A talk given at Zhejiang Univ. (Nov. 15, 2013)and Frontiers in Combinatorics (Beijing, Dec. 15, 2013)and Xiamen Univ. (Dec. 20, 2013)and the 2nd Workshhop on Combinatorics and Graph Theory(Changsha, June 2, 2014)and Institute of Math., Academia Sinica (Taipei, July 24, 2014)and Conf. on Combinatorics and its Appl. (Harbin, May 19, 2018)and St. Petersburg Dept. of Steklov Math. Instituteof Russian Academy of Sci. (July 19, 2018)

Determinants, Permutations and AdditiveCombinatorics

Zhi-Wei Sun

Nanjing University, Nanjing 210093, P. R. Chinazwsun@nju.edu.cn

http://math.nju.edu.cn/∼zwsun

July 19, 2018

Abstract

In this talk we will introduce some new problems and relatedprogress on determinants involving Legendre symbols, circularpermutations and additive combinatorics. For example, weconjecture that for any finite subset A of an additive cyclic groupG with |A| = n > 3, there is a circular permutation a1, . . . , an ofthe elements of A such that all the n sums

a1+a2+a3, a2+a3+a4, . . . , an−2+an−1+an, an−1+an+a1, an+a1+a2

are pairwise distinct. The speaker has proved this when G is theinfinite cyclic group Z.

2 / 41

Part I. Some new problems on determinants

3 / 41

Hankel-type determiants

For an n × n matrix A = (aij)16i ,j6n over the field of complexnumbers, we often write detA in the form |aij |16i ,j6n.

For a sequence a0, a1, a2, . . . of numbers, the determinants|ai+j |06i ,j6n are said to be of the Hankel type.

For bn =∑n

k=0

(nk

)ak (n = 0, 1, 2, . . .), it is known that

|ai+j |06i ,j6n = |bi+j |06i ,j6n.

For Catalan numbers Cn = 1n+1

(2nn

)(n = 0, 1, 2, . . .), the

Hankel-type determinant |Ci+j |06i ,j6n takes the value 1.

4 / 41

Legendre symbols

Let p be an odd prime and a ∈ Z. The Legendre symbol ( ap ) is

given by(a

p

)=

0 if p | a,1 if p - a and x2 ≡ a (mod p) for some x ∈ Z,−1 if p - a and x2 ≡ a (mod p) for no x ∈ Z.

It is well known that (abp ) = ( ap )(bp ) for any a, b ∈ Z. Also,(

−1

p

)= (−1)(p−1)/2 =

{1 if p ≡ 1 (mod 4),

−1 if p ≡ −1 (mod 4);(2

p

)= (−1)(p

2−1)/8 =

{1 if p ≡ ±1 (mod 8),

−1 if p ≡ ±3 (mod 8).

The Law of Quadratic Reciprocity: If p and q are distinct oddprimes, then (

p

q

)(q

p

)= (−1)

p−12· q−1

2 .

5 / 41

Chapman’s work on determinants with Legendre symbolentries

In 2004, R. Chapman [Acta Arith.] used quadratic Gauss sums todetermine the values of∣∣∣∣ ( i + j − 1

p

) ∣∣∣∣16i ,j6(p−1)/2

and

∣∣∣∣ ( i + j − 1

p

) ∣∣∣∣16i ,j6(p+1)/2

.

Since (p + 1)/2− i + (p + 1)/2− j − 1 ≡ −(i + j) (mod p), wesee that∣∣∣∣ ( i + j − 1

p

) ∣∣∣∣16i ,j6(p−1)/2

=

(−1

p

) ∣∣∣∣ ( i + j

p

) ∣∣∣∣16i ,j6(p−1)/2

and ∣∣∣∣ ( i + j − 1

p

) ∣∣∣∣16i ,j6(p+1)/2

=

∣∣∣∣ ( i + j

p

) ∣∣∣∣06i ,j6(p−1)/2

.

