Download Some sophisticated congruences involving Fibonacci numbers

A talk given at the National Center for Theoretical Sciences
(Hsinchu, Taiwan; July 20, 2011)
and Shanghai Jiaotong University (Nov. 4, 2011)
Some sophisticated congruences involving
Fibonacci numbers
Zhi-Wei Sun
Nanjing University
Nanjing 210093, P. R. China
[email protected]
http://math.nju.edu.cn/∼zwsun
Nov. 4, 2011
Abstract
The well-known Fibonacci numbers play important roles in many
areas of mathematics and they have very nice number-theoretic
properties. We will focus on some sophisticated congruences on
Fibonacci numbers including the recent determination of Fp−( p )
5
modulo p 3 , where p is an odd prime. We will also mention some
conjectures related to 5-adic valuations for further research.
2 / 42
Part A. On Fibonacci quotients and Wall-Sun-Sun primes
3 / 42
Fibonacci numbers and Lucas numbers
The Fibonacci sequence {Fn }n>0 is defined by
F0 = 0, F1 = 1, Fn+1 = Fn + Fn−1 (n = 1, 2, 3, . . .).
Thus
F0 = 0, F1 = F2 = 1, F3 = 2, F4 = 3, F5 = 5, F6 = 8,
F7 = 13, F8 = 21, F9 = 34, F10 = 55, F11 = 89, F12 = 144.
The Lucas numbers L0 , L1 , L2 , . . . are given by
L0 = 2, L1 = 1, Ln+1 = Ln + Ln−1 (n = 1, 2, 3, . . .).
Relations between Fibonacci numbers and Lucas numbers:
Ln = Fn−1 + Fn+1 and 5Fn = Ln−1 + Ln+1 .
4 / 42
More known results on Fibonacci numbers
An explicit expression.
b(n−1)/2c Fn =
X
k=0
n−1−k
.
k
Y. Bugeaud, M. Mignotte and S. Siksek [Ann. of Math.
163(2006)]: The only powers in the Fibonacci sequence are
F0 = 0, F1 = F2 = 1, F6 = 23 , and F12 = 122 .
Ke-Jian Wu and Z. W. Sun [Math. Comp. 78(2009)]: Let
a =312073868852745021881735221320236651673651
93670823768234185354856354918873864275
and
M =368128524439220711844024989130760705031462
29820861211558347078871354783744850778.
Then, for any x ≡ a (mod M), the number x 2 cannot be written
in the form Fn /2 ± p m with p a prime and m, n ∈ N = {0, 1, . . .}.
5 / 42
Lucas sequences
Let N = {0, 1, 2, . . .} and Z+ = {1, 2, 3, . . .}.
Fix A, B ∈ Z. The Lucas sequence un = un (A, B) (n ∈ N) and its
companion vn = vn (A, B) (n ∈ N) are defined as follows:
u0 = 0, u1 = 1, and un+1 = Aun − Bun−1 (n = 1, 2, 3, . . .);
v0 = 2, v1 = A, and vn+1 = Avn − Bvn−1 (n = 1, 2, 3, . . .).
Example : Fn = un (1, −1), Ln = vn (1, −1); un (2, 1) = n, vn (2, 1) = 2.
Universal Formulae: Let ∆ = A2 − 4B, and let
√
√
A+ ∆
A− ∆
α=
and β =
2
2
2
be the two roots of the equation x − Ax + B = 0. Then
( n n
α −β
X
if ∆ 6= 0,
k n−1−k
α−β
un =
α β
=
n(A/2)n−1 if ∆ = 0.
06k<n
And
vn = α n + β n .
6 / 42
Basic properties of Lucas sequences
As (αn − β n )(β n + β n ) = α2n − β 2n , we have
u2n = un vn .
Note also that
vn2 − ∆un2 =(αn + β n )2 − ((α − β)un )2
=(αn + β n )2 − (αn − β n )2 = 4(αβ)n = 4B n .
Lucas’ Theorem. Let A, B ∈ Z with (A, B) = 1. Then
(um , un ) = |u(m,n) |.
In particular,
m | n ⇒ um | un .
Example. F2n = Fn Ln , L2n − 5Fn2 = 4(−1)n , (Fm , Fn ) = F(m,n) .
7 / 42
Legendre symbols
Let p be an odd prime
given by

