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On a conjecture concerning integral real roots of certain cubic polynomials
Linear Algebra and its Applications 446 (2014) 265–268
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
On a conjecture concerning integral real roots of certain cubic polynomials ✩ Junhua He School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan, 611731, PR China
a r t i c l e
i n f o
Article history: Received 23 September 2013 Accepted 19 December 2013 Available online 7 January 2014 Submitted by R. Brualdi MSC: 05C50 14H52 Keywords: Unicyclic graphs Elliptic curve Rational solutions
a b s t r a c t In this paper, we determine all the rational pairs (x, n) such that fn (x) = 3x3 + (7 − 10n)x2 + 2 6n2 − 11n + 8 x − 4n3 − 6n2 − 10n + 24 = 0. It follows no integer conjecture of optimal problem.
that for each positive integer n 5, there is solution x for the polynomial. This confirms a of Li et al. concerning the uniqueness of a class graphs in the study of an extremal graph theory © 2013 Elsevier Inc. All rights reserved.
1. Introduction In [1], Li, Tam and Su considered the problem of maximizing (also, minimizing) the absolute values of the signless Laplacian coefficients among all unicyclic graphs of a given order. They found that optimal graph for the minimization problem is unique. Believing that the optimal graph for the maximization problem is also unique, they posed the following conjecture: ✩ Research supported by the fundamental research funds for Chinese universities under Grant ZYGX2012J120. E-mail address: [email protected].
0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.12.029
266
J. He / Linear Algebra and its Applications 446 (2014) 265–268
Conjecture 1. For every positive integer n 5, the unique real root of the cubic polynomial fn (x) = 3x3 + (7 − 10n)x2 + 2 6n2 − 11n + 8 x − 4n3 − 6n2 − 10n + 24 is never an integer. In this paper, we prove the following. Theorem 1. The rational solutions (x, n) of the equation fn (x) = 0 are: (2, 2), (3, 3), (4, 4), (2, 3), (3, 4), (8/3, 3), (2, 5/2), (3, 7/2). By the above theorem, Conjecture 1 is true. Thus, there is a unique optimal graph for the maximization problem on unicyclic graphs. 2. Proof of Theorem 1 For notational convenience, we let g(x, y) := fy (x) = 3x3 + (7 − 10y)x2 + 2 6y 2 − 11y + 8 x − 4y 3 − 6y 2 − 10y + 24 . We begin with the following. Lemma 1. The finite rational points of the elliptic curve E with equation Y 2 = X(X − 1)(X + 3) are: (0, 0), (1, 0), (−3, 0), (−1, 2), (−1, −2), (3, 6), (3, −6). Proof. It is well known [2] that the set of all the rational solutions of the equation together with an infinite point O forms the Mordell–Weil group E(Q) of E, substituting X by X1 − 1 we will obtain the minimal Weierstrass equation of E: E:
Y 2 = X13 − X12 − 4X1 + 4,
the computation by online database LMFDB [3] shows that the Mordell–Weil group E (Q) of E is isomorphic to Z4 × Z2 , and we have E (Q) = (1, 0), (2, 0), (−2, 0), (0, 2), (0, −2), (4, 6), (4, −6), O , hence the finite rational points of E are exactly those listed in the lemma.
2
Proof of Theorem 1. Let x = S + 3, y = S + T + 3. By a direct computation one can show that g(x, y) = S 3 + 2T S 2 + (2T − 1)S − 2T (T − 1)(2T − 1).
J. He / Linear Algebra and its Applications 446 (2014) 265–268
267
Denote h(S, T ) := S 3 + 2T S 2 + (2T − 1)S − 2T (T − 1)(2T − 1). Assume (s, t) is a rational solution of the equation h(S, T ) = 0. (A) If t = 0, then 0 = h(s, t) = h(s, 0) = s3 − s = (s + 1)s(s − 1), hence s = −1, 0 or 1. In this case (−1, 0), (0, 0), (1, 0) are the rational solutions of the equation h(S, T ) = 0. (B) If t = 0, then s = at for some a ∈ Q, hence we have 0 = h(s, t) = h(at, t) =
3 a + 2a2 − 4 t2 + 2(a + 3)t − (a + 2) t.