6 / 41

Chapman’s evil determinants

Conjecture (Chapman, 2003) Let p be an odd prime, and write

ε(2−( 2

p))h(p)

p = rp + sp√p with rp, sp ∈ Z,

where εp and h(p) denote the fundamental unit and the classnumber of the real quadratic field Q(

√p) respectively. Then∣∣∣∣ ( j − i

p

) ∣∣∣∣06i ,j6(p−1)/2

=

{−rp if p ≡ 1 (mod 4),

1 if p ≡ 3 (mod 4).

As Chapman could not solve this problem for several years, hecalled the determinant evil.

Chapman’s conjecture on his “evil” determinant was recentlyconfirmed by M. Vsemirnov [Linear Algebra Appl. 2012, and ActaArith. 2013] via matrix decomposition and quadratic Gauss sums.

7 / 41

Determining |( i+djp )|06i ,j6(p−1)/2 mod p

Theorem (Sun, 2013). Let p be an odd prime. For d ∈ Z define

R(d , p) :=

∣∣∣∣ ( i + dj

p

) ∣∣∣∣06i ,j6(p−1)/2

.

If p ≡ 1 (mod 4), then

R(d , p) ≡((

d

p

)d

)(p−1)/4 p − 1

2! (mod p).

When p ≡ 3 (mod 4), we have

R(d , p) ≡

{( 2p ) (mod p) if (dp ) = 1,

1 (mod p) if (dp ) = −1.

Also,

R(−d , p) ≡(

2

p

)R(d , p) (mod p),∣∣∣∣ ( i + dj + c

p

) ∣∣∣∣06i ,j6(p−1)/2

≡ R(d , p) (mod p) for all c ∈ Z.

8 / 41

Determining |( i2+dj2

p )|06i ,j6(p−1)/2 mod p

Theorem (Sun, 2013) Let p be an odd prime and let d ∈ Z.Define

T (d , p) :=

∣∣∣∣ ( i2 + dj2

p

) ∣∣∣∣06i ,j6(p−1)/2

.

Then (T (d , p)

p

)=

{( 2p ) if (dp ) = 1,

1 if (dp ) = −1.

Also,

T (−d , p) ≡(

2

p

)T (d , p) (mod p)

and∣∣∣∣ ( i2 + dj2 + c

p

) ∣∣∣∣06i ,j6(p−1)/2

≡ T (d , p) (mod p) for all c ∈ Z.

9 / 41

Some auxiliary results

A Known Result. Let P(z) =∑n−1

k=0 akzk be a polynomial with

complex number coefficients. Then we have

|P(xi + yj)|16i ,j6n = ann−1

n−1∏k=0

(n − 1

k

)×

∏16i<j6n

(xi − xj)(yj − yi ).

Zolotarev’s Theorem (1872). Let p be any odd prime, and leta ∈ Z with p - a. For m ∈ Z let {m}p denote the least nonnegativeresidue of an integer m modulo p. Then, the permutation{aj}p (j = 1, . . . , p − 1) of 1, . . . , p − 1 has the sign ( a

p ).

Theorem (H. Pan, 2006) Let p be an odd prime, and let a ∈ Zwith p - a. For each j = 1, . . . , (p − 1)/2 let σa(j) be the uniquer ∈ {1, . . . , (p − 1)/2} such that aj ≡ r or − r (mod p). Then,the sign of the permutation σa equals ( a

p )(p+1)/2.

10 / 41

Conjecture on |( i2+dj2+c

p )|16i ,j6(p−1)/2

Conjecture→Theorem (Sun, July 2013) Let p be an odd prime,and define

S(d , p) :=

∣∣∣∣ ( i2 + dj2

p

) ∣∣∣∣16i ,j6(p−1)/2

for d ∈ Z.

If (dp ) = 1, then (−S(d ,p)p ) = 1. If (dp ) = −1, then S(d , p) = 0.

Conjecture (Sun, August 2013) Let p be an odd prime, and letc , d ∈ Z with p - cd . Define

Sc(d , p) =

∣∣∣∣ ( i2 + dj2 + c

p

) ∣∣∣∣16i ,j6(p−1)/2

.