if p

0
a
= 1
if p

p

−1 if p
and a ∈ Z. The Legendre symbol ( pa ) is
| a,
- a and x 2 ≡ a (mod p) for some x ∈ Z,
- a and x 2 ≡ a (mod p) for no x ∈ Z.
a b
It is well known that ( ab
p ) = ( p )( p ) for any a, b ∈ Z. Also,
(
1
if p ≡ 1 (mod 4),
−1
= (−1)(p−1)/2 =
p
−1 if p ≡ −1 (mod 4);
(
1
if p ≡ ±1 (mod 8),
2
(p 2 −1)/8
= (−1)
=
p
−1 if p ≡ ±3 (mod 8);
(
1
if p ≡ ±1 (mod 5),
p
5
=
=
p
5
−1 if p ≡ ±2 (mod 5).
8 / 42
Congruence properties of Lucas sequences
Let p be an odd prime. Then
∆
up ≡
(mod p) and vp ≡ A (mod p).
p
In fact,
vp = αp + β p ≡ (α + β)p = Ap ≡ A (mod p);
also
∆up = (α − β)(αp − β p ) ≡ (α − β)p+1 = ∆(p+1)/2
(mod p)
2
and hence up ≡ ( ∆
p ) (mod p) if p - ∆. When ∆ = A − 4B ≡ 0
(mod p), we have
p−1 A2
A
∆
up ≡ up A,
=p
≡
(mod p).
4
2
p
9 / 42
Congruence properties of Lucas sequences
Theorem. If p is an odd prime not dividing B, then
up−( ∆ ) ≡ 0
(mod p).
(∗)
p
Proof. If p | ∆, then up−( ∆ ) = up ≡ ( ∆
p ) = 0 (mod p).
p
If ( ∆
p ) = −1, then
up−( ∆ ) = up+1 = Aup + vp ≡ A
p
∆
p
+A=0
(mod p).
If ( ∆
p ) = 1, then
up−( ∆ ) = up−1
p
and
2Bup−1 = Aup − vp ≡ A
∆
p
−A=0
(mod p).
10 / 42
Congruence properties of Lucas sequences
The congruence p | up−( ∆ ) is actually an extension of Fermat’s
p
little theorem. If a 6≡ 0, 1 (mod p), then
ap−1 − 1
= up−1 (a+1, a) = u (a+1)2 −4a (a+1, a) ≡ 0
p−
a−1
p
Example. If p is an odd prime, then
5
p
(mod p), Lp ≡ 1
Fp ≡
=
p
5
(mod p).
(mod p),
and
Fp−( p ) ≡ 0
5
(mod p).
For an odd prime p, we call the integer Fp−( p ) /p a Fibonacci
5
quotient.
11 / 42
D. D. Wall’s question
In 1960 D. D. Wall [Amer. Math. Monthly] studied Fibonacci
numbers modulo a positive integer systematically.
Let p be an odd prime and let n(p) be the smallest positive integer
n with p | Fn . Can Fn(p) be a multiple of p 2 ? Wall found no such
primes.
One can show that any positive integer d divides some positive
Fibonacci numbers. Let n(d) be the smallest positive integer such
that d | Fn(d) . Then Wall’s question is whether n(p) 6= n(p 2 ) for
any odd prime p.
12 / 42
Zhi-Hong Sun and Zhi-Wei Sun’s contribution
Theorem (Zhi-Hong Sun and Zhi-Wei Sun [Acta Arith.
60(1992)]). Let p 6= 2, 5 be a prime.
(i) We have n(p) = n(p 2 ) ⇐⇒ p 2 | Fp−( p ) .
5
(ii) We have
Fp−( p )
5
p
≡2
p−1
X
k=1
k≡−p (mod 5)
1
≡ −2
k
p−1
X
k=1
k≡2p (mod 5)
1
k
(mod p).
(iii) If p 2 - Fp−( p ) , then the first case of Fermat’s last theorem
5
holds for the exponent p, i.e., there are no positive integers x, y , z
with p - xyz such that x p + y p = z p .
(iv) If p ≡ 1, 9 (mod 20) and p = x 2 + 5y 2 with x, y ∈ Z, then
p | F(p−1)/4 ⇐⇒ 4 | xy .
13 / 42
Zhi-Hong Sun and Zhi-Wei Sun’s contribution
(v) We can determine F(p±1)/2 and L(p±1)/2 mod p in the
following way:
