −6bt     0 0 −4t −5at −6bt 0  Ac(t)1 =  .  0 −4t −5at −6bt 0 0     0 −5at −6bt 0 (4/3)t2 (2/3)at2  0 −6bt 0 0 (2/3)at2 (5/6)a2 − 2b t2

Suppose n ≥ 5. Then, by Proposition 4 and equation (5) in [O-S], we have

 n 0 ...... 0 2t at   0 −(n − 2)t −(n − 1)at −nbt    .  . . .   .   . −(n − 1)at −nbt 0    ......  . .   . −nbt 0 .  (1) Ac(t) =    ......   ......   . . .   .   . .   0 −(n − 2)t −(n − 1)at −nbt 0 0     2t −(n − 1)at −nbt 0 ... 0 2t2 at2  at −nbt 0 ...... 0 at2 (a2 − 2b)t2 and hence (2)  1 0 ...... 0 0 0   0 −(n − 2)t −(n − 1)at −nbt    . . .  .   .   . −(n − 1)at −nbt 0   . .  . . . .  . .   . −nbt 0 .  Ac(t)1 =   .  ......   ......   . . .   .   . .   0 −(n − 2)t −(n − 1)at −nbt 0 0   2
2 2
2   0 −(n − 1)at −nbt 0 ... 0 2 1 − n t 1 − n at  2
2  1
2 2 0 −nbt 0 ...... 0 1 − n at 1 − n a − 2b t

Here, similar to Ac(t)k (2 ≤ k ≤ n − 2), we inductively define the matrix W (t)k = (wij(t)k)1≤i,j≤n (2 ≤ k ≤ n − 2) as follows; 6 SHUICHI OTAKE AND TONY SHASKA

(1) W (t)1 = (wij(t)1)1≤i,j≤n, where qt i + j = n, 2 ≤ i, j ≤ n − 2,  rt i + j = n + 1, 2 ≤ i, j ≤ n − 1, wij(t)1 = st i + j = n + 2, 2 ≤ i, j ≤ n,  0 otherwise. t (2) W (t)k = (wij(t)k)1≤i,j≤n (2 ≤ k ≤ n − 2) = S(t)kW (t)k−1S(t)k, where  n   Y wkm(t)k−1  Rn n − k, m; − (2 ≤ k ≤ n0),  qt m=n−k+1    n   S(t)k = wkk(t)k−1 Y wkm(t)k−1 R n − k, k; − R n − k, m; −  n −2qt n qt  m=k+1   (n0 < k ≤ n − 2). Example 2. (1) Let n = 5. Then,  0 0 0 0 0   0 0 qt rt st    W (t)1 =  0 qt rt st 0 ,    0 rt st 0 0  0 st 0 0 0  0 0 0 0 0   0 0 qt 0 0   2  W (t)n =  0 qt rt (qs − r )t/q −rst/q , 0    0 0 (qs − r2)t/q −(2qs − r2)rt/q2 −(qs − r2)st/q2  0 0 −rst/q −(qs − r2)st/q2 rs2t/q2  0 0 0 0 0   0 0 qt 0 0    W (t)n−2 =  0 qt 0 0 0 .    0 0 0 −(2qs − r2)rt/q2 −(qs − r2)st/q2  0 0 0 −(qs − r2)st/q2 rs2t/q2 (2) Let n = 6. Then,  0 0 0 0 0 0   0 0 0 qt rt st     0 0 qt rt st 0  W (t)1 =  ,  0 qt rt st 0 0     0 rt st 0 0 0  0 st 0 0 0 0  0 0 0 0 0 0   0 0 0 qt 0 0     0 0 qt 0 0 0  W (t)n =  , 0  0 qt 0 (qs − r2)t/q −(2qs − r2)rt/q2 −(qs − r2)st/q2     0 0 0 −(2qs − r2)rt/q2 −(q2s2 − 3qr2s + r4)t/q3 (2qs − r2)rst/q3  0 0 0 −(qs − r2)st/q2 (2qs − r2)rst/q3 (qs − r2)s2t/q3  0 0 0 0 0 0   0 0 0 qt 0 0     0 0 qt 0 0 0  W (t)n−2 =  .  0 qt 0 0 0 0     0 0 0 0 −(q2s2 − 3qr2s + r4)t/q3 (2qs − r2)rst/q3  0 0 0 0 (2qs − r2)rst/q3 (qs − r2)s2t/q3 ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS7

In the same way, let us compute the matrix W (t)n−2 for any integer n ≥ 7.

Lemma 2. Suppose n ≥ 7. Let {xm}, {ym} be sequences defined by the next recurrence relations;

r s x = 0, x = −qt, x = rt, x = − x − x (m ≥ 1), 0 1 2 m+2 q m+1 q m s y = y = 0, y = − x (m ≥ 1) 0 1 m+1 q m and put

( n0 − 1 n : odd, n1 = n0 − 2 n : even.

Then, for any integer k such that 2 ≤ k ≤ n1, we have

0 (i, j) = (k, `) or (`, k)(n − k + 1 ≤ ` ≤ n),   x`−n+k+2 (i, j) = (k + 1, `) or (`, k + 1) (n − k ≤ ` ≤ n − 1),  y`−n+k+2 (i, j) = (k + 2, `) or (`, k + 2) (n − k ≤ ` ≤ n − 1), wij(t)k = −sx /q (i, j) = (k + 1, n) or (n, k + 1),  k  −syk/q (i, j) = (k + 2, n) or (n, k + 2),  wij(t)k−1 otherwise.