Then

(Sc(d , p)

p

)=

1 if ( cp ) = 1 & (dp ) = −1,

(−1p ) if ( cp ) = (dp ) = −1,

(−2p ) if (−cp ) = (dp ) = 1,

(−6p ) if (−cp ) = −1 & (dp ) = 1.11 / 41

Conjectures on |( i2+cij+dj2

n )|16i ,j6n−1

For any odd integer n > 1 and integers c and d , I introduced thenotations

(c , d)n :=

∣∣∣∣ ( i2 + cij + dj2

n

) ∣∣∣∣16i ,j6n−1

.

Conjecture (Sun, 2013). If d is nonzero, then there are infinitelymany odd primes p with (c, d)p = 0. When (c , d)p is nonzero, itsp-adic valuation ( i.e., p-adic order) must be even.

Theorem (Sun). Let c , d ∈ Z. Then (c , d)n = 0 for any positiveodd integer n with (dn ) = −1.

Conjecture (Sun). (i) For any odd integer n > 3 we have(2, 3)n ≡ 0 (mod n2).

(ii) For any odd integer n > 5 we have (6, 15)n ≡ 0 (mod n2).

12 / 41

Two more conjectures

Conjecture (Sun). (i) (6, 1)n = (3, 2)n = (4, 2)n = 0 for anypositive integer n ≡ 3 (mod 4).

(ii) (3, 3)n = 0 for any positive odd integer n 6≡ 1 (mod 12).

Remark. This has just been confirmed by F. Petrov, Z.-W. Sunand M. Vsemirnov jointly.

Conjecture (Sun, 2013). Let p = 2n + 1 be an odd prime. Then∣∣∣∣ ( i2 − n!j

p

) ∣∣∣∣ = 0 ⇐⇒ p ≡ 3 (mod 4).

Remark. Fedor Petrov has solved the mod p version of thisconjecture.

13 / 41

A curious conjectureTheorem (Sun). (i) For any odd prime p, we have∣∣∣∣( i+j

p )

i + j

∣∣∣∣16i ,j6(p−1)/2

≡

{( 2p ) (mod p) if p ≡ 1 (mod 4),

((p − 1)/2)! (mod p) if p ≡ 3 (mod 4).

(ii) Let p ≡ 3 (mod 4) be a prime. Then∣∣∣∣ 1

i2 + j2

∣∣∣∣16i ,j6(p−1)/2

≡(

2

p

)(mod p).

Conjecture (Sun, August 2013). (i) For any prime p > 3, we have∣∣∣∣(i2 + j2)

(i2 + j2

p

) ∣∣∣∣06i ,j6(p−1)/2

≡ 0 (mod p).

(Confirmed by D. Grinberg, G.-N. Han, Z.-W. Sun and L.-L. Zhao.)(ii) If p > 5 and p ≡ 1 (mod 4), then∣∣∣∣(i + j)

(i + j

p

) ∣∣∣∣16i ,j6(p−1)/2

and

∣∣∣∣(j − i)

(j − i

p

) ∣∣∣∣16i ,j6(p−1)/2

are quadratic residues modulo p. 14 / 41

A conjecture involving Domb numbers

Conjecture (Sun, August 2013). Define Domb numbers by

Dn :=n∑

k=0

(n

k

)2(2k

k

)(2(n − k)

n − k

)(n = 0, 1, . . .),

Then, for any prime p, we have

|Di+j |06i ,j6p−1 ≡

{(−1p )(4x2 − 2p) (mod p2) if p = x2 + 3y2,

0 (mod p2) if p ≡ 2 (mod 3).

We also have some other similar conjectures.

15 / 41

Another conjecture involving binary quadratic forms

Conjecture (Sun, August 2013). For n = 0, 1, 2, . . . let

hn =n∑

k=0

(n

k

)2

Ck with Ck =

(2kk

)k + 1

(the k-th Catalan number).

Let p be any odd prime. If p ≡ 1 (mod 3) and p = x2 + 3y2 withx , y ∈ Z and x ≡ 1 (mod 3), then

|hi+j |06i ,j6p−1 ≡ (−1)(p−1)/2(

2x − p

2x

)(mod p2).

