Proof of a Conjecture on Monomial Graphs

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Proof of a Conjecture on Monomial Graphs Xiang-dong Hou Department of Mathematics and Statistics University of South Florida Joint work with Stephen D. Lappano and Felix Lazebnik New Directions in Combinatorics IMS, Singapore, 2016-05-26 X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 1 / 30

outline The bipartite graph G q (f, g) and its background The conjecture Permutation polynomials Previous results Outline of a proof of the conjecture Open questions X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 2 / 30

The bipartite graph G q (f, g) and its background The conjecture Permutation polynomials Previous results Outline of a proof of the conjecture Open questions X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 3 / 30

the bipartite graph G q (f, g) Let F q be the finite field with q elements, q odd. Let f, g F q [X, Y ]. The graph G = G q (f, g) is an undirected bipartite graph with vertex partitions P = F 3 q and L = F 3 q, and edges defined as follows: a vertex (p) = (p 1, p 2, p 3 ) P is adjacent to a vertex [l] = [l 1, l 2, l 3 ] L if and only if p 2 + l 2 = f (p 1, l 1 ) and p 3 + l 3 = g(p 1, l 1 ). X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 4 / 30

some graph theory Let k 2, and g k (n) denote the greatest number of edges in a graph with n vertices and a girth at least 2k + 1. The function g k (n) has been studied extensively. Bondy and Simonovits 1974: g k (n) c k n 1+ 1 k for k 2. Lazebnik, Ustimenko and Woldar 1995: {c 2 g k (n) k n1+ 3k 3+ɛ if k 2, k 5, c 5 n1+ 1 5 if k = 5, where ɛ = 0 if k is odd, ɛ = 1 if k is even, and c k and c k are positive constants depending on k only. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 5 / 30

when k = 2, 3, 5 g k (n) {c k n1+ 2 3k 3+ɛ if k 2, k 5, c 5 n1+ 1 5 if k = 5, The only known values of k for which the lower bound for g k (n) is of magnitude n 1+1/k, which is the same as the magnitude of the upper bound, are k = 2, 3, and 5. The lower bound for k = 3 is given by the graph G q (XY, XY 2 ). In fact, G q (XY, XY 2 ) has girth 8. The lower bound for k = 2, 5 are given by graphs constructed in a similar manner. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 6 / 30

The bipartite graph G q (f, g) and its background The conjecture Permutation polynomials Previous results Outline of a proof of the conjecture Open questions X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 7 / 30

monomial graphs When f, g F q [X, Y ] are monomials, the graph G q (f, g) is called a monomial graph. We will only consider monomial graphs. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 8 / 30

the conjecture Conjecture 1 Let q be an odd prime power. Then every monomial graph of girth eight is isomorphic to G q (XY, XY 2 ). Theorem (Dmytrenko, Lazebnik, Williford 2007) Let q be odd. Every monomial graph of girth 8 is isomorphic to G q (XY, X k Y 2k ), where 1 k q 1 is an integer not divisible by p. The condition that G q (XY, X k Y 2k ) has girth 8 implies that certain polynomials are permutations of F q. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 9 / 30

The bipartite graph G q (f, g) and its background The conjecture Permutation polynomials Previous results Outline of a proof of the conjecture Open questions X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 10 / 30

permutation polynomial A permutation polynomial (PP) of F q is a polynomial f F q [X] such that the function defined by a f (a) is a bijection on F q. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 11 / 30

A k and B k For an integer 1 k q 1, let Theorem (DLW 2007) A k = X k[ (X + 1) k X k] F q [X], B k = [ (X + 1) 2k 1 ] X q 1 k 2X q 1 F q [X]. Let q be odd and 1 k q 1 be such that p k. If G q (XY, X k Y 2k ) has girth 8, then both A k and B k are PPs of F q. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 12 / 30

conjectures on A k and B k Conjecture A (DLW 2007) Let q be a power of an odd prime p and 1 k q 1. Then A k is a PP of F q if and only if k is a power of p. Conjecture B (DLW 2007) Let q be a power of an odd prime p and 1 k q 1. Then B k is a PP of F q if and only if k is a power of p. Either of Conjectures A and B implies Conjecture 1. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 13 / 30

The bipartite graph G q (f, g) and its background The conjecture Permutation polynomials Previous results Outline of a proof of the conjecture Open questions X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 14 / 30

prior status of Conjecture 1 For e > 1, gpf(e) = the greatest prime factor of e; gpf(1) = 1. Theorem (DLW 2007) Conjecture 1 is true if one of the following occurs. (i) q = p e, where p 5 and gpf(e) 3. (ii) 3 q 10 10. Theorem (Kronenthal 2012) For each prime r or r = 1, there is a positive integer p 0 (r) such that Conjecture 1 is true for q = p e with gfp(e) r and p p 0 (r). In particular, one can choose p 0 (5) = 7, p 0 (7) = 11, p 0 (11) = 13. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 15 / 30

prior status of Conjectures A and B Theorem (DLW 2007) Conjecture A is true for q = p. For each odd prime p, let α(p) be the smallest positive even integer a such that ( ) a ( 1) a/2 2 a (mod p). a/2 Theorem (Kronenthal 2012) Let p be an odd prime. If Conjecture B is true for q = p e, then it is also true for q = p em whenever m p 1 (p 1)/α(p). X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 16 / 30

current status of the conjectures Lappano, Lazebnik, H 2015 Conjecture A is true for q = p e, where p is an odd prime and gpf(e) p 1. Conjecture B is true for q = p e, where e > 0 is arbitrary and p is an odd prime satisfying α(p) > (p 1)/2. Conjecture 1 is true. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 17 / 30