Proof. Let us prove this lemma by induction on k. First, suppose k = 2. Then, by the definition of W (t)2, we have

 0 ······ ...... 0 0 0 0   .   . qt 0 0     .   . qt rt (qs − r2)t/q −rst/q     . . .   . . rt st −rst/q −s2t/q     . . . . .   . . . st 0 0 0  W (t)2 =   ,  ......   ......   . .   . . . .   ......   0 0 qt rt st .   . . .   . . . . .   0 qt rt st 0 .   . .   2 . . .   0 0 (qs − r )t/q −rst/q 0 .  0 0 −rst/q −s2t/q 0 ············ 0 which implies the claim of Lemma2 for k = 2 since

(qs − r2)t rst x = rt, x = , y = st, y = − . 2 3 q 2 3 q 8 SHUICHI OTAKE AND TONY SHASKA

Next, suppose Lemma2 is true for k = 2, ··· , m − 1 (m − 1 < n1). Then, since

 0 ······························ 0 0  . . . . .  . . . .   . . .     .   .   . qt 0 ··· 0 0     .   . qt x2 x3 ··· xm −sxm−1/q     . . .   . . rt y y ··· y −sy /q   2 3 m m−1   . . .   . . . . .   . st 0 0 ··· 0 0   ......   ......  W (t)m−1 =  . .  ,    ......   . . . . .   . qt rt st .   . . . .   ......   . qt x2 y2 0 .    . . . .  . . .   . . 0 x y 0 .   3 3   ......   ......   ......   . .   . . .   0 ··· 0 xm ym 0 .  0 ··· 0 −sxm−1/q −sym−1/q 0 ·················· 0 we have

 0 ······························ 0 0  . . . . .  . . . .   . . .     .   .   . qt 0 ··· 0 0     .   . qt 0 0 ··· 0 0     . . .   . . rt x0 x0 ··· x0 x0   2 3 m m+1   . . .   . . . . . 0 0 0 0   . st y2 y3 ··· ym ym+1   ......   ......  W (t)m =  . .  ,    ......   . . . . .   . qt rt st .   . . . .   . 0 0 . . . . .   . qt 0 x2 y2 .    . . . .  . . .   . . 0 0 x0 y0 .   3 3   ......   ......   ......   . .   0 0 . . .   0 ··· 0 0 xm ym .  0 0 0 ··· 0 0 xm+1 ym+1 ·················· 0 where x r s x0 = y − ` · rt = − x − x = x (2 ≤ ` ≤ m), ` ` qt q ` q `−1 `+1 x s y0 = − ` · st = − x = y (2 ≤ ` ≤ m), ` qt q ` `+1 s  r   s  s  r s  s x0 = − y + − · − x = − − x − x = − x m+1 q m−1 q q m−1 q q m−1 q m−2 q m  s  s  s y0 = − · − x = − y . m+1 q q m−1 q m

This completes the proof of Lemma2.  ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS9

Proposition 2. For any integer n ≥ 5, we have  qt i + j = n (2 ≤ i, j ≤ n − 2),  x (i, j) = (n − 1, n − 1),  n−1 wij(t)n−2 = yn−1 (i, j) = (n − 1, n) or (n, n − 1),  (−s/q)yn−2 (i, j) = (n, n),  0 otherwise.

Proof. Proposition2 has been proved for n = 5, 6 in Example2. Thus, we suppose n ≥ 7. First, by solving the recurrence relation given in Lemma2, we have

m m−1  (−1) mr 2  2m−1qm−2 t (r − 4qs = 0),  (3) xm = n √ m  √ mo (−1)m r+ r2−4qs − r− r2−4qs  √ t (r2 − 4qs 6= 0).  2mqm−2 r2−4qs Note that n m  mo (−1)m r + pr2 − 4qs − r − pr2 − 4qs t 2mqm−2pr2 − 4qs b(m−1)/2c (−1)mt X  m  = · 2 rm−2k−1(r2 − 4qs)k 2mqm−2 2k + 1 k=0 and hence, by putting r2 − 4qs = 0, we have

m m m n p 2   p 2  o (−1) r + r − 4qs − r − r − 4qs (−1)mmrm−1 t = t, 2mqm−2pr2 − 4qs 2m−1qm−2 the solution for the case of r2 − 4qs = 0. Here, suppose n is odd. Then, we have n1 = (n − 3)/2 and hence by Lemma2,

0 (i, j) = ((n − 3)/2, `) or (`, (n − 3)/2) ((n + 5)/2 ≤ ` ≤ n),  x (i, j) = ((n − 1)/2, `) or (`, (n − 1)/2) ((n + 3)/2 ≤ ` ≤ n − 1),  `−(n−1)/2  y`−(n−1)/2 (i, j) = ((n + 1)/2, `) or (`, (n + 1)/2) ((n + 3)/2 ≤ ` ≤ n − 1), wij (t)n1 = −sx(n−3)/2/q (i, j) = ((n − 1)/2, n) or (n, (n − 1)/2),  −sy /q (i, j) = ((n + 1)/2, n) or (n, (n + 1)/2),  (n−3)/2  wij (t)(n−5)/2 otherwise.

Therefore, we have the next expression of the matrix W (t)n1 ;

 0 ····················· 0 0   . . . . .   . . . .     .   .   . qt 0 ··· 0 0     .   . qt x2 x3 ··· x(n−1)/2 −sx(n−3)/2/q     .   . qt rt y y ··· y −sy /q   2 3 (n−1)/2 (n−3)/2  . W (t)n1 =  . .   . .   . qt x2 y2 0 .   . . .   . .   . 0 x y .   3 3   . . . . .   . . . . .   . . . . .     .   0 ··· 0 x(n−1)/2 y(n−1)/2 .  0 ··· 0 −sx(n−3)/2/q −sy(n−3)/2/q ············ 0 10 SHUICHI OTAKE AND TONY SHASKA

t Then, since W (t)n1+1 = S(t)n1+1W (t)n1 S(t)n1+1, where n   Y n + 1 w(n−1)/2,m(t)n1 S(t) = R , m; − n1+1 n 2 qt m=(n+3)/2

and wn−1,(n+1)/2(t)n1+1 = w(n+1)/2,n−1(t)n1+1 = x(n+1)/2, we have x x w (t) = − (n−1)/2 · y − (n−1)/2 · x n−1,n−1 n1+1 qt (n−1)/2 qt (n+1)/2 x  s  x = − (n−1)/2 · − x − (n−1)/2 · x qt q (n−3)/2 qt (n+1)/2  (n−1)/2  (n−1)/2 (−1)(n−1)/2 r + pr2 − 4qs − r − pr2 − 4qs 1 = − · t qt 2(n−1)/2q(n−5)/2pr2 − 4qs

 (n−3)/2  (n−3)/2 (−1)(n−3)/2 r + pr2 − 4qs − r − pr2 − 4qs  s  × − · t q 2(n−3)/2q(n−7)/2pr2 − 4qs

 (n−1)/2  (n−1)/2 (−1)(n−1)/2 r + pr2 − 4qs − r − pr2 − 4qs 1 − · t qt 2(n−1)/2q(n−5)/2pr2 − 4qs