(
0 (mod p)
if p ≡ 1 (mod 4),
F(p−( p ))/2 ≡
5 (p−3)/4
b(p+5)/10c
5
2(−1)
( p )5
(mod p) if p ≡ 3 (mod 4);
F(p+( p ))/2 ≡
5
L(p−( p ))/2 ≡
5
L(p+( p ))/2
5
(
(−1)b(p+5)/10c ( p5 )5(p−1)/4 (mod p)
if p ≡ 1 (mod 4),
(−1)b(p+5)/10c ( p5 )5(p−3)/4
(mod p) if p ≡ 3 (mod 4);
(
2(−1)b(p+5)/10c ( p5 )5(p−1)/4 (mod p) if p ≡ 1 (mod 4),
0 (mod p)
if p ≡ 3 (mod 4);
(
(−1)b(p+5)/10c 5(p−1)/4 (mod p)
if p ≡ 1 (mod 4),
≡
5 (p+1)/4
b(p+5)/10c
(−1)
( p )5
(mod p) if p ≡ 3 (mod 4).
The
obtained via expressing the sum
P theorem was
n
k≡r (mod 10) k in terms of Fibonacci and Lucas numbers.
14 / 42
Wall-Sun-Sun primes
R. Crandall, K. Dilcher and C. Pomerance [Math. Comp.
66(1997)] called those primes p satisfying Fp−(p/5) ≡ 0 (mod p 2 )
Wall-Sun-Sun primes. Heuristically there should be infinitely many
Wall-Sun-Sun primes though they are very rare.
Wall-Sun-Sun primes were also introduced in many papers and
books including the famous book R. E. Cradall and C. Pomerance,
Prime Numbers: A Computational Perspective, Springer, 2001.
Up to now no Wall-Sun-Sun primes have been found.
The current search record is due to F. G. Dorais and D. W. Klyve
(2010): There are no Wall-Sun-Sun primes below 9.7 × 1014 .
15 / 42
Connection to cyclotomic fields
S. Jakubec [Math. Comp. 67(1998)]: Let p = 2l + 1 ≡ 7
(mod 8) and q be odd primes with l a prime and p ≡ −5 (mod q).
Suppose that the order of q modulo l is (l − 1)/2. If q divides the
class number of the real cyclotomic field Q(ζp + ζp−1 ), then q must
be a Wall-Sun-Sun prime.
16 / 42
Bernoulli polynomials and Fibonacci quotients
Bernoulli numbers B0 , B1 , B2 , . . . are given by
n X
n+1
B0 = 1,
Bk = 0 (n = 1, 2, 3, . . .).
k
k=0
Bernoulli polynomials are those
n X
n
Bn (x) =
Bk x n−k (n = 0, 1, 2, . . .).
k
k=0
Theorem (A. Granville and Z. W. Sun [Pacific J. Math. 1996]).
Let p 6= 2, 5 be a prime. Then
ap 5 Fp−( p ) 5
a
5
− Bp−1 ≡
·
+ qp (5) (mod p)
Bp−1
5
5 4
p
4
for a = 1, 2, 3, 4, and for a = 1, 3, 7, 9 we have
a
ap 15 Fp−( p ) 5
5
Bp−1
−Bp−1 ≡
·
+ qp (5)+2qp (2) (mod p),
10
5
4
p
4
where qp (m) denotes the Fermat quotient (mp−1 − 1)/p.
17 / 42
Part B. On Fp−( p5 ) mod p 3 and
some super congruences involving Fibonacci numbers
18 / 42
A curious identity and its consequence
H. Pan and Z. W. Sun [Discrete Math. 2006]. If
l, m, n ∈ {0, 1, 2, . . .} then
l
X
l
m−k
2k
(−1)
k
n
k − 2l + m
k=0
l X
l
2k
n−l
=
.
k
n
m + n − 3k − l
m−k
k=0
On the basis of this identity, for d, r ∈ {0, 1, 2, . . .} the authors
constructed explicit F (d, r ) and G (d, r ) such that for any prime
p > max{d, r } we have
(
p−1
X
F (d, r ) (mod p) if p ≡ 1 (mod 3),
r
k Ck+d ≡
G (d, r ) (mod p) if p ≡ 2 (mod 3),
k=1
where Cn denotes the Catalan number
1 2n
n+1 n
.
