quadratic residues and quartic residues modulo primesmaths.nju.edu.cn/~zwsun/205i.pdf · 2020. 9....

Int. J. Number Theory 16 (2020), no. 8, 1833–1858.

QUADRATIC RESIDUES AND QUARTIC RESIDUES

MODULO PRIMES

ZHI-WEI SUN

Abstract. In this paper we study some products related to quadratic residuesand quartic residues modulo primes. Let p be an odd prime and let A be any

integer. We determine completely the product

fp(A) :=∏

16i,j6(p−1)/2

p-i2−Aij−j2

(i2 −Aij − j2)

modulo p; for example, if p ≡ 1 (mod 4) then

fp(A) ≡

−(A2 + 4)(p−1)/4 (mod p) if (A2+4p

) = 1,

(−A2 − 4)(p−1)/4 (mod p) if (A2+4p

) = −1,

where ( ·p

) denotes the Legendre symbol. We also determine

(p−1)/2∏i,j=1

p-2i2+5ij+2j2

(2i2 + 5ij + 2j2

)and

(p−1)/2∏i,j=1

p-2i2−5ij+2j2

(2i2 − 5ij + 2j2

)modulo p.

1. Introduction

For n ∈ Z+ = {1, 2, 3, . . .} and x = a/b with a, b ∈ Z, b 6= 0 and gcd(b, n) = 1,we let {x}n denote the unique integer r ∈ {0, . . . , n− 1} with r ≡ x (mod n) (i.e.,a ≡ br (mod n)). The well-known Gauss Lemma (see, e.g., [6, p. 52]) states thatfor any odd prime p and integer x 6≡ 0 (mod p) we have(

x

p

)= (−1)|{16k<p/2: {kx}p>p/2}|, (1.1)

where ( ·p ) is the Legendre symbol. This was extended to Jacobi symbols by M.

Jenkins [7] in 1867, who showed (by an elementary method) that for any positiveodd integer n and integer x with gcd(x, n) = 1 we have(x

n

)= (−1)|{16k<n/2: {kx}n>n/2}|, (1.2)

where ( ·n ) is the Jacobi symbol. In the textbook [9, Chapters 11-12], H. Rademachersupplied a proof of Jenkins’ result by using subtle properties of quadratic Gausssums.

Now we present our first new theorem.

Theorem 1.1. Let n be a positive odd integer, and let x ∈ Z with gcd(x(1−x), n) =1. Then

(−1)|{16k<n/2: {kx}n>k}| =

(2x(1− x)

n

). (1.3)

Key words and phrases. Quadratic residues, quartic residues, congruences, Lucas sequences.2020 Mathematics Subject Classification. Primary 11A15; Secondary 11A07, 11B39.

1

2 Z.-W. SUN

Also,

(−1)|{16k<n/2: {kx}n>n/2 & {k(1−x)}n>n/2}| =

(2

n

), (1.4)

(−1)|{16k<n/2: {kx}n<n/2 & {k(1−x)}n<n/2}| =

(2x(x− 1)

n

), (1.5)

and

(−1)|{16k<n/2: {kx}n>n/2>{k(1−x)}n}| =

(2x

n

). (1.6)

Let p be an odd prime, and let a, b, c ∈ Z and

Sp(a, b, c) :=∏

16i<j6p−1

p-ai2+bij+cj2

(ai2 + bij + cj2). (1.7)

Using Theorem 1.1, together with [15, Theorem 1.2], we completely determineSp(a, b, c) mod p in terms of Legendre symbols.

Theorem 1.2. Let p be an odd prime, and let a, b, c ∈ Z and ∆ = b2 − 4ac. Whenp - ac(a+ b+ c), we have

Sp(a, b, c) ≡

{(a(a+b+c)

p ) (mod p) if p | ∆,−(ac(a+b+c)∆

p ) (mod p) if p - ∆.(1.8)

In the case p | ac(a+ b+ c), we have

Sp(a, b, c) ≡

0 (mod p) if p | a, p | b and p | c,−(−ap ) (mod p) if p - a, p | b and p | c,−( bp ) (mod p) if p | a, p - b and p | c,−(−cp ) (mod p) if p | a, p | b and p - c,−( cp ) (mod p) if p | a, p - bc and p | b+ c,

−(ap ) (mod p) if p - ab, p | a+ b and p | c,−(−ap ) (mod p) if p - ac, p | a− c and p | a+ b+ c,

(−acp ) (mod p) if p - ac(a− c) and p | a+ b+ c,

(−a(a+b)p ) (mod p) if p - ab(a+ b) and p | c,

(−c(b+c)p ) (mod p) if p | a and p - bc(b+ c),

(1.9)

We will prove Theorem 1.1 and those parts of Theorem 1.2 not covered by [15,Theorem 1.2] in Section 2.

Let p be an odd prime. For a, b, c ∈ Z we introduce

Tp(a, b, c) :=

(p−1)/2∏i,j=1

p-ai2+bij+cj2

(ai2 + bij + cj2). (1.10)

Our following theorem determines Tp(1,−(a+b),−1) modulo p for all a, b ∈ Z withab ≡ −1 (mod p).

QUADRATIC RESIDUES AND QUARTIC RESIDUES MODULO PRIMES 3

Theorem 1.3. Let p be any odd prime, and let a, b ∈ Z with ab ≡ −1 (mod p).Set

{a, b}p :=

(p−1)/2∏i,j=1

i6≡aj,bj (mod p)

(i− aj)(i− bj), (1.11)

which is congruent to Tp(1,−(a+ b),−1) modulo p.(i) We have

−{a, b}p ≡

{(a−bp ) (mod p) if p ≡ 1 (mod 4) & p - (a− b),(a(a−b)

p ) = (a2+1p ) (mod p) if p ≡ 3 (mod 4).

(1.12)

(ii) If a ≡ b (mod p) and p ≡ 1 (mod 8), then

{a, b}p ≡ (−1)(p+7)/8 p− 1

2! (mod p).

If p ≡ 5 (mod 8) and a ≡ b ≡ (−1)k((p− 1)/2)! (mod p) with k ∈ {0, 1}, then

{a, b}p ≡ (−1)k+(p−5)/8 (mod p).

Our proof of Theorem 1.3 will be given in Section 3.For any A ∈ Z, we define the Lucas sequences {un(A)}n>0 and {vn(A)}n>0 by

u0(A) = 0, u1(A) = 1, and un+1(A) = Aun(A) + un−1(A) for n = 1, 2, 3, . . . ,

and

v0(A) = 2, v1(A) = A, and vn+1(A) = Avn(A) + vn−1(A) for n = 1, 2, 3, . . . .

It is well known that

un(A) =αn − βn

α− βand vn(A) = αn + βn

for all n ∈ N = {0, 1, 2, . . .}, where

α =A+√A2 + 4

2and β =

A−√A2 + 4

2.