 (n+1)/2  (n+1)/2 (−1)(n+1)/2 r + pr2 − 4qs − r − pr2 − 4qs × t 2(n+1)/2q(n−3)/2pr2 − 4qs

 n−2  p  (n−3)/2  p  = −s r + r2 − 4qs − (4qs) r + r2 − 4qs

n−2  (n−3)/2  p   p  n−2 n−4 2 − (4qs) r − r2 − 4qs + r − r2 − 4qs t 2 q (r − 4qs)

 n  p  (n−1)/2  p  + r + r2 − 4qs − (4qs) r + r2 − 4qs

n  (n−1)/2  p   p  n n−3 2 − (4qs) r − r2 − 4qs + r − r2 − 4qs t 2 q (r − 4qs)

(  p n−2  p 2  = r + r2 − 4qs r + r2 − 4qs − 4qs

) n−2  2    p   p  n n−3 2 + r − r2 − 4qs r − r2 − 4qs − 4qs t 2 q (r − 4qs)

 n−1  n−1 2pr2 − 4qs r + pr2 − 4qs − r − pr2 − 4qs = t 2nqn−3(r2 − 4qs)

 n−1  n−1 r + pr2 − 4qs − r − pr2 − 4qs

= t = xn−1. 2n−1qn−3pr2 − 4qs ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS11

Similarly, we have x −sy −sx w (t) = w (t) = − (n−1)/2 · (n−3)/2 − (n−3)/2 x n−1,n n1+1 n,n−1 n1+1 qt q q2t (n+1)/2  (n−1)/2  (n−1)/2 (−1)(n−1)/2 r + pr2 − 4qs − r − pr2 − 4qs s2 = − · t q3t 2(n−1)/2q(n−5)/2pr2 − 4qs  (n−5)/2  (n−5)/2 (−1)(n−5)/2 r + pr2 − 4qs − r − pr2 − 4qs × t 2(n−5)/2q(n−9)/2pr2 − 4qs  (n−3)/2  (n−3)/2 (−1)(n−3)/2 r + pr2 − 4qs − r − pr2 − 4qs s + · t q2t 2(n−3)/2q(n−7)/2pr2 − 4qs  (n+1)/2  (n+1)/2 (−1)(n+1)/2 r + pr2 − 4qs − r − pr2 − 4qs × t 2(n+1)/2q(n−3)/2pr2 − 4qs 2  s p n−3 (n−5)/2 p 2 = − · (r + r2 − 4qs) − (4qs) (r + r2 − 4qs) q   (n−5)/2 p 2 p n−3 n−3 n−5 2 − (4qs) (r − r2 − 4qs) + (r − r2 − 4qs) t 2 q (r − 4qs)  s p n−1 (n−3)/2 p 2 + · (r + r2 − 4qs) − (4qs) (r + r2 − 4qs) q   (n−3)/2 p 2 p n−1 n−1 n−4 2 − (4qs) (r − r2 − 4qs) + (r − r2 − 4qs) t 2 q (r − 4qs)

p n−2 p n−2 s (r + r2 − 4qs) − (r − r2 − 4qs) = · t q 2n−2qn−4pr2 − 4qs s = − x = y . q n−2 n−1

Moreover, since wn,(n+1)/2(t)n1+1 = w(n+1)/2,n(t)n1+1 = −sx(n−1)/2/q, we have

wn,n(t)n1+1 −sx −sy −sx −sx = − (n−3)/2 · (n−3)/2 − (n−3)/2 · (n−1)/2 q2t q q2t q  (n−3)/2  (n−3)/2 (−1)(n−3)/2 r + pr2 − 4qs − r − pr2 − 4qs s3 = · t q4t 2(n−3)/2q(n−7)/2pr2 − 4qs  (n−5)/2  (n−5)/2 (−1)(n−5)/2 r + pr2 − 4qs − r − pr2 − 4qs × t 2(n−5)/2q(n−9)/2pr2 − 4qs  (n−3)/2  (n−3)/2 (−1)(n−3)/2 r + pr2 − 4qs − r − pr2 − 4qs s2 − · t q3t 2(n−3)/2q(n−7)/2pr2 − 4qs  (n−1)/2  (n−1)/2 (−1)(n−1)/2 r + pr2 − 4qs − r − pr2 − 4qs × t 2(n−1)/2q(n−5)/2pr2 − 4qs 12 SHUICHI OTAKE AND TONY SHASKA

3  s p n−4 (n−5)/2 p = − · (r + r2 − 4sq) − (4qs) (r + r2 − 4qs) q2   (n−5)/2 p p n−4 n−4 n−6 2 − (4qs) (r − r2 − 4qs) + (r − r2 − 4qs) t 2 q (r − 4qs)

2  s p n−2 (n−3)/2 p + · (r + r2 − 4sq) − (4qs) (r + r2 − 4qs) q2   (n−3)/2 p p n−2 n−2 n−5 2 − (4qs) (r − r2 − 4qs) + (r − r2 − 4qs) t 2 q (r − 4qs)

p n−3 p n−3 s2 (r + r2 − 4sq) − (r − r2 − 4qs) = · t q2 2n−3qn−5pr2 − 4qs

s  s  s = − − x = − y . q q n−3 q n−2

Therefore, we can express the matrix W (t)n0 (= W (t)n1+1) as follows;

 0 ····················· 0 0  . . . .  . . . . .   . . .     .   . qt 0 ··· 0 0     .   . qt 0 0 ··· 0 0     .  W (t) =  . qt ∗ ∗ ∗ · · · ∗ ∗  . n0    .   . qt 0 ∗ ∗ ∗ · · · ∗ ∗     . . .   . . 0 0 ∗ ∗ ∗ · · · ∗ ∗     ......   ......     0 ··· 0 0 ∗ ∗ ∗ · · · xn−1 yn−1  0 ··· 0 0 ∗ ∗ ∗ · · · yn−1 (−s/q)yn−2

Then, by the definition of the matrix Wk (n0 < k ≤ n − 2), we have

 0 ·················· 0 0  . . . .  . . . . .   . . .     .   . qt 0 ··· 0 0     .   . qt 0 0 ··· 0 0     .  W (t)n−2 =  . qt 0 0 0 ··· 0 0  ,    . . .   . . 0 0 0 0 ··· 0 0     ......   ......   ......   0 0 0 0 0 ··· x y   n−1 n−1  0 ··· 0 0 0 0 ··· yn−1 (−s/q)yn−2 which completes the proof of Proposition2 for odd n. ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS13

Next, suppose n is even. Then, we have n1 = (n − 4)/2 and hence by Lemma2, we have

0 (i, j) = ((n − 4)/2, `) or (`, (n − 4)/2) ((n + 6)/2 ≤ ` ≤ n),   x`−n/2 (i, j) = ((n − 2)/2, `) or (`, (n − 2)/2) ((n + 4)/2 ≤ ` ≤ n − 1),  y`−n/2 (i, j) = (n/2, `) or (`, n/2) ((n + 4)/2 ≤ ` ≤ n − 1), wij(t)n = 1 −sx /q (i, j) = ((n − 2)/2, n) or (n, (n − 2)/2),  (n−4)/2  −sy(n−4)/2/q (i, j) = (n/2, n) or (n, n/2),  wij(t)(n−6)/2 otherwise.