19 / 42
Some congruences involving central binomial coefficients
Let p be an odd prime.
H. Pan and Z. W. Sun [Discrete Math. 2006].
p−1 X
2k
p−d
≡
(mod p) (d = 0, . . . , p),
k +d
3
k=0
p−1 2k X
k
≡0 (mod p) for p > 3.
k
k=1
Sun & R. Tauraso [AAM 45(2010); IJNT 7(2011)].
a −1 a
pX
2k
p
≡
(mod p 2 ),
k
3
k=0
p−1 2k X
8
k
≡ p 2 Bp−3 (mod p 3 ) for p > 3,
k
9
k=1
p−1
X
Fp−( p )
(−1)k 2k
5
≡−5
(mod p) (p 6= 5).
k
k
p
k=1
20 / 42
Some auxiliary identities
Sun and Tauraso [Adv. in Appl. Math. 2010]: Let n ∈ Z+ and
d ∈ N. Then
X 2k d X 2n n + d − k +
=
,
k +d
k
3
3
06k<n
X
06k<n
06k<n+d
k+d
(−1)
2k
k +d
+ F2d =
X
06k<n+d
k
(−1)
2n
F2(n+d−k) ,
k
and
X 2n
X (−1)k+d 2k d
+
(−1)k L2(n+d−k)
k
k +d
k
0<k<n
06k<n+d
2n − 1
n+d
= L2d − (−1)
2
− δd,0 .
n+d −1
21 / 42
On
Pp−1
2k
k=0 k
/mk mod p 2
Z. W. Sun [Sci. China Math. 53(2010)]: Let p be an odd prime
and let m ∈ Z with p - m. Then
p−1
X
k=0
2k
k
mk
≡
m2 − 4m
p
+ up−( m2 −4m ) (m − 2, 1) (mod p 2 ).
p
In particular,
p−1
X
k=0
p−1
X
k=0
(−1)k
2k
k
2k
2k
k
≡(−1)(p−1)/2 (mod p 2 ),
≡
p 5
1 − 2Fp−( p )
5
(mod p 2 ) (p 6= 5).
(Note that (−1)n−1 un (−3, 1) = un (3, 1) = F2n = Fn Ln .)
22 / 42
Two conjectures on Fibonacci and Lucas numbers
Conjecture (Sun and Tauraso [Adv. in Appl. Math. 2010]) Let
p 6= 2, 5 be a prime and let a ∈ Z+ . Then
a −1
pX
k=0
(−1)k
2k
k
≡
pa
5
1 − 2Fpa −( pa )
(mod p 3 ).
5
Conjecture (Roberto Tauraso, Jan. 2010). For any prime p > 5
we have
p−1
X
Lk
≡ 0 (mod p).
k2
k=1
These two conjectures are very sophisticated and difficult to prove.
Theorem (Hao Pan and Z. W. Sun, arXiv:1010.2489). The two
conjectures are true!
23 / 42
Another congruence for Fp−( p5 ) mod p 3
Theorem (Z. W. Sun, arXiv:0911.3060) Let p 6= 2, 5 be a prime
and let a ∈ Z+ . Then
!
a
(p a −1)/2
2k
Fpa −( pa )
X
p
k
5
≡
1+
(mod p 3 ).
5
2
(−16)k
k=0
The proof is also very sophisticated and quite difficult. Below is a
key lemma.
Lemma. Let p 6= 2, 5 be an odd prime. For any a ∈ Z+ we have
a
Lpa − 1
p
1
−
Fpa + 1 ≡ − Fp2a −( pa ) (mod p 4 ).
5
5
2
5
P
(2)
If we set Hk = 0<j6k 1/j 2 for k = 0, 1, 2, . . ., then
!
p−1
p 5 Fp−( p ) 2
X
2k
(2)
5
(−1)k
Hk ≡
(mod p).
k
5 2
p
k=0
24 / 42
A further conjecture
Conjecture (Sun, 2010). Let p be an odd prime and let a ∈ Z+ .
(i) If p a ≡ 1, 2 (mod 5), or a > 1 and p 6≡ 3 (mod 5),
b4p a /5c
X
5
k 2k
(−1)
(mod p 2 ).
≡
pa
k
k=0
If
pa
≡ 1, 3 (mod 5), or a > 1 and p 6≡ 2 (mod 5), then
b3p a /5c
X
5
k 2k
(−1)
(mod p 2 ).
≡
pa
k
k=0
(ii) If p ≡ 1, 7 (mod 10) or a > 2, then
b7p a /10c
2k
X
5
k
≡
(mod p 2 ).
k
pa
(−16)
k=0
If p ≡ 1, 3 (mod 10) or a > 2, then
b9p a /10c
2k
X
5
k
≡
(mod p 2 ).
k
pa
(−16)
k=0
25 / 42
Some other congruences involving Fibonacci numbers
Theorem (Sun). For any prime p > 5, we have