Thus (A±√A2 + 4

2

)n=vn(A)± un(A)

√A2 + 4

2for all n ∈ N. (1.13)

Now we state our fourth theorem which determines Tp(1,−A,−1) for any oddprime p and integer A.

Theorem 1.4. Let p be an odd prime and let A ∈ Z.(i) Suppose that p | (A2 + 4). Then p ≡ 1 (mod 4), A/2 ≡ (−1)k((p −

1)/2)! (mod p) for some k ∈ {0, 1}, and

Tp(1,−A,−1) ≡

{(−1)(p+7)/8((p− 1)/2)! (mod p) if p ≡ 1 (mod 8),

(−1)k+(p−5)/8 (mod p) if p ≡ 5 (mod 8).(1.14)

(ii) When (A2+4p ) = 1, we have

Tp(1,−A,−1) ≡

{−(A2 + 4)(p−1)/4 (mod p) if p ≡ 1 (mod 4),

−(A2 + 4)(p+1)/4u(p−1)/2(A)/2 (mod p) if p ≡ 3 (mod 4).

(1.15)

4 Z.-W. SUN

(iii) When (A2+4p ) = −1, we have

Tp(1,−A,−1) ≡

{(−A2 − 4)(p−1)/4 (mod p) if p ≡ 1 (mod 4),

(−A2 − 4)(p+1)/4u(p+1)/2(A)/2 (mod p) if p ≡ 3 (mod 4).

(1.16)

We will prove Theorem 1.4 in Section 4.Let p be a prime with p ≡ 1 (mod 4). Then (p−1

2 !)2 ≡ −1 (mod p) by Wilson’s

theorem. We may write p = x2 + y2 with x, y ∈ Z, x ≡ 1 (mod 4) and y ≡p−1

2 !x (mod p). Recall that an integer a not divisible by p is a quartic residue

modulo p (i.e., z4 ≡ a (mod p) for some z ∈ Z) if and only if a(p−1)/4 ≡ 1 (mod p).Dirichlet proved that 2 is a quartic residue modulo p if and only if 8 | y (see, e.g.,[6, p. 64, Exer. 28]). On the other hand, we have∣∣∣∣{1 6 k <

p

4:

(k

p

)= 1

}∣∣∣∣ ≡ 0 (mod 2) ⇐⇒ y ≡ (−1)(p−1)/4 − 1 (mod 8)

as discovered by K. Burde [2] and re-proved by K. S. Williams [16]. In view ofWilliams and J. D. Currie [17, (1.4)], we have

2(p−1)/4 ≡ (−1)|{16k<p4 : ( kp )=−1}| ×

{1 (mod p) if p ≡ 1 (mod 8),p−1

2 ! (mod p) if p ≡ 5 (mod 8).

By Dirichlet’s class number formula (see, e.g., L.E. Dickson [4, p. 101]),

p− 1

2− 4

∣∣∣∣{1 6 k <p

4:

(k

p

)= −1

}∣∣∣∣ = h(−4p),

where h(d) with d ≡ 0, 1 (mod 4) not a square denotes the class number of thequadratic field with discriminant d. In 1905, Lerch (see, e.g., [5]) proved that

h(−3p) = 2∑

16k<p/3

(k

p

).

By [17, Lemma 14],

(−3)(p−1)/4 ≡

{(−1)h(−3p)/4 (mod p) if p ≡ 1 (mod 12),

(−1)(h(−3p)−2)/4 p−12 ! (mod p) if p ≡ 5 (mod 12).

Thus, if p ≡ 1 (mod 12) then

(−3)(p−1)/4 ≡ (−1)12

∑(p−1)/3k=1 (( kp )−1)+ p−1

6 = (−1)|{16k<p3 : ( kp )=−1}| (mod p);

similarly, if p ≡ 5 (mod 12) then

(−3)(p−1)/4 ≡ (−1)|{16k<p3 : ( kp )=−1}| p− 1

2! (mod p).

From Theorem 1.4, we deduce the following result which will be proved in Section5.

Theorem 1.5. Let p be an odd prime.


(i) We have

Tp(1,−1,−1) ≡

−5(p−1)/4 (mod p) if p ≡ 1, 9 (mod 20),

(−5)(p−1)/4 (mod p) if p ≡ 13, 17 (mod 20),

(−1)b(p−10)/20c (mod p) if p ≡ 3, 7 (mod 20),

(−1)b(p−5)/10c (mod p) if p ≡ 11, 19 (mod 20).

(1.17)

(ii) We have

Tp(1,−2,−1) ≡

−2(p−1)/4 (mod p) if p ≡ 1 (mod 8),

2(p−1)/4 (mod p) if p ≡ 5 (mod 8),

(−1)(p−3)/8 (mod p) if p ≡ 3 (mod 8),

(−1)(p−7)/8 (mod p) if p ≡ 7 (mod 8).

(1.18)

Now we state our sixth theorem.

Theorem 1.6. Let p > 3 be a prime and let δ ∈ {±1}. If p ≡ 1 (mod 4), then

Tp(2, 5δ, 2) ≡ (−1)b(p+11)/12c (mod p). (1.19)

When p ≡ 3 (mod 4), we have

Tp(2, 5δ, 2) ≡(

6

p

)δ2δ

3δ

((p− 3)/2

(p− 3)/4

)−2δ

(mod p). (1.20)

Note that there is no simple closed form for(

(p−3)/2(p−3)/4

)modulo a prime p ≡

3 (mod 4). For a prime p ≡ 1 (mod 4) with p = x2+y2 (x, y ∈ Z and x ≡ 1 (mod 4)),

Gauss showed the congruence(

(p−1)/2(p−1)/4

)≡ 2x (mod p), and S. Chowla, B. Dwork

and R. J. Evans [3] used Gauss and Jacobi sums to prove further that((p− 1)/2

(p− 1)/4

)≡ 2p−1 + 1

2

(2x− p

2x

)(mod p2),

which was first conjectured by F. Beukers. (See also [1, Chapter 9] for furtherrelated results.)

Though we have made some numerical tests via a computer, we are unable tofind general patterns for Tp(a, b, c) modulo p, where p is an arbitrary odd primeand a, b, c are arbitrary integers.

Let p be an odd prime. It is known (cf. [15, (1.6) and (1.7)]) that∏16i<j6(p−1)/2

p-i2+j2

(i2 + j2) ≡

{(−1)b(p−5)/8c (mod p) if p ≡ 1 (mod 4),

(−1)b(p+1)/8c (mod p) if p ≡ 3 (mod 4).

From this we immediately get∏16i<j6(p−1)/2

p-i2+j2

(i2 + j2

p

)=

{1 if p ≡ 1 (mod 4),

(−1)b(p+1)/8c if p ≡ 3 (mod 4).