If p ≡ 2 (mod 3) then

|hi+j |06i ,j6p−1 ≡ (−1)(p+1)/2 3p( (p+1)/2(p+1)/6)

) (mod p2).

16 / 41

One more conjecture involving binary quadratic forms

Conjecture (Sun, August 2013). For n = 0, 1, 2, . . . let

wn =

bn/3c∑k=0

(−1)k3n−3k(

n

3k

)(2k

k

)(3k

k

)and Wn = |wi+j |06i ,j6n.

(i) For any prime p ≡ 1 (mod 3), write 4p = x2 + 27y2 withx , y ∈ Z and x ≡ 1 (mod 3), then

Wp−1 ≡ (−1)(p+1)/2(x − p

x

)(mod p2).

(ii) Wn = 0 if and only if n ≡ 1 (mod 3).

Remark. C. Krattenthaler has confirmed that Wn = 0 for anyn ≡ 1 (mod 3).

17 / 41

Joint work with Bao-Xuan Zhu [Int. J. Number Theory,14(2018), 1265-1277]

From the abstract: In this paper we confirm several conjecturesof Z.-W. Sun on Hankel-type determinants for some combinatorialsequences including Franel numbers, Domb numbers and Aperynumbers. For any nonnegative integer n, define

fn :=n∑

k=0

(n

k

)3

, Dn :=n∑

k=0

(n

k

)2(2k

k

)(2(n − k)

n − k

),

bn :=n∑

k=0

(n

k

)2(n + k

k

), An :=

n∑k=0

(n

k

)2(n + k

k

)2

.

For n = 0, 1, 2, . . ., we show that 6−n|fi+j |0≤i ,j≤n and12−n|Di+j |0≤i ,j≤n are positive odd integers, and 10−n|bi+j |0≤i ,j≤nand 24−n|Ai+j |0≤i ,j≤n are always integers.

18 / 41

Some open conjectures on Hankel-type determinantsRecall the two kinds of Apery numbers

bn :=n∑

k=0

(n

k

)2(n + k

k

)and An :=

n∑k=0

(n

k

)2(n + k

k

)2

(n = 0, 1, . . .).

Conjecture 1 (Sun, August 14, 2013). We have |bi+j |06i ,j6n > 0and |Ai+j |06i ,j6n > 0 for all n = 0, 1, 2, . . ..

Conjecture 2 (Sun, August 2013). For any positive integer n, wehave

|B2i+j |06i ,j6n < 0 and |E 2

i+j |06i ,j6n > 0,

where the Bernoulli numbers B0,B1,B2, . . . and the Euler numbersE0,E1,E2, . . . are given by

x

ex − 1=∞∑n=0

Bnxn

n!and

2

ex + e−x=∞∑n=0

Enxn

n!.

19 / 41

A conjecture involving the Mobius function

Mobius function:

µ(n) =

1 if n = 1,

(−1)r if n is a product of r distinct primes,

0 if n is not squarefree.

Conjecture (Sun, August 2013). For any positive integer n, wehave

|µ(i + j − 1)|16i ,j6n 6= 0 and |µ2(i + j)|16i ,j6n 6= 0.

Conjecture (Sun, August 2013). Let f (p) take 1 or 0 according asp is prime or not. Then, for any integer n > 15 we have

|f (i + j)|16i ,j6n 6= 0.

20 / 41

Part II. Some open problems in additive combinatorics

21 / 41

Filz’s Prime Circle Problem

Filz’s Conjecture (1982). For any n = 2, 4, 6, . . ., there is acircular permutation i1, . . . , in of 1, . . . , n such that all the nadjacent sums

i1 + i2, i2 + i3, . . . , in−1 + in, in + i1

are prime.

Filz’s Conjecture in Graph-Theoretic Language. Let n be apositive even integer. Construct a simple graph G with vertices1, . . . , n such that the vertices i and j are adjacent if and only ofi + j is prime. Then G is a Hamiltonian graph.