Thus, we have the next expression of the matrix W (t)n1 ;

 0 ························ 0 0  . . . .  . . . . .   . . .     .   .   . qt 0 ··· 0 0   .   .   . qt x2 x3 ··· x(n−2)/2 −sx(n−4)/2/q     .   . qt rt y y ··· y −sy /q   2 3 (n−2)/2 (n−4)/2   .  W (t)n1 =  .  .  . qt rt st 0 0 ··· 0 0     . .   . qt x2 y2 0 0 .     . . . .   . . .   . 0 x3 y3 0 .   ......   ......   ......     .   0 ··· 0 x(n−2)/2 y(n−2)/2 0 .  0 ··· 0 −sx(n−4)/2/q −sy(n−4)/2/q 0 ············ 0

Then, for any integer ` ((n + 4)/2 ≤ ` ≤ n), let us put

t V (t)` = (vij(t)`)1≤i,j≤n = R(t)`W (t)n1 R(t)`, where `   Y w(n−2)/2,m(t)n1 R(t) = R (n + 2)/2, m; − . ` n qt m=(n+4)/2

Note that we have V (t)n = W (t)n1+1. Claim . For any integer `0 ((n + 4)/2 ≤ `0 ≤ n − 1), we have

0 (i, j) = ((n − 2)/2, `) or (`, (n − 2)/2) ((n + 4)/2 ≤ ` ≤ `0),   0 x`−(n−2)/2 (i, j) = (n/2, `) or (`, n/2) ((n + 2)/2 ≤ ` ≤ ` ),  y (i, j) = ((n + 2)/2, `) or (`, (n + 2)/2) ((n + 2)/2 ≤ ` ≤ `0),  `−(n−2)/2
2 0 vij (t)`0 = −x`−n/2/qt y2 (i, j) = (`, `) ((n + 4)/2 ≤ ` ≤ ` ),  (−x /qt)  m−n/2  0  ×y`−(n−2)/2 (i, j) = (`, m) or (m, `) ((n + 4)/2 ≤ ` < m ≤ ` ),  wij (t)(n−4)/2 otherwise. 14 SHUICHI OTAKE AND TONY SHASKA

Proof of Claim. To ease notation, let us put k = (n − 2)/2. Then, by definition,

 0 ························ 0 0  . . . .  . .   . . .     .   . qt 0 ··· 0 0     .   . qt 0 x ··· x −sx /q   3 k k−1   .   .   . qt x2 x3 y3 ··· yk −syk−1/q   .  V (t)(n+4)/2 =  .   . qt x2 y2 y3 0 ··· 0 0   . .   . 2 .   . qt 0 x3 y3 (−x2/qt) y2 .   . . .   . . . .   . 0 x3 y3 0 .     ......   ......     .   0 ··· 0 xk yk 0 .  0 ··· 0 −sxk−1/q −syk−1/q 0 ············ 0 and we get Claim for V (t)(n+4)/2. Here, suppose Claim is true for V (t)`0−1 and hence  0 ····················· 0 0 ···  . . . .  . .   . . .     .   .   . qt 0 0 ··· 0 x`0−k−1 ···     .   . qt x2 x3 x4 ··· x`0−k−1 y`0−k−1 ···     .   0   . qt x2 y2 y3 y4 ··· y` −k−1 0 ···  V (t) 0 =  . . .  . ` −1  . . 2   . 0 x3 y3 (−x2/qt) y2 (−x3/qt) y3 ··· (−x`0−k−2/qt)y3 0 ···   .   . 2   . 0 x y (−x /qt) y (−x /qt) y ··· (−x 0 /qt)y 0 ···   4 4 3 3 3 2 ` −k−2 4   ......   ......   ......   2   0 0 x`0−k−1 y`0−k−1 (−x`0−k−2/qt)y3 (−x`0−k−2/qt)y4 ··· (−x`0−k−2/qt) y2 0 ···     0 ··· x`0−k−1 y`0−k−1 0 0 0 ··· 0 0 ···   ......  ...... Thus, by a direct computation, we have

 0 ····················· 0 0 ···  . . . .  . .   . . .     .   .   . qt 0 0 ··· 0 x`0−k ···     .   . qt x2 x3 x4 ··· x`0−k y`0−k ···     .   0   . qt x2 y2 y3 y4 ··· y` −k 0 ···  V (t) 0 =  . . .  `  . . 2   . 0 x3 y3 (−x2/qt) y2 (−x3/qt) y3 ··· (−x`0−k−1/qt)y3 0 ···   .   . 2   . 0 x y (−x /qt) y (−x /qt) y ··· (−x 0 /qt)y 0 ···   4 4 3 3 3 2 ` −k−1 4   ......   ......   ......   2   0 0 x`0−k y`0−k (−x`0−k−1/qt)y3 (−x`0−k−1/qt)y4 ··· (−x`0−k−1/qt) y2 0 ···     0 ··· x`0−k y`0−k 0 0 0 ··· 0 0 ···   ......  ......