As the product∏

16i<j6(p−1)/2(i2 − j2) modulo p was determined via [15, (1.5)],

we also know the value of the product∏16i<j6(p−1)/2

(i2 − j2

p

)=

∏16i<j6(p−1)/2

(i− jp

)(i+ j

p

)

6 Z.-W. SUN

Motivated by this, we obtain the following result.

Theorem 1.7. Let p > 3 be a prime and let δ ∈ {±1}. Then

∏16i<j6(p−1)/2

(j + δi

p

)=

(−1)|{0<k<

p4 : ( kp )=δ}| if p ≡ 1 (mod 4),

(−1)(p−3)/8 if p ≡ 3 (mod 8),

(−1)(p+1)/8+(h(−p)+1)/2 if p ≡ 7 (mod 8).

(1.21)

We will prove Theorems 1.6 and 1.7 in Section 6, and pose ten conjectures inSection 7.

2. Proofs of Theorems 1.1 and 1.2

Proof of Theorem 1.1. For each k = 1, . . . , (n− 1)/2, we have⌊kx

n

⌋−⌊k(x− 1)

n

⌋=

{0 if {kx}n > k,

1 if {kx}n < k.

Thus ∣∣∣{1 6 k <n

2: {kx}n > k

}∣∣∣ =n− 1

2−

(n−1)/2∑k=1

(⌊kx

n

⌋−⌊k(x− 1)

n

⌋)and hence

(−1)|{16k<n/2: {kx}n>k}|(−1

n

)=(−1)

∑(n−1)/2k=1 bkx/nc+

∑(n−1)/2k=1 bk(x−1)/nc

=(−1)(∑(n−1)/2k=1 (2x−1)k−

∑(n−1)/2k=1 {kx}n−

∑(n−1)/2k=1 {k(x−1)}n)/n

=(−1)(n2−1)/8(−1)∑(n−1)/2k=1 {kx}n+

∑(n−1)/2k=1 {k(x−1)}n .

As {kx}n ≡ 1 + n− {kx}n (mod 2) for all k = 1, . . . , (n− 1)/2, we have

(n−1)/2∑k=1

{kx}n ≡(n−1)/2∑k=1

{kx}n<n/2

{kx}n +

(n−1)/2∑k=1

{kx}n>n/2

(1 + (n− {kx}n))

=

(n−1)/2∑k=1

{kx}n>n/2

1 +

(n−1)/2∑r=1

r

=∣∣∣{1 6 k <

n

2: {kx}n >

n

2

}∣∣∣+n2 − 1

8(mod 2).

and hence

(−1)∑(n−1)/2k=1 {kx}n =

(xn

)( 2

n

)=

(2x

n

)with the help of (1.2). Similarly,

(−1)∑(n−1)/2k=1 {k(x−1)}n =

(xn

)( 2

n

)=

(2(x− 1)

n

).


In view of the above, we obtain

(−1)|{16k<n/2: {kx}n>k}| =

(−2

n

)(−1)

∑(n−1)/2k=1 {kx}n(−1)

∑(n−1)/2k=1 {k(x−1)}n

=

(−2

n

)(2x

n

)(2(x− 1)

n

)=

(2x(1− x)

n

).

This proves (1.3).When 1 6 k < n/2 and {kx}n < n/2, we clearly have

{kx}n > k ⇐⇒ {k(1− x)}n >n

2.

Thus ∣∣∣{1 6 k <n

2: {kx}n > k

}∣∣∣− ∣∣∣{1 6 k <n

2: {kx}n >

n

2

}∣∣∣=∣∣∣{1 6 k <

n

2: {kx}n <

n

2< {k(1− x)}n

}∣∣∣=∣∣∣{1 6 k <

n

2: {k(1− x)}n >

n

2

}∣∣∣−∣∣∣{1 6 k <

n

2: {kx}n >

n

2& {k(1− x)}n >

n

2

}∣∣∣ ,and hence by (1.2) and (1.3) we get

(−1)|{16k<n/2: {kx}n>n/2 & {k(1−x)}n>n/2}|

=(−1)|{16k<n/2: {kx}n>k}|(xn

)(1− xn

)=

(2

n

).

This proves (1.4).In view of (1.2) and (1.4), we also have

(−1)|{16k<n/2: {kx}n>n/2>{k(1−x)}n}|

=(−1)|{16k<n/2: {kx}n>n/2}|−||{16k<n/2: {kx}n>n/2 & {k(1−x)}n>n/2}|

=(xn

)( 2

n

)=

(2x

n

).

This proves (1.6).By (1.4) and (1.6), we have(

2

n

)(−1)|{16k<n/2: {kx}n<n/2 & {k(1−x)}n<n/2}|

=(−1)|{16k<n/2: ({kx}n−n/2)({k(1−x)}n−n/2)>0}|

=(−1)(n−1)/2−|{16k<n/2: {kx}n>n/2>{k(1−x)}n or {k(1−x)}n>n/2>{kx}n}|

=

(−1

n

)(2x

n

)(2(1− x)

n

)=

(x(x− 1)

n

).

So (1.5) also holds.The proof of Theorem 1.1 is now complete. �For any odd prime p and rational p-adic integer x, we define

Np(x) := |{1 6 k < p/2 : {kx}p > k}|. (2.1)

8 Z.-W. SUN

Proof of Theorem 1.2. By parts (ii)-(iv) of Sun [15, Theorem 1.2], the desiredresult holds except for the cases

I. p - ac(a− c) and p | a+ b+ c,

II. p - ab(a+ b) and p | c,III. p | a and p - bc(b+ c).

In case I, by [15, Theorem 1.2(iii)] we have

Sp(a, b, c) ≡ (−1)Np(a/c)

(2c(a− c)

p

)(mod p).

As

(−1)Np(a/c) =

(2a(c− a)

p

)by Theorem 1.1, we obtain

Sp(a, b, c) ≡(

2a(c− a)

p

)(2c(a− c)

p

)=

(−acp

)(mod p).

In case II, by [15, Theorem 1.2(iv)] we have

Sp(a, b, c) ≡ (−1)Np(−a/b)(

2

p

)(mod p).

As

(−1)Np(−a/b) =

(−2a(a+ b)

p

)by Theorem 1.1, we get

Sp(a, b, c) ≡(−2a(a+ b)

p

)(2

p

)=

(−a(a+ b)

p

)(mod p).

In case III, by [15, Theorem 1.2(iv)] we have

Sp(a, b, c) ≡ (−1)Np(−c/b)(

2

p

)(mod p).

Since (−1)Np(−c/b) = (−2c(b+c)p ) by Theorem 1.1, we deduce that

Sp(a, b, c) ≡(−2c(b+ c)

p

)(2

p

)=

(−c(b+ c)

p

)(mod p).