A Similar Observation (conjectured by Sun and proved by Y.-G.Chen). For any integer n > 4, there is a circular permutationi1, . . . , in of 1, . . . , n such that

|i1 − i2|, |i2 − i3|, . . . , |in−1 − in|, |in − i1|

are prime.22 / 41

Two conjectures involving |i ± j | and |i2 − j2|

Conjecture (Sun, 2013-09-09). For any positive integer n, thereexists a circular permutation i0, i1, . . . , in of 0, 1, . . . , n such that allthe 2n + 2 numbers

|i0 ± i1|, |i1 ± i2|, . . . , |in−1 ± in|, |in ± i0|

are of the form (p − 1)/2, where p is an odd prime.

Conjecture (Sun, 2013-09-10). For any positive integer n 6= 2, 4,there is a circular permutation i0, i1, . . . , in of 0, . . . , n such that allthe n + 1 numbers

|i20 − i21 |, |i21 − i22 |, . . . , |i2n−1 − i2n |, |i2n − i20 |

are of the form (p − 1)/2, where p is an odd prime.

23 / 41

Circular permutations related to twin primes

Conjecture (Sun, 2013-09-08) For any positive integer n, there isa circular permutation i0, i1, . . . , in of 0, 1, . . . , n such that all then + 1 adjacent sums i0 + i1, i1 + i2, . . . , in−1 + in, in + i0 belong tothe set

{k ∈ Z+ : 6k − 1 and 6k + 1 are twin primes}.Remark. This conjecture implies the twin prime conjecture.Qing-Hu Hou has verified it for all n 6 100. For n = 10 we maytake the circular permutation (0, 5, 2, 3, 9, 1, 6, 4, 8, 10, 7).

Let a(n) denote the number of undirected circular permutationi0, i1, . . . , in of 0, 1, . . . , n meeting the requirement. I found thevalues of a(1), . . . , a(9): 1, 1, 1, 2, 2, 2, 5, 2, 12. Later MaxAlekseyev computed a(n) for n = 10, . . . , 25.

39, 98, 526, 2117, 6663, 15043, 68403, 791581, 4826577,

19592777, 102551299, 739788968, 4449585790,

36547266589, 324446266072, 2743681178070.24 / 41

Theorem (Sun, 2013-08) Let a1, . . . , an be a monotonic sequenceof n distinct real numbers. Then there is a permutation b1, . . . , bnof a1, . . . , an with b1 = a1 such that

|b1 − b2|, |b2 − b3|, . . . , |bn−1 − bn|are pairwise distinct.

Proof. If a1 > a2 > . . . > an, then −a1 < −a2 < . . . < −an. Sowe may assume that a1 < a2 < . . . < an without loss of generality.If n = 2k is even, then the permutation

(b1, . . . , bn) = (a1, a2k , a2, a2k−1, . . . , ak−1, ak+2, ak , ak+1)

meets our purpose since

a2k − a1 > a2k − a2 > a2k−1 − a2 > . . . > ak+2 − ak > ak+1 − ak .

When n = 2k − 1 is odd, the permutation

(b1, . . . , bn) = (a1, a2k−1, a2, a2k−2, . . . , ak−1, ak+1, ak)

meets the requirement since

a2k−1−a1 > a2k−1−a2 > a2k−2−a2 > . . . > ak+1−ak−1 > ak+1−ak .25 / 41

Corollary. There is a circular permutation q1, . . . , qn of the first nprimes p1, . . . , pn with q1 = p1 = 2 and qn = pn such that the ndistances

|q1 − q2|, |q2 − q3|, . . . , |qn−1 − qn|, |qn − q1|

are pairwise distinct.

Conjecture (Sun, 2013-09-01). Let a1, a2, . . . , an be n distinct realnumbers. Then there is a permutation b1, . . . , bn of a1, . . . , anwith b1 = a1 such that the n − 1 numbers

|b1 − b2|, |b2 − b3|, . . . , |bn−1 − bn|

are pairwise distinct.

Francesco Monopoli [Electron. J. Combin. 22(2015), no. 3,#P3.20]: The conjecture holds if the set A = {a1, a2, . . . , an}forms an arithmetic progression.