if p ≡ ±1 (mod 5),
p−1
0 (mod p)
X Fk 2k
≡ 1 (mod p)
if p ≡ ±13 (mod 30),

12k k

k=0
−1 (mod p) if p ≡ ±7 (mod 30).
Theorem (Sun). Let p 6= 2, 5 be a prime. Then
p−1
p X
2k
(mod p 2 ),
F2k
≡(−1)bp/5c 1 −
5
k
k=0
p−1
p X
2k
F2k+1
≡(−1)bp/5c
(mod p 2 ),
k
5
k=0
(p−1)/2
X F2k 2k ≡(−1)(p−1)/2+bp/5c (mod p 2 ),
16k k
k=0
(p−1)/2
p
X F2k+1 2k (p−1)/2+bp/5c 5 + ( 5 )
≡(−1)
(mod p 2 ).
k
4
16k
k=0
26 / 42
A conjecture on 5-adic valuations
Conjecture (Sun). For any n ∈ Z+ the number
n−1
2k
(−1)bn/5c−1 X
F2k+1
sn :=
2n
2
k
(2n + 1)n n k=0
is a 5-adic integer and furthermore


6 (mod 25) if



4 (mod 25) if
sn ≡

1 (mod 25) if



9 (mod 25) if
n ≡ 0 (mod 5),
n ≡ 1 (mod 5),
n ≡ 2, 4 (mod 5),
n ≡ 3 (mod 5).
Also, if a, b ∈ Z+ and a > b then the sum
5a −1
1 X
2k
F2k+1
2a
5
k
k=0
modulo 5b only depends on b.
27 / 42
A conjecture on q-Fibonacci numbers
Recall that the usual q-analogue of n ∈ N is given by
X
1 − qn
=
qk
[n]q =
1−q
06k<n
which tends to n as q → 1. For any n, k ∈ N with n > k,
Q
n
0<r 6n [r ]q
Q
= Q
k q
( 0<s6k [s]q )( 0<t6n−k [t]q )
is a natural extension of the usual binomial coefficient kn . A
q-analogue of Fibonacci numbers introduced by I. Schur is defined
as follows:
F0 (q) = 0, F1 (q) = 1, and Fn+1 (q) = Fn (q) + q n Fn−1 (q) (n > 0).
Conjecture (Sun) Let a and m be positive integers. Then we have
the following congruence in the ring Z[q]:
5aX
m−1
2k
q −2k(k+1)
F
(q) ≡ 0 (mod [5a ]2q ).
k q 2k+1
k=0
28 / 42
Four series involving Fibonacci and Lucas numbers
In Oct. 2010 Sun observed the following identities:
∞
X
L2k
4π 2
π2
√ ,
,
=
2k
2
2
5
25
5
k
k
k
k
k=1
k=1
∞
∞
2k
2k
X
X
2π
2π
k F2k+1
k L2k+1
= √ ,
=
.
k
5
(2k + 1)16k
(2k
+
1)16
5 5
∞
X
F2k
=
2k
k=0
k=0
√
In fact, they can be obtained by putting x = ( 5 ± 1)/2 in the
known identities
∞
∞
2k
x 2k+1
X
X
x
x
x 2k
k
arcsin =
and
= 2 arcsin2 .
2k
k
2
2
(2k + 1)4 2
k2 k
k=0
k=1
Note that
π
sin
=
10
√
5−1
3π
and sin
=
4
10
√
5+1
.
4
29 / 42
Corresponding conjectural congruences
Conjecture (Sun, 2010). Let p 6= 2, 5 be a prime and set
q := Fp−( p ) /p. Then
5
p
p
(p−3)/2
p−1
X
F2k
k2
k=1
2k
k
p−1
X
L2k
k 2 2k
k
k=1
2k
k
≡−
p 3
5
5
q + p q2
2
4
(mod p 2 ),
15
5
≡ − q − p q 2 (mod p 2 ),
2
4
1
F2k+1
5 2
(p+1)/2 p
≡(−1)
q+ pq
(mod p 2 ),
5
2
8
(2k + 1)16k
k=0
(p−3)/2
2k
X
5 2
(p+1)/2 5
k L2k+1
≡(−1)
q+ pq
(mod p 2 ).
2
8
(2k + 1)16k
X
k=0
In Oct. 2011 K. Hessami Pilehrood and T. Hessami Pilehrood
[arXiv:1110.5308] proved the last two congruences.
30 / 42
A conjecture related to p = x 2 + 15y 2 and p = 3x 2 + 5y 2