In view of the above, we have completed the proof of Theorem 1.2. �

3. Proof of Theorem 1.3

Lemma 3.1. Let p be any odd prime. Then(p− 1

2!

)2

≡ (−1)(p+1)/2 (mod p), (3.1)

and ∏16i<j6(p−1)/2

(j2 − i2) ≡

{−((p− 1)/2)! (mod p) if p ≡ 1 (mod 4),

1 (mod p) if p ≡ 3 (mod 4).(3.2)

Remark 3.2. (3.1) is an easy consequence of Wilson’s theorem. (3.2) is also known,see [15, (1.5)] and its few-line proof there.


Lemma 3.3. Let p be a prime with p ≡ 1 (mod 4). Then∣∣∣∣∣{

1 6 k 6p− 1

2:

{k × p− 1

2!

}p

>p

2

}∣∣∣∣∣ =p− 1

4. (3.3)

Proof. Let a = ((p − 1)/2)!. Then a2 ≡ −1 (mod p) by (3.1). For any k =1, . . . , (p − 1)/2, there is a unique integer k∗ ∈ {1, . . . , (p − 1)/2} congruent to akor −ak modulo p. Note that k∗ 6= k since a 6≡ ±1 (mod p). If {ak}p > p/2 then{ak∗}p = {a(−ak)}p = k < p/2; if {ak}p < p/2 then {ak∗}p = {a(ak)}p = p− k >p/2. So, exactly one of {ak}p and {ak∗}p is greater than p/2. Therefore (3.3)holds. �

Remark 3.4. In view of Gauss’ Lemma, Lemma 3.3 is stronger than the fact that

( ((p−1)/2)!p ) = ( 2

p ) for any prime p ≡ 1 (mod 4) (cf. [14, Lemma 2.3]).

Proof of Theorem 1.3. Observe that

(p−1)/2∏i,j=1

p-i2−j2

(i2 − j2) =∏

16i<j6(p−1)/2

(i2 − j2)(j2 − i2)

=(−1)((p−1)/2

2 )∏

16i<j6(p−1)/2

(j2 − i2)2

and hence

(p−1)/2∏i,j=1

p-i2−j2

(i2 − j2) ≡ −(

2

p

)(mod p) (3.4)

with the help of Lemma 3.1.When a ≡ ±1 (mod p), we have b ≡ ∓1 6≡ a (mod p) and hence

{a, b}p ≡ −(

2

p

)=

{−(a−bp ) (mod p) if p ≡ 1 (mod 4),

−(a2+1p ) (mod p) if p ≡ 3 (mod 4).

So (1.12) holds in the case a ≡ ±1 (mod p).Below we assume that a 6≡ ±1 (mod p). As ab ≡ −1 (mod p), we have

{a, b}p =

(p−1)/2∏i,j=1

i6≡aj,bj (mod p)

(i− aj)×(p−1)/2∏i,j=1

j 6≡ai,bi (mod p)

(j − bi)

≡(p−1)/2∏i,j=1

p-(i−aj)(j+ai)

(i− aj)×(p−1)/2∏i,j=1

p-(i+aj)(j−ai)

i+ aj

a(mod p).

(3.5)

(i) If p ≡ 3 (mod 4), then (−1p ) = −1 and hence a 6≡ b (mod p). Now we prove

(1.12) under the assumption a 6≡ b (mod p). Note that a2 6≡ ±1 (mod p).

10 Z.-W. SUN

For 1 6 i, j 6 (p−1)/2, we cannot have i2−a2j2 ≡ j2−a2i2 ≡ 0 (mod p). Thus

∣∣∣∣{(i, j) : 1 6 i, j 6p− 1

2& p - (i+ aj)(j − ai)

}∣∣∣∣=

(p− 1

2

)2

−∣∣∣∣{(i, j) : 1 6 i, j 6

p− 1

2& p | i+ aj

}∣∣∣∣−∣∣∣∣{(i, j) : 1 6 i, j 6

p− 1

2& p | j − ai

}∣∣∣∣=

(p− 1

2

)2

−∣∣∣∣{1 6 j 6

p− 1

2: {aj}p >

p

2

}∣∣∣∣−∣∣∣∣{1 6 i 6

p− 1

2: {ai}p <

p

2

}∣∣∣∣=(p− 1)

p− 1

4− p− 1

2=p− 1

2· p+ 1

2− (p− 1)

and hence

a|{(i,j): 16i,j6(p−1)/2 & p-(i+aj)(j−ai)}| ≡ (a(p−1)/2)(p+1)/2 ≡(a

p

)(p+1)/2

(mod p).

Combining this with (3.5), we obtain

(a

p

)(p+1)/2

{a, b}p

≡(p−1)/2∏i,j=1

p-(i2−a2j2)(j2−a2i2)

(i2 − a2j2)×(p−1)/2∏i,j=1

i≡aj (mod p)

(i+ aj)×(p−1)/2∏i,j=1

i≡−aj (mod p)

(i− aj)

×(p−1)/2∏i,j=1

j−ai≡−a(i+bj)≡0 (mod p)

(i− aj)×(p−1)/2∏i,j=1

j+ai≡a(i−bj)≡0 (mod p)

(i+ aj)

≡(p−1)/2∏i,j=1

p-i2−a2j2

(i2 − a2j2)

/ (p−1)/2∏i,j=1

p|j2−a2i2

(i2 − a2j2)

× (−1)|{16j6(p−1)/2: {aj}p>p/2}|(p−1)/2∏j=1

(2aj)

× (−1)|{16j6(p−1)/2: {bj}p>p/2}|(p−1)/2∏j=1

(bj + aj) (mod p)


Thus, by using (3.4) and Gauss’ Lemma, we see that(a

p

)(p+1)/2

{a, b}p ≡(p−1)/2∏i,k=1

p-i2−k2

(i2 − k2)

/ (p−1)/2∏i=1

(i2 − a2(a2i2))

×(ab

p

)(2a(a+ b))(p−1)/2

(p− 1

2!

)2

≡−(

2

p

)(−1

p

)(2(a2 − 1)(1− a4)

p

)=−

(a2 + 1

p

)= −

(a2 − ab

p

)(mod p).

This proves (1.12).(ii) Now we come to prove part (ii) of Theorem 1.3. Assume that a ≡ b (mod p).