Since t = 0, we have 3 a + 2a2 − 4 t2 + 2(a + 3)t − (a + 2) = 0. Hence a3 + 2a2 − 4 = 0 or t=
−(a + 3) ±
(a + 3)2 + (a3 + 2a2 − 4)(a + 2) . a3 + 2a2 − 4
Notice that ±1, ±2, ±4 are the only possible rational roots of a3 +2a2 −4 = 0, but none of them satisfies the equation, so a3 +2a2 −4 = 0 and hence (a+3)2 +(a3 +2a2 −4)(a+2) = b2 for some b ∈ Q because t ∈ Q. Thus we have 2 b2 = (a + 3)2 + a3 + 2a2 − 4 (a + 2) = (a + 1)2 − 1/2 + 3/4. Let c := (a + 1)2 − 1/2, we have (b + c)(b − c) = 3/4. Let k := b + c ∈ Q∗ , then we have 2 3 b − c = 4k and c = 12 [(b + c) − (b − c)] = 4k8k−3 . Thus we have (a + 1)2 = c +
8k 3 + 8k 2 − 6k 1 = . 2 16k2
Therefore, 2 4k(a + 1) = 8k3 + 8k 2 − 6k = 2k(2k − 1)(2k + 3). Let A = 2k, B = 4k(a + 1). Then (A, B) is a finite rational point of the elliptic curve Y 2 = X(X − 1)(X + 3). By Lemma 1 we have 2k = 0, ±1 or ±3, as a result, k = ± 12 , ± 32 because k ∈ Q∗ . Notice that A = 2k,
B = 4k(a + 1), t=
a=
−(a + 3) ± b = 0, a3 + 2a2 − 4
B − 1, 4k s = at.
b=
4k 2 + 3 , 8k
J. He / Linear Algebra and its Applications 446 (2014) 265–268
268
If k = − 12 , then A = −1 and by Lemma 1 B = ±2, so a = 0 or −2, b = −1. If a = 0, then t=
−3 ± 1 −a − 3 ± b = = 1 or a3 + 2a2 − 4 −4
1 . 2
t=
−1 ± 1 −a − 3 ± b = = 0 or a3 + 2a2 − 4 −4
1 . 2
If a = −2, then
Notice that t = 0 and s = at, therefore (s, t) = (0, 1), (0, 1/2) or (−1, 1/2). Similarly, if k = 3/2, we have (s, t) = (0, 1), (0, 1/2) or (−1, 1/2). If k = 1/2 or −3/2, we have (s, t) = (−1, 1) or (−1/3, 1/3). Combining the results above, we see that (−1, 0), (0, 0), (1, 0), (−1, 1), (0, 1), (−1/3, 1/3), (−1, 1/2), (0, 1/2) are exactly the rational solutions of equation h(S, T ) = 0. Furthermore, we have h(s, t) = 0 if and only if g(s + 3, s + t + 3) = 0 by definition of h(S, T ). Hence the rational solutions (x, y) of equation g(x, y) = 0 are: (2, 2), (3, 3), (4, 4), (2, 3), (3, 4), (8/3, 3), (2, 5/2), (3, 7/2).
2
Acknowledgements The research was done while the author was visiting the College of William and Mary. The author sincerely thanks Professor C.K. Li for drawing his attention to the conjecture, and for some helpful discussion. He also would like to thank Professor Dietrich Burde for his suggestion to compute the group of rational points of an elliptic curve by online database LMFDB. Finally, he thanks the referee for his/her valuable suggestions on the manuscript. References [1] H.-H. Li, B.-S. Tam, Li Su, On the signless Laplacian coefficients of unicyclic graphs, Linear Algebra Appl. 439 (2013) 2008–2028. [2] J.H. Silverman, The Arithmetic of Elliptic Curves, Springer, 1986. [3] The LMFDB Collaboration, Elliptic Curves/Q Database, Home page of Elliptic Curves/Q for Elliptic Curve 24.a4, http://www.lmfdb.org/EllipticCurve/Q?label=[0%2C2%2C0%2C-3%2C0]&jump= label+or+isogeny+class/, 2013.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
On a conjecture concerning integral real roots of certain cubic polynomials ✩ Junhua He School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu, Sichuan, 611731, PR China
a r t i c l e
i n f o
Article history: Received 23 September 2013 Accepted 19 December 2013 Available online 7 January 2014 Submitted by R. Brualdi MSC: 05C50 14H52 Keywords: Unicyclic graphs Elliptic curve Rational solutions
a b s t r a c t In this paper, we determine all the rational pairs (x, n) such that fn (x) = 3x3 + (7 − 10n)x2 + 2 6n2 − 11n + 8 x − 4n3 − 6n2 − 10n + 24 = 0. It follows no integer conjecture of optimal problem.