0 and we get Claim by induction on ` .  ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS15

By above Claim,

 0 ····················· 0 0  . . . .  . . . . .   . . .     .   . qt 0 0 ··· 0 −sx /q   k−1   .   .   . qt x2 x3 x4 ··· xk+1 −syk−1/q     .   . qt x2 y2 y3 y4 ··· yk+1 0  V (t)n−1 =    . . .   . . 2   . 0 x3 y3 (−x2/qt) y2 (−x3/qt) y3 ··· (−xk/qt)y3 0   .   . 2   . 0 x4 y4 (−x3/qt) y3 (−x3/qt) y2 ··· (−xk/qt)y4 0     ......   ......   2   0 0 xk+1 yk+1 (−xk/qt)y3 (−xk/qt)y4 ··· (−xk/qt) y2 0  0 · · · −sxk−1/q −syk−1/q 0 0 0 ··· 0 0 and hence, by the definition of V (t)n, we have

W (t)n1+1 = V (t)n =  0 ····················· 0 0  . . . .  . . . . .   . . .     .   .   . qt 0 0 ··· 0 0   .   .   . qt x2 x3 x4 ··· xk+1 −sxk/q     .   . qt x y y y ··· y −sy /q   2 2 3 4 k+1 k  .  . . .   . . 2 2 2   . 0 x3 y3 (−x2/qt) y2 (−x3/qt) y3 ··· (−xk/qt)y3 (xk−1/q t )y2y3   .   . 2 2 2   . 0 x4 y4 (−x3/qt) y3 (−x3/qt) y2 ··· (−xk/qt)y4 (xk−1/q t )y2y4     ......   ......     0 0 x y (−x /qt)y (−x /qt)y ··· (−x /qt)2y (x /q2t2)y y   k+1 k+1 k 3 k 4 k 2 k−1 2 k+1  2 2 2 2 2 2 2 2 2 3 0 ··· 0 −sxk/q −syk/q (xk−1/q t )y2y3 (xk−1/q t )y2y4 ··· (xk−1/q t )y2yk+1 (xk−1/q t ) y2

Therefore, by the definition of the matrix W (t)n1+2 = W (t)n0 , we have

wn−1,n−1(t)n0 2  x 2 x
= − (n−2)/2 y − n/2 qt 2 qt

 n−2  n−2  s r + pr2 − 4qs + r − pr2 − 4qs − 2(4qs)(n−2)/2 = t 2n−2qn−4 (r2 − 4qs)  n  n r + pr2 − 4qs + r − pr2 − 4qs − 2(4qs)n/2 − t 2nqn−3 (r2 − 4qs)  n−2   2 r + pr2 − 4qs 4qs − r + pr2 − 4qs = t 2nqn−3 (r2 − 4qs)  n−2   2 r − pr2 − 4qs 4qs − r − pr2 − 4qs + t 2nqn−3 (r2 − 4qs)  n−2     r + pr2 − 4qs −2pr2 − 4qs r + pr2 − 4qs = t 2nqn−3 (r2 − 4qs)  n−2     r − pr2 − 4qs 2pr2 − 4qs r − pr2 − 4qs + t 2nqn−3 (r2 − 4qs)

= xn−1. 16 SHUICHI OTAKE AND TONY SHASKA

Similarly, we have x sx x w (t) = w (t) = (n−4)/2 y y + (n−2)/2 n/2 n−1,n n0 n,n−1 n0 q2t2 2 n/2 q2t x  s  sx x = (n−4)/2 y − x + (n−2)/2 n/2 q2t2 2 q (n−2)/2 q2t  n−3  n−3  − r + pr2 − 4qs + r − pr2 − 4qs − 2r (4qs)(n−4)/2 s2 = − · t q3 2n−3qn−7 (r2 − 4qs)  n−1  n−1  − r + pr2 − 4qs + r − pr2 − 4qs − 2r (4qs)(n−2)/2 s + · t q2 2n−1qn−5 (r2 − 4qs)   n−3  n−3  s 4qs r + pr2 − 4qs + 4qs r − pr2 − 4qs − 2r (4qs)(n−2)/2 = t 2n−1qn−3 (r2 − 4qs)  n−1  n−1  s r + pr2 − 4qs + r − pr2 − 4qs − 2r (4qs)(n−2)/2 − t 2n−1qn−3 (r2 − 4qs)  n−3     s r + pr2 − 4qs −2pr2 − 4qs r + pr2 − 4qs = t 2n−1qn−3 (r2 − 4qs)  n−3     s r − pr2 − 4qs 2pr2 − 4qs r − pr2 − 4qs + t 2n−1qn−3 (r2 − 4qs)  n−2  n−2 p 2 p 2 s r + r − 4qs − r − r − 4qs = − · t = yn−1, q 2n−2qn−4pr2 − 4qs

2  x 2 s2 x
w (t) = (n−4)/2 y3 − (n−2)/2 n,n n0 q2t2 2 q3t  n−4  n−4  s3 r + pr2 − 4qs + r − pr2 − 4qs − 2(4qs)(n−4)/2 = t 2n−4qn−4(r2 − 4qs)  n−2  n−2  s2 r + pr2 − 4qs + r − pr2 − 4qs − 2(4qs)(n−2)/2 − t 2n−2qn−3(r2 − 4qs)   n−4  n−4  s2 4qs r + pr2 − 4qs + 4qs r − pr2 − 4qs − 2(4qs)(n−2)/2 = t 2n−2qn−3(r2 − 4qs)  n−2  n−2  s2 r + pr2 − 4qs + r − pr2 − 4qs − 2(4qs)(n−2)/2 − t 2n−2qn−3(r2 − 4qs)  n−4     s2 r + pr2 − 4qs −2pr2 − 4qs r + pr2 − 4qs = t 2n−2qn−3 (r2 − 4qs)  n−4     s2 r − pr2 − 4qs 2pr2 − 4qs r − pr2 − 4qs + t 2n−2qn−3 (r2 − 4qs)  n−3  n−3 − r + pr2 − 4qs − r − pr2 − 4qs  s   s  s = − · − · t = − yn−2. q q 2n−3qn−5pr2 − 4qs q ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS17

Therefore, we can express the matrix W (t)n0 as follows;  0 ·················· 0 0  . . . .  . . . . .   . . .     .   . qt 0 ··· 0 0     .   . qt 0 0 ··· 0 0     .  . W (t)n0 =  . qt 0 ∗ ∗ · · · ∗ ∗     . . .   . . 0 0 ∗ ∗ · · · ∗ ∗     ......   ......   ......   0 0 0 ∗ ∗ · · · x y   n−1 n−1  0 ··· 0 0 ∗ ∗ · · · yn−1 (−s/q)yn−2

Then, by the definition of the matrix Wk (n0 < k ≤ n − 2), we have  0 ·················· 0 0  . . . .  . . . . .   . . .     .   . qt 0 ··· 0 0     .   . qt 0 0 ··· 0 0     .  W (t)n−2 =  . qt 0 0 0 ··· 0 0  ,    . . .   . . 0 0 0 0 ··· 0 0     ......   ......   ......   0 0 0 0 0 ··· x y   n−1 n−1  0 ··· 0 0 0 0 ··· yn−1 (−s/q)yn−2 which completes the proof of Proposition2 for even n. 