Then a2 ≡ ab ≡ −1 (mod p) and hence p ≡ 1 (mod 4). As j ± ai ≡ ±a(i ∓aj) (mod p), by (3.5) we have

{a, b}p ≡a−|{(i,j): 16i,j6(p−1)/2 & p-i+aj}|(p−1)/2∏i,j=1p-i−aj

(i− aj)×(p−1)/2∏i,j=1p-i+aj

(i+ aj)

≡a−(p−1)2/4+|{(i,j): 16i,j6(p−1)/2 & p|i+aj}|(p−1)/2∏i,j=1

p-i2−(aj)2

(i2 − (aj)2)

×(p−1)/2∏i,j=1p|i−aj

(i+ aj)×(p−1)/2∏i,j=1p|i+aj

(i− aj)

≡a|{(i,j): 16i,j6(p−1)/2 & p|i+aj}|(p−1)/2∏i,k=1

p-i2−k2

(i2 − k2)

× (−1)|{16j6(p−1)/2: {aj}p>p/2}|(p−1)/2∏j=1

(2aj). (mod p)

Applying (3.4) and Gauss’ Lemma, from the above we get

{a, b}p ≡a|{16j6(p−1)/2: {aj}p>p/2}| ×(−(

2

p

))(a

p

)(2a)(p−1)/2 p− 1

2!

≡− p− 1

2!× a|{16j6(p−1)/2: {aj}p>p/2}| (mod p).

(3.6)

As a2 ≡ −1 ≡ ((p − 1)/2)!)2, we have a ≡ (−1)k((p − 1)/2)! (mod p) for somek ∈ {0, 1}. In view of Lemma 3.3,∣∣∣∣∣

{1 6 j 6

p− 1

2:

{−p− 1

2! j

}p

>p

2

}∣∣∣∣∣=

∣∣∣∣∣{

1 6 j 6p− 1

2:

{p− 1

2! j

}p

<p

2

}∣∣∣∣∣ =p− 1

4.

12 Z.-W. SUN

Hence

a|{16j6(p−1)/2: {aj}p>p/2}|

=a(p−1)/4 = (−1)k(p−1)/4

(p− 1

2!

)(p−1)/4

≡

{(p−1

2 !)2(p−1)/8 ≡ (−1)(p−1)/8 (mod p) if p ≡ 1 (mod 8),

(−1)k p−12 !(p−1

2 !)2(p−5)/8 ≡ (−1)k+(p−5)/8 p−12 ! (mod p) if p ≡ 5 (mod 8).

Combining this with (3.6) we immediately obtain the desired results in Theorem1.3(ii).

In view of the above, we have completed the proof of Theorem 1.3. �


Proof of Theorem 1.4(i). As A2 ≡ −4 (mod p), we have (−1p ) = 1 and hence p ≡

1 (mod 4). Since ((p− 1)/2)2 ≡ −1 ≡ (A/2)2 (mod p), for some k ∈ {0, 1} we haveA/2 ≡ (−1)k((p−1)/2)! (mod p). Choose a, b ∈ Z with a ≡ b ≡ A/2 (mod p). Notethat {a, b}p ≡ Tp(1,−A,−1) (mod p). Applying Theorem 1.3(ii) we immediatelyget the desired (1.14). �

For any odd prime p and integer A with ∆ = A2 + 4 6≡ 0 (mod p), it is known(cf. [13, Lemma 2.3]) that

u(p−( ∆p )/2(A)v(p−( ∆

p ))/2(A) = up−( ∆p )(A) ≡ 0 (mod p).

Lemma 4.1. Let A ∈ Z and let p be an odd prime not dividing ∆ = A2 + 4. Then

p | v(p−( ∆p ))/2(A) ⇐⇒

(−1

p

)= −1. (4.1)

Remark 4.2. This is a known result, see, e.g., [10, Chapter 2, (IV.23)].

Proof of Theorem 1.4(ii). Let ∆ = A2 + 4. As (∆p ) = 1, we have d2 ≡ ∆ for

some d ∈ Z with p - d. Choose integers a and b such that

a ≡ A+ d

2(mod p) and b ≡ A− d

2(mod p).

Then a+b ≡ A (mod p) and ab ≡ (A2−d2)/4 ≡ −1 (mod p). Thus, for any i, j ∈ Zwe have

i2 −Aij − j2 ≡ (i− aj)(i− bj) (mod p).

If p ≡ 1 (mod 4), then

(A2 + 4)(p−1)/4 ≡ d(p−1)/2 ≡ (a− b)(p−1)/2 ≡(a− bp

)(mod p).

If p ≡ 3 (mod 4), then(2d(A+ d)

p

)=

(ad

p

)=

(a(a− b)

p

).


For any positive integer n, clearly

un(A) =1√∆

((A+√

∆

2

)n−

(A−√

∆

2

)n)

=1

2n−1

b(n−1)/2c∑k=0

(n

2k + 1

)An−1−2k∆k

≡ 1

2n−1

b(n−1)/2c∑k=0

(n

2k + 1

)An−1−2kd2k

=1

d

((A+ d

2

)n−(A− d

2

)n)(mod p)

and

vn(A) =

(A+√

∆

2

)n+

(A−√

∆

2

)n=

1

2n−1

bn/2c∑k=0

(n

2k

)An−2k∆k

≡ 1

2n−1

bn/2c∑k=0

(n

2k

)An−2kd2k =

(A+ d

2

)n+

(A− d

2

)n(mod p).

In the case p ≡ 3 (mod 4), we have p | v(p−1)/2(A) by Lemma 4.1, and hence

(2δ(A+ d)

p

)≡d(p−1)/2

(A+ d

2

)(p−1)/2

≡ d(p−1)/2 v(p−1)/2(A) + du(p−1)/2(A)

2

≡(d2)(p+1)/4 u(p−1)/2(A)

2≡ ∆(p+1)/4u(p−1)/2(A)

2(mod p).

In view of the above, we obtain (1.15) from Theorem 1.3(i). �

Lemma 4.3. Let p be an odd prime, and let A ∈ Z with ∆ = A2 + 4 6≡ 0 (mod p).If p ≡ 1 (mod 4), then

u(p+( ∆p ))/2(A) ≡ ±∆(p−1)/4 (mod p). (4.2)


u(p−( ∆p ))/2(A) ≡ ±2∆(p−3)/4 (mod p). (4.3)

Remark 4.4. This is a known result, see, e.g., [11, Theorem 4.1].

Proof of Theorem 1.4(iii). Let

∆ = A2 + 4, α =A+√

∆

2and β =

A−√

∆

2.

14 Z.-W. SUN

As x2−Ax−1 = (x−α)(x−β), both α and β are algebraic integers. Observe that

(p−1)/2∏i,j=1

(i2 −Aij − j2)

=

(p−1)/2∏i,j=1

(i− αj)(i− βj) =

(p−1)/2∏i,j=1

(i− αj)×(p−1)/2∏i,j=1

(j − βi)

=

(p−1)/2∏i,j=1

(i− αj)×(p−1)/2∏i,j=1

αj + i

α= α−((p−1)/2)2

(p−1)/2∏i,j=1

(i2 − α2j2)

=α−((p−1)/2)2(p−1)/2∏i,j=1

(−j2)

(α2 − i2

j2

)

=(−α)−((p−1)/2)2

(p− 1

2!