26 / 41

Circular permutations of quadratic residues (I)

Conjecture (Sun, 2013-09). For any prime p = 2n + 1 > 13, thereis a circular permutation a1, . . . , an of the (p − 1)/2 = n quadraticresidues modulo p such that all the n adjacent sums

a1 + a2, a2 + a3, . . . , an−1 + an, an + a1

are quadratic residues (or quadratic nonresidues) modulo p. Also,for any prime p = 2n + 1 > 5, there is a circular permutationb1, . . . , bn of the (p − 1)/2 = n quadratic residues modulo p suchthat all the n adjacent differences

b1 − b2, b2 − b3, . . . , bn−1 − bn, bn − b1

are quadratic residues (or quadratic nonresidues) modulo p.

Later this was confirmed by N. Alon and J. Bourgain [Geom.Funct. Anal. 24(2014), 721-739]. Below is a lemma.

B. Jacobson (JCTB 1980): For each k > 1, any 2-connectedk-regular graph with at most 3k vertices is Hamiltonian.

27 / 41

Circular permutations of quadratic residues (I)

Conjecture (Sun, 2013-09). For any prime p = 2n + 1 > 13, thereis a circular permutation a1, . . . , an of the (p − 1)/2 = n quadraticresidues modulo p such that all the n adjacent sums

a1 + a2, a2 + a3, . . . , an−1 + an, an + a1

are quadratic residues (or quadratic nonresidues) modulo p. Also,for any prime p = 2n + 1 > 5, there is a circular permutationb1, . . . , bn of the (p − 1)/2 = n quadratic residues modulo p suchthat all the n adjacent differences

b1 − b2, b2 − b3, . . . , bn−1 − bn, bn − b1

are quadratic residues (or quadratic nonresidues) modulo p.

Later this was confirmed by N. Alon and J. Bourgain [Geom.Funct. Anal. 24(2014), 721-739]. Below is a lemma.

B. Jacobson (JCTB 1980): For each k > 1, any 2-connectedk-regular graph with at most 3k vertices is Hamiltonian.

28 / 41

Alon and Bourgain’s general result

Theorem (Alon & Bourgain). There exists a constant c > 0 suchthat for any prime power q and For any multiplicative subgroup Aof the finite field Fq with

|A| = d ≥ cq3/4√

(log q)(log log log q)

log log q,

there is a numbering a1, a2, . . . , ad of the elements of A such that

a1 + a2, a2 + a3, . . . , ad−1 + ad , ad + a1

all belong to A.

Their tools include algebraic graph theory and probability method.

A Key Lemma (Krivelevich & Sudakov). Let G be a d-regulargraph with n vertices. If n is large enough, and the absolute valueof each nontrivial eigenvalue of the adjacency matrix of G issmaller than d(log log n)2/(1000(log n) log log log n), then G is aHamiltonian graph.

29 / 41

Circular permutations of quadratic residues (II)

Theorem (Sun, Oct. 2013). Let Fq be a finite field withq = 2n + 1 > 266 elements. Set

S = {a2 : a ∈ F∗q = Fq \ {0}} and T = F ∗q \ S .

Then, there is a circular permutation a1, . . . , an of all the elementsof S such that

{a1 + a2, a2 + a3, . . . , an−1 + an, an + a1} = S (or T ).

Also, there is a circular permutation a1, . . . , an of all the elementsof S such that

{a1 − a2, a2 − a3, . . . , an−1 − an, an − a1} = S (or T ).

30 / 41

Circular permutations of quadratic residues (II)

Proof. Let ε ∈ {±1}, and R = S or T . Choose an element a ∈ T .By a result of Wenbao Han [Acta Math. Sinica 32(1989)], thereexists a primitive root g of Fq such that 1 + εg2 (or a + εag2) isalso a primitive root and hence an element of T . So there is aprimitive root g with 1 + εg2 ∈ R. Set ai = g2i for i = 1, . . . , n.Then

{a1, a2, . . . , an} = S and {a1 + εa2, . . . , an + εa1} = R.