In the end, let us apply Proposition2 to the matrix Ac(t)1 (n ≥ 5). Then, by the definition of the matrix Ac(t)k (2 ≤ k ≤ n − 2) and Proposition2, we have  1 0 ...... 0 0 0   0 −(n − 2)t 0 0 

 . .   . . . . .   .   . 0 0    ......  . . .   . . .  (4) Ac(t)n−2 =   ,  ......   ......   . . .   . . .   ......   0 −(n − 2)t 0 0     0 0 0 ...... 0 a(c) (t) a(c) (t)   n−1,n−1 n−2 n−1,n n−2  (c) (c) 0 0 0 ...... 0 an,n−1(t)n−2 an,n(t)n−2 where  (c) a (t) = 2 (1 − 2/n) t2 +x ¯  n−1,n−1 n−2 n−1  (c) (c) 2 an−1,n(t)n−2 = an,n−1(t)n−2 = (1 − 2/n) at − (nbx¯n−2)/(n − 2)   (c)  2 2 2 2 2 an,n(t)n−2 = (1 − 1/n) a − 2b t + (n b x¯n−3)/(n − 2) . Here, for any integer m ≥ 0, we denote (−1)m (Am − Bm) x¯m = t, 2m{−(n − 2)}m−2p(n − 1)2a2 − 4n(n − 2)b

A = −(n − 1)a + p(n − 1)2a2 − 4n(n − 2)b,

B = −(n − 1)a − p(n − 1)2a2 − 4n(n − 2)b. 18 SHUICHI OTAKE AND TONY SHASKA

4. Proof of Theorems By a direct computation, we have " (c) (c) # an−1,n−1(t)n−2 an−1,n(t)n−2 det (c) (c) an,n−1(t)n−2 an,n(t)n−2  2  = 1 − (a2 − 4b)t4 n 2(n − 2) n2b2x¯  n − 1  2(n − 2) nbx¯ + t2 · n−3 + a2 − 2b t2 · x¯ + at2 · n−2 n (n − 2)2 n n−1 n n − 2 n2b2 n2b2 + x¯ x¯ − x¯2 (n − 2)2 n−1 n−3 (n − 2)2 n−2 (5) 2 ( 2  2 ) (n − 2)(a − 4b) 2nb x¯ (n − 1)a − 2nb x¯n−1 = t4 + n−3 + + 2abx¯ t2 n n − 2 n n−2 n2b2 + x¯ x¯ − x¯2
. (n − 2)2 n−1 n−3 n−2 Here, let us put ( ) 2nb2x¯ (n − 1)a2 − 2nb x¯ α(t) = n−3 + n−1 + 2abx¯ t2, n − 2 n n−2 n2b2 β(t) = x¯ x¯ − x¯2
. (n − 2)2 n−1 n−3 n−2 Lemma 3. Suppose n ≥ 5 and put n − k − 3 n(n − 1) 5n2 − (6k + 23)n + 10k + 24 n − k − 3 S =(n − 1)3 a4 − a2b k k n − k − 3 k n − k − 4 + 4n2(n − 2) b2. k Then, we have  m0 n+k k n−2k−4 k n−2k−4 k X (−1) n (n − 1) (n − 2) a b Sk  t3 (a 6= 0, or a = 0, n : even), α(t) = n(n − 2)n−3 k=0  0 (a = 0, n : odd). nn−1bn−1 β(t) = − t2. (n − 2)n−3 To carry out these computations, we need some combinatorial identities. Lemma 4. [Rio, Chapter 2, Problem 18 (c), Chapter 6, Problem 18 (a)] Let N (≥ 0) be an integer and put m = bN/2c, m = dN/2e = b(N +1)/2c. Then, we have m X N + 1j N − k = 2N−2k (k ∈ , 0 ≤ k ≤ m), 2j + 1 k k Z j=k

m X N + 1j  N + 1 − k N − k = 2N−2k 2 − 2j k k k j=k

N + 1 − k N − k = 2N−2k + (k ∈ , 0 ≤ k ≤ m). k k − 1 Z ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS19

Proof. We omit the proof of the first identity and let us prove the second one. By using the convention  n  = 0 (n ∈ , m ∈ ), n + m Z≥0 Z≥1 we have m m X N + 1j X N + 1j = 2j + 1 k 2j + 1 k j=k j=k and hence m m m m X N + 1j X N + 1j X N + 1j X N + 1j + = + 2j k 2j + 1 k 2j k 2j + 1 k j=k j=k j=k j=k m X N + 1 N + 1 j = + 2j 2j + 1 k j=k m X N + 2j = 2j + 1 k j=k for any k (0 ≤ k ≤ m). Thus, by using the first identity, we get the second one.  Proof of Lemma3. We first prove the second equality. Then, by the definition of x¯m, we have n2b2 −An−1Bn−3 − An−3Bn−1 + 2An−2Bn−2 β(t) = · t2 (n − 2)2 22n−4(n − 2)2n−8 {(n − 1)2a2 − 4n(n − 2)b} n o n2b2 (4n(n − 2)b)n−3 A2 + B2
− 2 (4n(n − 2)b)n−2 = − t2 22n−4(n − 2)2n−6 {(n − 1)2a2 − 4n(n − 2)b}

22n−6nn−1(n − 2)n−3bn−1 · 4 (n − 1)2a2 − 4n(n − 2)b = − t2 22n−4(n − 2)2n−6 {(n − 1)2a2 − 4n(n − 2)b}

nn−1bn−1 = − t2, (n − 2)n−3 which is the second equality. For the first equality, let us first suppose a = 0. Then, we have n o (−1)m {−4n(n − 2)b}m/2 − (−1)m {−4n(n − 2)b}m/2 x¯m = t 2m {−(n − 2)}m−2 {−4n(n − 2)b}1/2  2 {−4n(n − 2)b}(m−1)/2  t (m : odd), = 2m(n − 2)m−2 0 (m : even)  (−nb)(m−1)/2  t (m : odd), = (n − 2)(m−3)/2 0 (m : even) 20 SHUICHI OTAKE AND TONY SHASKA and hence we have

2nb2x¯  α(t) = n−3 − 2bx¯ t2 n − 2 n−1  2nb2 (−nb)(n−4)/2 2b(−nb)(n−2)/2   · t − t t2 (n : even), = n − 2 (n − 2)(n−6)/2 (n − 2)(n−4)/2 0 (n : odd) (−1)n/24n(n−2)/2bn/2  t3 (n : even), = (n − 2)(n−4)/2 0 (n : odd), which is the claim of Lemma3 for the case a = 0 since for even n (n ≥ 6),

Pm0 (−1)n+knk(n − 1)n−2k−4(n − 2)kan−2k−4bkS k=0 k t3 n(n − 2)n−3 (−1)(3n−4)/2n(n−4)/2(n − 2)(n−4)/2b(n−4)/2 · 4n2(n − 2)b2 = t3 n(n − 2)n−3 (−1)n/24n(n−2)/2bn/2 = t3. (n − 2)(n−4)/2

Next, we suppose a 6= 0 and let us put

F = (n − 1)2a3 − 2n(2n − 3)ab, G = (n − 1)a2 − 2nb, H = (n − 1)2a2 − 4n(n − 2)b.