)2 p−12

(p−1)/2∏j=1

(p−1)/2∏i=1

(α2 − i2

j2

)and hence

(p−1)/2∏i,j=1

(i2 −Aij − j2) ≡ β((p−1)/2)2(p−1)/2∏k=1

(α2 − k2)(p−1)/2 (mod p)

(in the ring of all algebraic integers) since for any j, k = 1, . . . , (p− 1)/2 there is aunique i ∈ {1, . . . , (p− 1)/2} congruent to jk or −jk modulo p. Note that

(p−1)/2∏x=1

(x− k2) ≡ x(p−1)/2 − 1 (mod p).

Thus(p−1)/2∏i,j=1

(i2 −Aij − j2) ≡β((p−1)/2)2

((α2)(p−1)/2 − 1)(p−1)/2

=β((p−1)/2)2 (αp−1 − 1

)(p−1)/2

=(

(−α)(p−1)/2 − β(p−1)/2)(p−1)/2

(mod p).

Case 1. p ≡ 1 (mod 4).In this case,

(p−1)/2∏i,j=1

(i2 −Aij − j2) ≡(α(p−1)/2 − β(p−1)/2

α− β

)(p−1)/2

(α− β)(p−1)/2

=u(p−1)/2(A)(p−1)/2∆(p−1)/4 (mod p).

As (∆p ) = −1, by Lemma 4.3 we have u(p−1)/2(A) ≡ ±∆(p−1)/4 (mod p) and hence

u(p−1)/2(A)(p−1)/2 ≡ ∆p−1

2 ·p−1

4 ≡ (−1)(p−1)/4 (mod p).

Therefore(p−1)/2∏i,j=1

(i2 −Aij − j2) ≡ (−∆)(p−1)/4 (mod p).


Case 2. p ≡ 3 (mod 4).In this case,

(p−1)/2∏i,j=1

(i2 −Aij − j2) ≡−(α(p−1)/2 + β(p−1)/2

)(p−1)/2

=− v(p−1)/2(A)(p−1)/2 (mod p).

It is easy to see that 2vn−1(A) = ∆un(A)−Avn(A) for all n = 1, 2, 3, . . .. Hence

2v(p−1)/2(A) = ∆u(p+1)/2(A)−Av(p+1)/2(A) ≡ ∆u(p+1)/2(A) (mod p)

since v(p+1)/2(A) = v(p−( ∆p ))/2(A) ≡ 0 (mod p) by Lemma 4.1. Thus

(p−1)/2∏i,j=1

(i2 −Aij − j2) ≡−(

∆

2u(p+1)/2(A)

)(p−1)/2

≡u(p+1)/2(A)

2

(u(p+1)/2(A)

2

)(p−3)/2

(mod p).

As (∆p ) = −1, by Lemma 4.3 we have u(p+1)/2(A) ≡ ±2∆(p−3)/4 (mod p), and

hence

(p−1)/2∏i,j=1

(i2 −Aij − j2) ≡u(p+1)/2(A)

2∆

p−34 ·

p−32 =

u(p+1)/2(A)

2∆

p+12 ·

p+14 −(p−1)

≡u(p+1)/2(A)

2(−∆)(p+1)/4 (mod p).

In view of the above, we have completed the proof of Theorem 1.4(iii). �


Proof Theorem 1.5. Note that {un(1)}n>0 is just the Fibonacci sequence {Fn}n>0.By [12, Corollary 2(iii)], if p ≡ 3 (mod 4) and ( 5

p ) = 1, then

u(p−1)/2(1) = F(p−1)/2 ≡ −2(−1)b(p−5)/10c5(p−3)/4 (mod p)

and hence

−5(p+1)/4u(p−1)/2(1)

2≡ (−1)b(p−5)/10c5(p+1)/4+(p−3)/4 ≡ (−1)b(p−5)/10c (mod p).

Similarly, if p ≡ 3 (mod 4) and ( 5p ) = −1, then

u(p+1)/2(1) = F(p+1)/2 ≡ 2(−1)b(p−5)/10c5(p−3)/4 (mod p)

and hence

5(p+1)/4u(p+1)/2(1)

2≡ (−1)b(p−5)/10c5(p+1)/4+(p−3)/4 ≡ (−1)b(p+5)/10c (mod p).

Therefore Theorem 1.4 with A = 1 yields (1.17).When p ≡ 1 (mod 4), we obviously have

8(p−1)/4 = 2(p−1)/2+(p−1)/4 ≡(

2

p

)2(p−1)/4 = (−2)(p−1)/4 (mod p).

16 Z.-W. SUN


u(p−1)/2(2) ≡ (−1)(p+1)/82(p−3)/4 (mod p)

by [11, (1.7)], and hence

−8(p+1)/4u(p−1)/2(2)

2≡− 2(3p+3)/4(−1)(p+1)/82(p−7)/4 ≡ (−1)(p−7)/8 (mod p).

Similarly, if p ≡ 3 (mod 8), then

u(p+1)/2(2) ≡ (−1)(p+5)/82(p−3)/4 (mod p)

by [11, (1.7)], and hence

(−8)(p+1)/4u(p+1)/2(2)

2≡− 2(3p+3)/4(−1)(p+5)/82(p−7)/4 ≡ (−1)(p−3)/8 (mod p).

So Theorem 1.4 with A = 2 yields (1.18). This ends the proof. �

6. Proofs of Theorems 1.6 and 1.7

For n = 1, 2, 3, . . ., we adopt the notation

n!! :=

b(n−1)/2c∏k=0

(n− 2k).

Lemma 6.1. Let p > 3 be a prime. Then

p− 1

2!!

(p−1)/2∏i,j=1p-2i+j

(2i+ j) ≡(−2

p

)p− 3

2!!

(p−1)/2∏i,j=1p-2i−j

(2i− j) ≡ ±1 (mod p). (6.1)

Proof. Set

Ap :=p− 1

2!!

(p−1)/2∏i,j=1p-2i+j

(2i+ j) and Bp :=p− 3

2!!

(p−1)/2∏i,j=1p-2i−j

(2i− j).

Then

ApBp((p− 1)/2)!

≡(p−1)/2∏i,j=1p-2i+j

(2i+ j)×(p−1)/2∏i,j=1

p-2i+p−j

(2i+ p− j) =

(p−1)/2∏i=1

(1

2i

p−1∏j=0p-2i+j

(2i+ j)

)

≡(p−1)/2∏i=1

(p− 1)!

2i≡

(−2p )

((p− 1)/2)!(mod p)

and hence

ApBp ≡(−2

p

)(mod p). (6.2)


On the other hand,

Bp((p− 3)/2)!!