Note that

g2i + εg2(i+1) = g2i (1 + εg2) ∈ R for all i = 1, . . . , n.

Remark. 266 in the theorem can be reduced to 13 via acomplicated analysis.

31 / 41

A conjecture on primitive roots

Conjecture (joint with Q.-H. Hou, 2013-09-05) Let Fq be thefinite field with q > 7 elements. Then there is a numberinga1, . . . , aq of the elements of Fq such that all the q sums

a1 + a2, a2 + a3, . . . , aq−1 + aq, aq + a1

are generators of the cyclic group F∗q = Fq \ {0} (i.e., primitiveelements of Fq).

Remark. We have verified this for all primes q < 545.

32 / 41

Two more conjectures on primitive roots

Conjecture (Sun, 2013-09-17) Let Fq be a finite field with q > 7elements and let a0 be any element of Fq. Then there is a circularpermutation a1, . . . , aq−1 of all the nonzero elements of Fq suchthat all the q − 1 elements

a0 + a1a2, a0 + a2a3, . . . , a0 + aq−2aq−1, a0 + aq−1a1

are primitive roots of the filed Fq.

Conjecture (Sun, 2013-04-24) For any prime q, there is apartition number p(n) < q which is a primitive root modulo q,where p(n) denotes the number of ways to write n as a sum ofpositive integers with the order of addends ignored.

33 / 41

Two conjectures related to coprime properties

Conjecture (Sun, 2013-09-07) For any positive integer n 6= 2, 4,there exists a permutation i0, i1, . . . , in of 0, 1, . . . , n with i0 = 0and in = n such that all the n + 1 adjacent sums

i0 + i1, i1 + i2, . . . , in−1 + in, in + i0

are coprime to both n − 1 and n + 1.

Remark. I have proved this for any positive odd integer n.

Conjecture (Sun, 2013-10-04). Let n > 1 be odd. Then there is areduced system {a1, . . . , aϕ(n)} of residues modulo n such that

{a1 − a2, a2 − a3, . . . , aϕ(n) − a1}

is also a reduced system of residues modulo n.

Remark. I have proved this for any odd prime power n.

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A conjecture on n real numbers

Conjecture (Sun, 2013-09-22) Let A be a set of n > 2 distinctnonzero real numbers. Then there is a circular permutationa1, a2, ..., an of all the elements of A such that the n adjacent sums

a1 + a2, a2 + a3, ..., an−1 + an, an + a1

are pairwise distinct, and that the n adjacent products

a1a2, a2a3, ..., an−1an, ana1

are also pairwise distinct, except for the following three cases:

(a) |A| = 4 and A has the form {±s,±t}.

(b) |A| = 5 and A has the form {r ,±s,±t}.

(c) |A| = 6 and A has the form {±r ,±s,±t}.

Remark. For the set A = {1, 2, . . . , n} with n an odd prime power,the natural order of the elements of A meets the requirement. Wealso have a similar conjecture involving differences and products.

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On polynomials over finite fields

It is known that∑

a∈Fqak = 0 for all k = 0, 1, . . . , q − 2, where Fq

denotes the finite field of q elements.

Conjecture (Sun, 2013-09-30). (i) For any finite field Fq and apolynomial P(x) ∈ Fq[x ] of degree smaller than q − 1, if P(x) isnot of the form c − x , then there is a circular permutationa1, . . . , aq of all the elements of Fq with

{P(a1) + a2, P(a2) + a3, . . . , P(aq−1) + aq, P(aq) + a1} = Fq.

(ii) Let F be any field with |F | > 7, and let A be a finite subset ofF with |A| = n > 2. Let P(x) ∈ F [x ] with deg(P) < p − 1 if F isof prime characteristic p. If P(x) is not of the form c − x , thenthere is a circular permutation a1, . . . , an of all the elements of Asuch that the n sums

P(a1) + a2, P(a2) + a3, . . . , P(an−1) + an, P(an) + a1

are pairwise distinct.36 / 41

A conjecture related to Snevily’s conjectureSnevily’s Conjecture (proved by Arsovski in 2011). Let G be anyabelian group of odd order, and let A and B be finite subsets of Gwith |A| = |B| = n. Then there is a numbering a1, . . . , an of the nelements of A and a numbering b1, . . . , bn of the n elements of Bsuch that a1 + b1, a2 + b2, . . . , an + bn are pairwise distinct.