that for each positive integer n 5, there is solution x for the polynomial. This confirms a of Li et al. concerning the uniqueness of a class graphs in the study of an extremal graph theory © 2013 Elsevier Inc. All rights reserved.
1. Introduction In [1], Li, Tam and Su considered the problem of maximizing (also, minimizing) the absolute values of the signless Laplacian coefficients among all unicyclic graphs of a given order. They found that optimal graph for the minimization problem is unique. Believing that the optimal graph for the maximization problem is also unique, they posed the following conjecture: ✩ Research supported by the fundamental research funds for Chinese universities under Grant ZYGX2012J120. E-mail address: [email protected].
0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.12.029
266
J. He / Linear Algebra and its Applications 446 (2014) 265–268
Conjecture 1. For every positive integer n 5, the unique real root of the cubic polynomial fn (x) = 3x3 + (7 − 10n)x2 + 2 6n2 − 11n + 8 x − 4n3 − 6n2 − 10n + 24 is never an integer. In this paper, we prove the following. Theorem 1. The rational solutions (x, n) of the equation fn (x) = 0 are: (2, 2), (3, 3), (4, 4), (2, 3), (3, 4), (8/3, 3), (2, 5/2), (3, 7/2). By the above theorem, Conjecture 1 is true. Thus, there is a unique optimal graph for the maximization problem on unicyclic graphs. 2. Proof of Theorem 1 For notational convenience, we let g(x, y) := fy (x) = 3x3 + (7 − 10y)x2 + 2 6y 2 − 11y + 8 x − 4y 3 − 6y 2 − 10y + 24 . We begin with the following. Lemma 1. The finite rational points of the elliptic curve E with equation Y 2 = X(X − 1)(X + 3) are: (0, 0), (1, 0), (−3, 0), (−1, 2), (−1, −2), (3, 6), (3, −6). Proof. It is well known [2] that the set of all the rational solutions of the equation together with an infinite point O forms the Mordell–Weil group E(Q) of E, substituting X by X1 − 1 we will obtain the minimal Weierstrass equation of E: E:
Y 2 = X13 − X12 − 4X1 + 4,
the computation by online database LMFDB [3] shows that the Mordell–Weil group E (Q) of E is isomorphic to Z4 × Z2 , and we have E (Q) = (1, 0), (2, 0), (−2, 0), (0, 2), (0, −2), (4, 6), (4, −6), O , hence the finite rational points of E are exactly those listed in the lemma.
2
Proof of Theorem 1. Let x = S + 3, y = S + T + 3. By a direct computation one can show that g(x, y) = S 3 + 2T S 2 + (2T − 1)S − 2T (T − 1)(2T − 1).
J. He / Linear Algebra and its Applications 446 (2014) 265–268
267
Denote h(S, T ) := S 3 + 2T S 2 + (2T − 1)S − 2T (T − 1)(2T − 1). Assume (s, t) is a rational solution of the equation h(S, T ) = 0. (A) If t = 0, then 0 = h(s, t) = h(s, 0) = s3 − s = (s + 1)s(s − 1), hence s = −1, 0 or 1. In this case (−1, 0), (0, 0), (1, 0) are the rational solutions of the equation h(S, T ) = 0. (B) If t = 0, then s = at for some a ∈ Q, hence we have 0 = h(s, t) = h(at, t) =
3 a + 2a2 − 4 t2 + 2(a + 3)t − (a + 2) t.