=

(p−1)/2∏i,j=1

p-2(p+1

2−i)−j

(2

(p+ 1

2− i)− j)

≡(p−1)/2∏i,j=1

p-2i+j−1

(1− j − 2i) =

(p−1)/2∏i=1

−2i

(−(p− 1)/2− 2i)∗

(p−1)/2∏j=1p-j+2i

(−j − 2i)

≡ (−2)(p−1)/2((p− 1)/2)!∏16i<p/24i6=p+1

((p+ 1)/2− 2i)

(p−1)/2∏i,j=1p-2i+j

(2i+ j)

× (−1)∑(p−1)/2i=1 ((p−1)/2−|{16j6(p−1)/2: p|2i+j}|) (mod p),

where k∗ is 1 or k according as p | k or not. Note that∏16i<p/4

(p+ 1

2− 2i

)=p− 3

2!!.

Therefore(−2

p

)BpAp≡ ((p− 1)/2)!

((p− 1)/2)!!×

∏(p+1)/4<i<p/2

1

((p+ 1)/2− 2i)

× (−1)((p−1)/2)2−|{16i6(p−1)/2: 2i>p/2}|

≡ ((p− 3)/2)!!(−1)((p−1)/2)2−((p−1)/2−b(p−1)/4c)

(−1)b(p−1)/4c∏(p+1)/4<i<p/2(2i− (p+ 1)/2)

= 1 (mod p),

and hence

Ap ≡(−2

p

)Bp (mod p) (6.3)

which gives the first congruence in (6.1).Combining (6.2) and (6.3), we see that A2

p ≡ B2p ≡ 1 (mod p). So (6.1) does

hold. This ends the proof. �

Proof of Theorem 1.6. As 2i2 + δ5ij + 2j2 = (i+ δ2j)(2i+ δj), we have

(p−1)/2∏i,j=1

p-2i2+δ5ij+2j2

(2i2 + δ5ij + 2j2) =

(p−1)/2∏i,j=1

p-(i+δ2j)(2i+δj)

(i+ δ2j)×(p−1)/2∏i,j=1

p-(i+δ2j)(2i+δj)

(2i+ δj)

=

(p−1)/2∏i,j=1

p-(i+δ2j)(2i+δj)

(i+ δ2j)×(p−1)/2∏i,j=1

p-(i+δ2j)(2i+δj)

(2j + δi)

=

(p−1)/2∏i,j=1p-i+δ2j

δ(i+ δ2j)2 ×(p−1)/2∏i,j=1p|j+δ2i

1

δ(i+ δ2j)2.

18 Z.-W. SUN

(Note that i + δ2j ≡ j + δ2i ≡ 0 (mod p) for no i, j = 1, . . . , (p − 1)/2.) Thus,applying Lemma 6.1 and (3.1) we get

(p−1)/2∏i,j=1

p-2i2+δ5ij+2j2

(2i2 + δ5ij + 2j2)

≡δ|(i,j): 16i,j6(p−1)/2 & p-i+δ2j}|

((p− 2 + δ)/2)!!2×

(p−1)/2∏i=1

{δ2i}p>p/2

1

δ(i+ δ2(p− 2δi))2

≡δ((p−1)/2)2−|{16j6(p−1)/2: {δ2j}p>p/2}|

((p− 2 + δ)/2)!!2×

(p−1)/2∏i=1

{δ2i}p>p/2

1

δ(i− 4i)2

≡ δ(p−1)2/4

((p− 2 + δ)/2)!!2×

(p−1)/2∏i=1

(3i)−1−δ ×bp/4c∏i=1

(3i)2δ

≡ δ(p−1)2/4

((p− 2 + δ)/2)!!2× 32δbp/4cbp/4c!2δ

((p− 1)/2)!1+δ

≡δ(p−1)2/4(−1)p+1

2 ·1+δ

2 32δbp/4c(

bp/4c!δ

((p− 2 + δ)/2)!!

)2

=(−1)(p+δ)/232δbp/4c(

bp/4c!δ

((p− 2 + δ)/2)!!

)2

(mod p).

Case 1. p ≡ 1 (mod 4).In this case,(

bp/4c!δ

((p− 2 + δ)/2)!!

)2

=

{1/2(p−1)/2 if δ = 1,

2(p−1)/2/((p− 1)/2)!2 if δ = −1.

Thus, with the help of (3.1) we have

(−1)(p+δ)/232δbp/4c(

bp/4c!δ

((p− 2 + δ)/2)!!

)2

≡(−1)(1+δ)/2

(3

p

)δ (2

p

)−δδ = −

(6

p

)= (−1)b(p+11)/12c (mod p),

and hence (1.19) follows from the above.Case 2. p ≡ 3 (mod 4).In this case, with the aid of (3.1) we have

(−1)(p+δ)/232δbp/4c(

bp/4c!δ

((p− 2 + δ)/2)!!

)2

≡(−1)(δ−1)/23δ((p−1)/2−1)

(2(p−3)/4

(p− 1)/2×(

(p−3)/2(p−3)/4

))2δ (p− 1

2!

)δ−1

≡δ(

3

p

)3−δ2δ(p+1)/2

((p− 3)/2

(p− 3)/4

)−2δ

≡(

6

p

)δ2δ

3δ

((p− 3)/2

(p− 3)/4

)−2δ

(mod p),

and hence (1.20) holds.


In view of the above, we have completed the proof. �

Proof of Theorem 1.7. Let n = (p− 1)/2. Clearly,∏16i<j6n

(j − i) =

n∏k=1

k|{16i6n: i+k6n}| =

n∏k=1

kn−k

and hence ∏16i<j6n

(j − ip

)=

(n!

p

)n n∏k=12-k

(k

p

). (6.4)

Case 1. p ≡ 1 (mod 4).In this case, n is even, and hence by (6.4) and [15, (3.5)] we have∏

16i<j6n

(j − ip

)=

((n− 1)!!

p

)= (−1)|{0<k<

p4 : ( kp )=−1}|.

By (3.2) and [14, Lemma 2.3],∏16i<j6n

(j2 − i2

p

)=

(−n!

p

)=

(2

p

)= (−1)(p−1)/4.

Thus we also have∏16i<j6n

(j + i

p

)= (−1)(p−1)/4(−1)|{0<k<

p4 : ( kp )=−1}| = (−1)|{0<k<

p4 : ( kp )=1}|.

So (1.21) holds in this case.Case 2. p ≡ 3 (mod 4).In this case, n is odd and∏

16i<j6n

(j2 − i2

p

)=

(1

p

)= 1

by (3.2). So it suffices to prove (1.21) for δ = −1.In view of (6.4), we have∏

16i<j6n

(j − ip

)=

(n!

p

)(n!!

p

)=

((n− 1)!!

p

). (6.5)

If p ≡ 3 (mod 8), then by [15, (3.6)] we have((n− 1)!!

p

)= (−1)b(p+1)/8c = (−1)(p−3)/8

and hence (1.21) holds for δ = −1. When p ≡ 7 (mod 8), we have(n!

p

)=

((−1)(h(−p)+1)/2

p

)= (−1)(h(−p)+1)/2 and

(n!!

p

)= (−1)(p+1)/8

by Mordell [8] and Sun [15, (3.6)] respectively, hence (1.21) with δ = −1 followsfrom (6.5).