Conjecture (Sun, 2013-09-03) Let A be an n-subset of a finiteadditive abelian group G with 2 - n or n - |G |.(i) There always exists a numbering a1, a2, . . . , an of all the nelements of A such that the n sums

a1 + a2, a2 + a3, . . . , an−1 + an, an + a1

are pairwise distinct.(ii) In the case 3 < n < |G |, there is a numbering a1, a2, . . . , an ofall the n elements of A such that the n differences

a1 − a2, a2 − a3, . . . , an−1 − an, an − a1

are pairwise distinct.37 / 41

A conjecture involving ai + 2ai+1

Conjecture (2013-09-20) Let G be an additive abelian group.

(i) If G is finite with |G | 6≡ 0 (mod 3), then for any finite subset Aof G with |A| = n > 3, there is a numbering a1, . . . , an of all theelements of A such that the n sums

a1 + 2a2, a2 + 2a3, . . . , an−1 + 2an, an + 2a1

are pairwise distinct.

(ii) Let A and B be finite subsets of G with |A| = |B| = n. Thenwe can write A = {a1, . . . , an} and B = {b1, . . . , bn} so that either

a1 + 2b1, a2 + 2b2, . . . , an−1 + 2bn−1, an + 2bn

are pairwise distinct, or

2a1 + b1, 2a2 + b2, . . . , 2an−1 + bn−1, 2an + bn

are pairwise distinct.Remark. When A = {a1, . . . , an} is an abelian group of the form(Z/3Z)r , a1 + 2a2 = a1 − a2, . . ., an + 2a1 = an − a1 cannot bepairwise distinct. Part (i) holds with G torsion-free.

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A theorem involving three subsets

Motivated by Snevily’s conjecture, we obtained the following resultinvolving three subsets.

Theorem (Sun, [Math. Res. Lett. 15(2008)]). Let G be anabelian group with cyclic torsion group, and let A,B,C be subsetsof G with |A| = |B| = |C | = n. Then, there is a numberinga1, . . . , an of the n elements of A, a numbering b1, . . . , bn of theelements of B, and a numbering c1, . . . , cn of the elements of Csuch that a1 + b1 + c1, . . . , an + bn + cn are pairwise distinct.

Corollary. Let N be any positive integer. For the N ×N ×N Latincube over Z/NZ formed by the Cayley addition table, eachn × n × n subcube with n 6 N contains a Latin transversal.

Conjecture (Sun, [Math. Res. Lett. 15(2008)]). Every n × n × nLatin cube contains a Latin transversal.

Remark. In 1967 Ryser conjectured that every Latin square of oddorder has a Latin transversal.

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A conjecture for abelian groups with no involutionAn element of a group is called an involution if its order is two.

Conjecture (Sun, 2013-09-04). Let G be an additive abeliangroup. If G is cyclic or G contains no involution, then for anyfinite subset A of G with |A| = n > 3, there is a numberinga1, . . . , an of all the n elements of A such that

a1+a2+a3, a2+a3+a4, . . . , an−2+an−1+an, an−1+an+a1, an+a1+a2

are pairwise distinct.

Remark. For a finite abelian group G = {a1, a2, . . . , an}, it is easyto see that 2(a1 + · · ·+ an) = 0.

Theorem (Sun, 2013-09-19). The conjecture holds for anytorsion-free abelian group G .

Remark. (1) I became too tired and ill immediately after I spentthe whole day to finish the proof of this theorem.(2) The conjecture is even open for cyclic groups of odd primeorders.

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For sources of my conjectures, you may visit

http://math.nju.edu.cn/∼zwsunhttp://arxiv.org/abs/1308.2900

http://arxiv.org/abs/1309.1678.http://arxiv.org/abs/1405.0290.

You are welcome to solve myconjectures!

Thank you!

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