Since t = 0, we have 3 a + 2a2 − 4 t2 + 2(a + 3)t − (a + 2) = 0. Hence a3 + 2a2 − 4 = 0 or t=
−(a + 3) ±
(a + 3)2 + (a3 + 2a2 − 4)(a + 2) . a3 + 2a2 − 4
Notice that ±1, ±2, ±4 are the only possible rational roots of a3 +2a2 −4 = 0, but none of them satisfies the equation, so a3 +2a2 −4 = 0 and hence (a+3)2 +(a3 +2a2 −4)(a+2) = b2 for some b ∈ Q because t ∈ Q. Thus we have 2 b2 = (a + 3)2 + a3 + 2a2 − 4 (a + 2) = (a + 1)2 − 1/2 + 3/4. Let c := (a + 1)2 − 1/2, we have (b + c)(b − c) = 3/4. Let k := b + c ∈ Q∗ , then we have 2 3 b − c = 4k and c = 12 [(b + c) − (b − c)] = 4k8k−3 . Thus we have (a + 1)2 = c +
8k 3 + 8k 2 − 6k 1 = . 2 16k2
Therefore, 2 4k(a + 1) = 8k3 + 8k 2 − 6k = 2k(2k − 1)(2k + 3). Let A = 2k, B = 4k(a + 1). Then (A, B) is a finite rational point of the elliptic curve Y 2 = X(X − 1)(X + 3). By Lemma 1 we have 2k = 0, ±1 or ±3, as a result, k = ± 12 , ± 32 because k ∈ Q∗ . Notice that A = 2k,
B = 4k(a + 1), t=
a=
−(a + 3) ± b = 0, a3 + 2a2 − 4
B − 1, 4k s = at.
b=
4k 2 + 3 , 8k
J. He / Linear Algebra and its Applications 446 (2014) 265–268
268
If k = − 12 , then A = −1 and by Lemma 1 B = ±2, so a = 0 or −2, b = −1. If a = 0, then t=
−3 ± 1 −a − 3 ± b = = 1 or a3 + 2a2 − 4 −4
1 . 2
t=
−1 ± 1 −a − 3 ± b = = 0 or a3 + 2a2 − 4 −4
1 . 2
If a = −2, then
Notice that t = 0 and s = at, therefore (s, t) = (0, 1), (0, 1/2) or (−1, 1/2). Similarly, if k = 3/2, we have (s, t) = (0, 1), (0, 1/2) or (−1, 1/2). If k = 1/2 or −3/2, we have (s, t) = (−1, 1) or (−1/3, 1/3). Combining the results above, we see that (−1, 0), (0, 0), (1, 0), (−1, 1), (0, 1), (−1/3, 1/3), (−1, 1/2), (0, 1/2) are exactly the rational solutions of equation h(S, T ) = 0. Furthermore, we have h(s, t) = 0 if and only if g(s + 3, s + t + 3) = 0 by definition of h(S, T ). Hence the rational solutions (x, y) of equation g(x, y) = 0 are: (2, 2), (3, 3), (4, 4), (2, 3), (3, 4), (8/3, 3), (2, 5/2), (3, 7/2).
2
Acknowledgements The research was done while the author was visiting the College of William and Mary. The author sincerely thanks Professor C.K. Li for drawing his attention to the conjecture, and for some helpful discussion. He also would like to thank Professor Dietrich Burde for his suggestion to compute the group of rational points of an elliptic curve by online database LMFDB. Finally, he thanks the referee for his/her valuable suggestions on the manuscript. References [1] H.-H. Li, B.-S. Tam, Li Su, On the signless Laplacian coefficients of unicyclic graphs, Linear Algebra Appl. 439 (2013) 2008–2028. [2] J.H. Silverman, The Arithmetic of Elliptic Curves, Springer, 1986. [3] The LMFDB Collaboration, Elliptic Curves/Q Database, Home page of Elliptic Curves/Q for Elliptic Curve 24.a4, http://www.lmfdb.org/EllipticCurve/Q?label=[0%2C2%2C0%2C-3%2C0]&jump= label+or+isogeny+class/, 2013.