Combining the above, we have finished the proof of (1.21). �

20 Z.-W. SUN

7. Some related conjectures

Motivated by our results in Section 1, here we pose 10 conjectures for furtherresearch. We have verified all the following conjectures for primes p < 13000.

Conjecture 7.1. Let p > 3 be a prime and let δ ∈ {±1}. Then∏16i<j6(p−1)/2

p-2i2+δ5ij+2j2

(2i2 + δ5ij + 2j2

p

)=

1

2

(δ

p

)((−1

p

)+

(2

p

)+

(6

p

)−(p

3

)).

(7.1)

Conjecture 7.2. Let p > 3 be a prime. Then∏16i<j6(p−1)/2

p-i2−ij+j2

(i2 − ij + j2

p

)=

{−1 if p ≡ 5, 7 (mod 24),

1 otherwise.(7.2)

Also, ∏16i<j6(p−1)/2

p-i2+ij+j2

(i2 + ij + j2

p

)=

{−1 if p ≡ 5, 11 (mod 24),

1 otherwise.(7.3)

Conjecture 7.3. Let p > 3 be a prime. Then∏16i<j6(p−1)/2

p-i2−3ij+j2

(i2 − 3ij + j2

p

)=

{−1 if p ≡ 7, 19 (mod 20),

1 otherwise.(7.4)

Also, ∏16i<j6(p−1)/2

p-i2+3ij+j2

(i2 + 3ij + j2

p

)=

{−1 if p ≡ 19, 23, 27, 31 (mod 40),

1 otherwise.(7.5)

Recall that for any prime p ≡ 3 (mod 4) the class number h(−p) of the imaginaryquadratic field Q(

√−p) is odd by [8].

Conjecture 7.4. Let δ ∈ {±1}.(i) For any prime p ≡ 1 (mod 12), we have

Tp(1, 4δ, 1) ≡ −3(p−1)/4 (mod p). (7.6)

(ii) Let p > 3 be a prime. Then∏16i<j6(p−1)/2

p-i2+δ4ij+j2

(i2 + δ4ij + j2

p

)

=

1 if p ≡ 1 (mod 24),

(−1)|{0<k<p4 : ( kp )=−1}| if p ≡ 17 (mod 24),

δ(−1)|{0<k<p12 : ( kp )=−1}|−1 if p ≡ 7 (mod 24),

δ(−1)|{0<k<p12 : ( kp )=−1}|+h(−p)−1

2 if p ≡ 19 (mod 24).

(7.7)

Conjecture 7.5. Let p > 3 be a prime. Then∏16i6j6(p−1)/2

p-4i2+j2

(4i2 + j2

p

)=

{1 if p ≡ 1, 7, 9, 19 (mod 20),

−1 otherwise.


Conjecture 7.6. Let p > 3 be a prime. Then

(−1)|{16k<p/3: ( kp )=−1}|(p−1)/2∏i,j=1p-3i+j

(3i+ j

p

)=

{1 if p ≡ ±1 (mod 12),

(−1)bp/12c if p ≡ ±5 (mod 12),

and

(p−1)/2∏i,j=1p-3i−j

(3i− jp

)=

(−1)|{16k<p/3: ( kp )=−1}|+(p−1)/12 if p ≡ 1 (mod 12),

(−1)|{16k<p/3: ( kp )=−1}|−1 if p ≡ 5 (mod 12),

(−1)|{16k<p/6: ( kp )=1}|+(p+1)/4 if p ≡ 7 (mod 12),

−1 if p ≡ 11 (mod 12).

.


(p−1)/2∏i,j=1p-4i+j

(4i+ j

p

)=

1 if p ≡ 1 (mod 4),

(−1)(h(−p)−1)/2+bp/8c if p ≡ 3 (mod 8),

(−1)|{16k<p/4: ( kp )=−1}| if p ≡ 7 (mod 8),

and(p−1)/2∏i,j=1p-4i−j

(4i− jp

)=

{(−1)(p−1)/4 if p ≡ 1 (mod 4),

(−1)bp/8c if p ≡ 3 (mod 4).


(−1)|{16k<p/10: ( kp )=−1}|(p−1)/2∏i,j=1p-5i+j

(5i+ j

p

)

=

(−1)b(p+1)/10c if p ≡ ±1,±3 (mod 20),

(−1)bp/20c if p ≡ ±7 (mod 20),

(−1)b(p+9)/20c if p ≡ ±9 (mod 20),

and

(p−1)/2∏i,j=1p-5i−j

(5i− jp

)=

(−1)(h(−p)+1)/2 if p ≡ 3, 7 (mod 20),

(−1)(p+9)/20 if p ≡ −9 (mod 20),

(−1)|{16k<p/10: ( kp )=−1}|+(h(−p)−1)/2 if p ≡ −1 (mod 20),

(−1)|{16k<p/10: ( kp )=−1}|+b(p+3)/20c if p ≡ 1,−3 (mod 20),

(−1)|{16k<p/10: ( kp )=−1}|+b(p−3)/10c if p ≡ −7, 9 (mod 20).

Conjecture 7.9. For any prime p > 3, we have

(p−1)/2∏i,j=1p-6i+j

(6i+ j

p

)=

(−1)|{16k<p/12: ( kp )=−1}| if p ≡ 1 (mod 24),

(−1)|{p+3

4 6k6b p+13 c: ( kp )=−1}| if p ≡ 5,−7,−11 (mod 24),

(−1)(h(−p)+1)/2+b(p+1)/24c if p ≡ −1,−5 (mod 24),

(−1)bp/24c−1 if p ≡ 7, 11 (mod 24),


(6i− jp

)= (−1)|{

p+24 <k< p

3 : ( kp )=1}|.

22 Z.-W. SUN


(−1)|{16k<p/4: ( kp )=−1}|(p−1)/2∏i,j=1p-8i+j

(8i+ j

p

)=

{(−1)(p+1)/8 if p ≡ −1 (mod 8),

1 otherwise,


(8i− jp

)=

(−1)|{16k<p/4: ( kp )=1}| if p ≡ 1 (mod 4),

(−1)(h(−p)+1)/2+(p−3)/8 if p ≡ 3 (mod 8),

−1 if p ≡ 7 (mod 8).

Acknowledgments. The research is supported by the Natural Science Foundationof China (Grant No. 11971222). The author would like to thank the the referee forhelpful comments.

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Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republic

of China

E-mail address: [email protected]

quadratic residues and quartic residues modulo primesmaths.nju.edu.cn/~zwsun/205i.pdf · 2020. 9....

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