The bipartite graph G q (f, g) and its background The conjecture Permutation polynomials Previous results Outline of a proof of the conjecture Open questions X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 18 / 30

strategy Let q be odd and 1 k q 1. We show that if both A k and B k are PPs of F q, then k is a power of p. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 19 / 30

Hermite s criterion f F q [X] is a PP of F q if and only if { f (x) s 0 if 0 s q 2, = 1 if s = q 1. x F q X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 20 / 30

power sums of A k and B k For each integer a > 0, let a {1,..., q 1} be such that a a (mod q 1); we also define 0 = 0. For 1 s q 1, x F q A k (x) s = ( 1) s+1 s ( )( ) s (ki) ( 1) i i (2ks), i=0 ( s B k (x) s = ( 2) s 2 i ( 1) j i x F q i,j )( i j )( ) (2kj) (ki). X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 21 / 30

criteria for A k and B k to be PP Theorem (i) A k is a PP of F q if and only if gcd(k, q 1) = 1 and ( )( ) s (ki) ( 1) i i (2ks) = 0 for all 1 s q 2. i (ii) B k is a PP of F q if and only if gcd(k, q 1) = 1 and ( )( ) s (2ki) ( 1) i i (ks) = ( 2) s for all 1 s q 2. i Too much information, too little readily useful. Need to choose suitable s such that useful information can be extracted from the above equations. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 22 / 30

notation Assume that 1 k q 1 with gcd(k, q 1) = 1. k, b {1,..., q 1} are such that q 1 a :=. k k k 1 (mod q 1), bk 1 (mod q 1). q 1 c :=. k X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 23 / 30

choosing s Theorem (i) A k is a PP of F q if and only if gcd(k, q 1) = 1 and ( )( ) s (ki) ( 1) i i (2ks) = 0 for all 1 s q 2. i (ii) B k is a PP of F q if and only if gcd(k, q 1) = 1 and ( )( ) s (2ki) ( 1) i i (ks) = ( 2) s for all 1 s q 2. i X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 24 / 30

choosing s Theorem (i) A k is a PP of F q if and only if gcd(k, q 1) = 1 and ( )( ) s (ki) ( 1) i i (2ks) = 0 for all 1 s q 2. i (ii) B k is a PP of F q if and only if gcd(k, q 1) = 1 and ( )( ) s (2ki) ( 1) i i (ks) = ( 2) s for all 1 s q 2. i Choose s = a, a 1, b, q 1 ck, q 1 (c 1)k, etc. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 24 / 30

choosing s Theorem (i) A k is a PP of F q if and only if gcd(k, q 1) = 1 and ( )( ) s (ki) ( 1) i i (2ks) = 0 for all 1 s q 2. i (ii) B k is a PP of F q if and only if gcd(k, q 1) = 1 and ( )( ) s (2ki) ( 1) i i (ks) = ( 2) s for all 1 s q 2. i Choose s = a, a 1, b, q 1 ck, q 1 (c 1)k, etc. All but a few terms vanish in the above equations. Useful information is obtained... X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 24 / 30

facts about A k and B k Lemma 1 Assume that A k is a PP of F q. Then all the base p digits of k are 0 or 1. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 25 / 30

facts about A k and B k Lemma 1 Assume that A k is a PP of F q. Then all the base p digits of k are 0 or 1. Lemma 2 Assume that all the base p digits of k are 0 or 1 and k is not a power of p, then c 0 (mod p). X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 25 / 30

facts about A k and B k Lemma 1 Assume that A k is a PP of F q. Then all the base p digits of k are 0 or 1. Lemma 2 Assume that all the base p digits of k are 0 or 1 and k is not a power of p, then c 0 (mod p). Lemma 3 Assume that q is odd, 1 < k q 1, and both A k and B k are PPs of F q. Then c is even and 2 2ck = ( ) ( 2(q 1) 2ck q 1 ck + ( 1) q 1 2 + c 2(q 1) 2ck 2 +1 )( ) 2c 1 2 (q 1) ( c 2 1)k. c + 2 X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 25 / 30

proof of Conjecture 1 Assume to the contrary that Conjecture 1 is false. Then there exists 1 k q 1, which is not a power of p, such that both A k and B k are PPs of F q. Lemma 1 and 2 imply that c 0 (mod p). Then ( ) 2c = 0. c + 2 Since q 1 ck p 1 (mod p), the sum (q 1 ck ) + (q 1 ck ) has a carry in base p at p 0, implying that ( ) 2(q 1) 2ck q 1 ck = 0. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 26 / 30

proof of Conjecture 1 conclusion Combing 2 2ck = ( ) ( 2(q 1) 2ck q 1 ck + ( 1) q 1 2 + c 2(q 1) 2ck 2 +1 )( ) 2c 1 2 (q 1) ( c 2 1)k, c + 2 and gives a contradiction. ( ) 2c = 0 = c + 2 ( ) 2(q 1) 2ck q 1 ck X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 27 / 30

The bipartite graph G q (f, g) and its background The conjecture Permutation polynomials Previous results Outline of a proof of the conjecture Open questions X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 28 / 30

open questions Conjecture A Let q be a power of an odd prime p and 1 k q 1. Then A k is a PP of F q if and only if k is a power of p. Conjecture B Let q be a power of an odd prime p and 1 k q 1. Then B k is a PP of F q if and only if k is a power of p. Conjecture A is true for q = p. Conjecture B has not been established for q = p. Although Conjectures A and B were originally stated for an odd characteristic, their status also appears to be unsettled for p = 2. X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 29 / 30

end Thank You! X. Hou (Universities of South Florida) a conjecture on monomial graphs IMS-2016-05-26 30 / 30