Then, we have

2nb2x¯ (n − 1)a2 − 2nb x¯ n−3 + n−1 + 2abx¯ n − 2 n n−2 2nb2 An−3 − Bn−3
(n − 1)a2 − 2nb An−1 − Bn−1
= √ t + √ t n − 2 2n−3(n − 2)n−5 H n 2n−1(n − 2)n−3 H

2ab(An−2 − Bn−2) + √ t 2n−2(n − 2)n−4 H

8n2(n − 2)b2(An−3 − Bn−3) (n − 1)a2 − 2nb (An−1 − Bn−1) = √ t + √ t 2n−1n(n − 2)n−3 H 2n−1n(n − 2)n−3 H 4n(n − 2)ab(An−2 − Bn−2) + √ t 2n−1n(n − 2)n−3 H 8n2(n − 2)b2 + (n − 1)a2 − 2nb A2 + 4n(n − 2)abA = √ An−3t 2n−1n(n − 2)n−3 H 8n2(n − 2)b2 + (n − 1)a2 − 2nb B2 + 4n(n − 2)abB − √ Bn−3t 2n−1n(n − 2)n−3 H ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS21 and 8n2(n − 2)b2 + (n − 1)a2 − 2nb A2 + 4n(n − 2)abA n √ o = 8n2(n − 2)b2 + (n − 1)a2 − 2nb 2(n − 1)2a2 − 4n(n − 2)b − 2(n − 1)a H n √ o + 4n(n − 2)ab −(n − 1)a + H √ = −2 (n − 1)2a3 − 2n(2n − 3)ab H + 2 (n − 1)a2 − 2nb H √ = −2F H + 2GH,

8n2(n − 2)b2 + (n − 1)a2 − 2nb B2 + 4n(n − 2)abB n √ o = 8n2(n − 2)b2 + (n − 1)a2 − 2nb 2(n − 1)2a2 − 4n(n − 2)b + 2(n − 1)a H n √ o + 4n(n − 2)ab −(n − 1)a − H √ = 2 (n − 1)2a3 − 2n(2n − 3)ab H + 2 (n − 1)a2 − 2nb H, √ = 2F H + 2GH. Hence, by putting n on−3 √ An−3 = −(n − 1)a + p(n − 1)2a2 − 4n(n − 2)b = I + J H, we have 2nb2x¯ (n − 1)a2 − 2nb x¯ n−3 + n−1 + 2abx¯ n − 2 n n−2 n √ o  √  n √ o  √  −2F H + 2GH I + J H − 2F H + 2GH I − J H = √ t 2n−1n(n − 2)n−3 H −FI + GHJ = t. 2n−3n(n − 2)n−3 Then, by using Lemma4, we have

m0   X n − 3 n−2`−3 ` I = {−(n − 1)a} (n − 1)2a2 − 4n(n − 2)b 2` `=0 m0   ( `   ) X n − 3 n−2`−3 X ` `−k k = {−(n − 1)a} (n − 1)2a2 {−4n(n − 2)b} 2` k `=0 k=0 m m X0 X0 n − 3` = (−1)n+k+122knk(n − 1)n−2k−3(n − 2)kan−2k−3bk 2` k k=0 `=k m X0 n − k − 3 n − k − 4 = 2n−2k−4 + k k − 1 k=0 × (−1)n+k+122knk(n − 1)n−2k−3(n − 2)kan−2k−3bk m X0 n − 3 n − k − 3 = (−1)n+k+12n−4nk(n − 1)n−2k−3(n − 2)k an−2k−3bk n − k − 3 k k=0 22 SHUICHI OTAKE AND TONY SHASKA and

m0   X n − 3 n−2`−4 ` J = {−(n − 1)a} (n − 1)2a2 − 4n(n − 2)b 2` + 1 `=0 m0   ( `   ) X n − 3 n−2`−4 X ` k = {−(n − 1)a} {(n − 1)2a2}`−k {−4n(n − 2)b} 2` + 1 k `=0 k=0 m m X0 X0  n − 3 ` = (−1)n+k22knk(n − 1)n−2k−4(n − 2)kan−2k−4bk 2` + 1 k k=0 `=k m X0 n − k − 4 = 2n−2k−4 (−1)n+k22knk(n − 1)n−2k−4(n − 2)kan−2k−4bk k k=0 m X0 n − k − 4 = (−1)n+k2n−4nk(n − 1)n−2k−4(n − 2)k an−2k−4bk, k k=0 which implies,

− FI + GHJ = −{(n − 1)2a3 − 2n(2n − 3)ab}I + {(n − 1)3a4 − 2n(n − 1)(3n − 5)a2b + 8n2(n − 2)b2}J

m0 X = (−1)n+k2n−4nk(n − 1)n−2k−4(n − 2)kan−2k−4bk k=0  n − 3 n − k − 3 n − 3 n − k − 3 × (n − 1)3 a4 − 2n(n − 1)(2n − 3) a2b n − k − 3 k n − k − 3 k n − k − 4 n − k − 4 n − k − 4  + (n − 1)3 a4 − 2n(n − 1)(3n − 5) a2b + 8n2(n − 2) b2 k k k m0 X = (−1)n+k2n−3nk(n − 1)n−2k−4(n − 2)kan−2k−4bk k=0  n − k − 3 n(n − 1) 5n2 − (6k + 23)n + 10k + 24 n − k − 3 × (n − 1)3 a4 − a2b k n − k − 3 k n − k − 4  + 4n2(n − 2) b2 k Therefore, we finally obtain

( 2  2 ) 2nb x¯ (n − 1)a − 2nb x¯n−1 −FI + GHJ α(t) = n−3 + + 2abx¯ t2 = t3 n − 2 n n−2 2n−3n(n − 2)n−3

m0 n+k k n−2k−4 k n−2k−4 k X (−1) n (n − 1) (n − 2) a b Sk 3 = t , n(n − 2)n−3 k=0 where

!  2 ! n − k − 3 n(n − 1) 5n − (6k + 23)n + 10k + 24 n − k − 3 S = (n − 1)3 a4 − a2b k k n − k − 3 k ! n − k − 4 + 4n2(n − 2) b2, k which completes the proof of Lemma3.  ON THE DISCRIMINANT OF A CERTAIN QUARTINOMIAL AND ITS TOTALLY COMPLEXNESS23

Proof of Therem2. Theorem 3 has been proved for n = 3, 4 and hence let us assume n ≥ 5. Then, by the definition of Ac(t)n−2 and equations (4), (5), we have

∆ (fc(t; x)) = det Ac(t) = n · det Ac(t)n−2 " (c) (c) # d(n−3)/2e n−3 n−3 an−1,n−1(t)n−2 an−1,n(t)n−2 = n · (−1) (n − 2) t · det (c) (c) . an,n−1(t)n−2 an,n(t)n−2 Therefore, by Lemma3, we have ( ) m1 n−1 n−2 2 2 n n−1 ∆ (fc(t; x)) = (−1) t (n − 2) (a − 4b)t + γct − n b , where  m0 X n+k k n−2k−4 k n−2k−4 k  (−1) n (n − 1) (n − 2) a b Sk (a 6= 0, or a = 0, n : even), γc = k=0 0 (a = 0, n : odd), which completes the proof.  n−2 2 2 n n−1 Proof of Theorem3. Put Q(t) = (n − 2) (a − 4b)t + γct − n b and suppose b = (n − 1)2a2/4n(n − 2). Then, by equation (5) and Lemma3, we have Q(t)

( 2 ( 2  2 ) n−3 (n − 2)(a − 4b) 2 2nb x¯n−3 (n − 1)a − 2nb x¯n−1 = n(n − 2) t + + + 2abx¯n−2 n n − 2 n ) nn−1bn−1 − (n − 2)n−3  (n − 1)2a2  = (n − 2)n−2 a2 − t2 n(n − 2) ( 2  2 ) n−3 2nb x¯n−3 (n − 1)a − 2nb x¯n−1 n n−1 + n(n − 2) + + 2abx¯n−2 − n b n − 2 n (n − 2)n−3a2 = − t2 n ( 2  2 ) n−3 2nb x¯n−3 (n − 1)a − 2nb x¯n−1 n n−1 + n(n − 2) + + 2abx¯n−2 − n b . n − 2 n Here, by equation (3), we have

( 2  2 ) n−3 2nb x¯n−3 (n − 1)a − 2nb x¯n−1 n(n − 2) + + 2abx¯n−2 n − 2 n ( 2n (n − 1)4a4 (−1)n−3(n − 3){−(n − 1)a}n−4 = n(n − 2)n−3 · · t n − 2 16n2(n − 2)2 2n−4 {−(n − 2)}n−5  (n − 1)a2 (n − 1)2a2  (−1)n−1(n − 1){−(n − 1)a}n−2 + − · t n 2n(n − 2) 2n−2 {−(n − 2)}n−3 ) (n − 1)2a3 (−1)n−2(n − 2){−(n − 1)a}n−3 + · t 2n(n − 2) 2n−3 {−(n − 2)}n−4  (−1)n(n − 1)n(n − 3)an (−1)n(n − 1)n(n − 3)an (−1)n(n − 1)n−1an  = n(n − 2)n−3 t + t − t 2n−1n(n − 2)n−2 2n−1n(n − 2)n−2 2n−2n(n − 2)n−4 (−1)n(n − 1)n−1an = − t, 2n−2(n − 2) 24 SHUICHI OTAKE AND TONY SHASKA which implies (n − 2)n−3a2 (−1)n(n − 1)n−1an Q(t) = − t2 − t − nnbn−1, n 2n−2(n − 2)  (−1)n(n − 1)n−1an 2  (n − 2)n−3a2  ∆ (Q(t)) = − − 4 · − · −nnbn−1
2n−2(n − 2) n (n − 1)2n−2a2n 4(n − 2)n−3a2 (n − 1)2a2 n−1 = − · nn · = 0 22n−4(n − 2)2 n 4n(n − 2) and hence, by completing the square, we have (n − 2)n−3a2  (−1)nn(n − 1)n−1an−2 2 Q(t) = − t + . n 2n−1(n − 2)n−2

Thus, if n is even, we have αc = 0 and hence by Theorem1, we have Nfc(ξ;x) = 0 for any positive real number ξ since x2 + ax + b has no real root in this case. Then, by considering the graph of the function y = xn + ξ(x2 + ax + b)(ξ > 0), we can 2 2 conclude that Nfc(ξ;x) = 0 whenever (n − 1) a /4n(n − 2) ≤ b.  References [B-B-G] B. Beresnevich; V. Bernik; F. G¨otze.Integral polynomials with small discriminants and resultants. Adv. Math. 298 (2016), 393-412. [D-S] K. Dilcher, Karl, K. B. Stolarsky. Resultants and discriminants of Chebyshev and related polynomials. Trans. Amer. Math. Soc. 357 (2005), no. 3, 965-981. [Fro] G. Frobenius. Ueber das Tr¨agheitsgesetzder quadratischen Formen. J. Reine Angew. Math. 114 (1895), 187-230. [Fuh] P. A. Fuhrmann. A polynomial approach to linear algebra. Second edition. Universitext. Springer, New York, 2012. [G-D] G. R. Greenfield; D. Drucker. On the discriminant of a trinomial. Linear Algebra Appl. 62 (1984), 105-112. [Ked] K. S. Kedlaya. A construction of polynomials with squarefree discriminants. Proc. Amer. Math. Soc. 140 (2012), no. 9, 3025-3033. [O-S] S. Otake; T. Shaska. Some remarks on the non-real roots of polynomials. arXiv:1802.02708. [Rio] J. Riordan. Combinatorial identities. John Wiley & Sons, Inc., New York-London-Sydney 1968. [Swa] R. G. Swan. Factorization of polynomials over finite fields. Pacific J. Math. 12 (1962), 1099-1106.

Department of Applied Mathematics, Waseda University, Japan E-mail address: [email protected]

Department of Mathematics and Statistics, Oakland University, Rochester, MI, 48309. E-mail address: [email protected]