Conjecture (Z. W. Sun, Sept. 18, 2011). Let p > 5 be a prime.
(i) If p ≡ 1, 4 (mod 15) and p = x 2 + 15y 2 (x, y ∈ Z) with
x ≡ 1 (mod 3), then
p−1
X
k
k=0
p−1
X
k=0
and
p−1
X
k=0
2k
k
3k
k
27k
2k 3k
k
k
27k
(3k + 2)
Fk ≡
2 p
− 2x (mod p 2 ),
15 x
Lk ≡4x −
2k
k
3k
k
27k
p
(mod p 2 )
x
Lk ≡ 4x
(mod p 2 ).
31 / 42
A conjecture related to p = x 2 + 15y 2 and p = 3x 2 + 5y 2
(ii) If p ≡ 2, 8 (mod 15) and p = 3x 2 + 5y 2 (x, y ∈ Z) with
y ≡ 1 (mod 3), then
p−1
X
2k
k
3k
k
27k
k=0
Fk ≡
p
− 4y
5y
(mod p 2 )
and
p−1
X
k
k=0
2k
k
27k
3k
k
Fk ≡
p−1
X
k
k=0
2k
k
27k
3k
k
4
Lk ≡ y
3
(mod p 2 ).
Remark. Sun has many other similar conjectures.
32 / 42
Part C. Proof of
Pp−1 Lk
k=1 k 2
≡ 0 (mod p) for any prime p > 5
33 / 42
Granville’s work
Let p be an odd prime. Glaisher proved that
p−1
1 X 2k
qp (2) ≡ −
2
k
(mod p).
k=1
A. Granville [Integers 4(2004)] confirmed the following conjecture
of L. Skula:
p−1 k
X
2
qp (2)2 ≡ −
(mod p).
k2
k=1
Define
p−1
X xk
x p + (1 − x)p − 1
and G (x) =
.
q(x) =
p
k2
k=1
Granville showed that if p > 3 then
G (x) ≡G (1 − x) + x p G (1 − x −1 ) (mod p),
q(x)2 ≡ − 2x p G (x) − 2(1 − x p )G (1 − x) (mod p).
34 / 42
A lemma and a proposition
Combining the last two congruences we obtain
Lemma. Let p > 3 be a prime. Then
x p + (1 − x)p − 1
p
2
≡ −2
p−1
X
(1 − x)k
k=1
k2
−2x 2p
p−1
X
(1 − x −1 )k
k=1
k2
(mod p)
Proposition. Let A and B be nonzero integers, and let α and β be
the two roots of the equation x 2 − Ax + B = 0. Let p be an odd
prime not dividing AB. Then
vp (A, B) − Ap
p
2
≡ −2A2
p−1
p−1
X
X
αk
α2k
2p
−2β
(mod p),
Ak k 2
(−B)k k 2
k=1
k=1
and
p−1 k k
p−1
X
X
vp (A, B) − Ap 2
αk
A α
p
2p
≡ −2Aα
− 2β
(mod p).
k
2
p
A k
Bkk2
k=1
k=1
35 / 42
Proof of the proposition
By the lemma and Fermat’s little theorem,
1 x p + (A − x)p − Ap 2
A2
p
p
(x/A) + (1 − x/A)p − 1 2
≡
p
p−1
p−1
x 2p X
X (1 − x/A)k
(1 − A/x)k
−
2
(mod p).
≡−2
k2
A
k2
k=1
k=1
βp
αp
Note that vp (A, B) =
+
=
+ (A − β)p and αβ = B. So
we have
p−1
X
vp (A, B) − Ap 2
(A − β)k
2
≡ − 2A
p
Ak k 2
− 2β
βp
k=1
p−1
X
2p
k=1
(1 − Aα/B)k
(mod p)
k2
and hence the first desired congruence holds since Aα − B = α2 .
36 / 42
Proof of the proposition (continued)
On the other hand,
αp (Ap − vp (A, B)) = αp (Ap − αp − β p ) = (B + α2 )p + (−α2 )p − B p
and hence
p
2
2p A − vp (A, B)
α
p
2
2
p
(−α ) + (B − (−α2 ))p − B p
=
p
≡ − 2B
= − 2B
2
p−1
X
(1 − (−α2 )/B)k
2
k=1
p−1
X
k=1
≡ − 2(αβ)2p
k2
2 2p
− 2(−α )
p−1
X
(1 − B/(−α2 ))k
k=1
k2
p−1
X (Aα)k
(Aα)k
4p
−
2α
Bkk2
α2k k 2
k=1
p−1
X
k=1
(Aα)k
Bkk2
− 2Aα3p
p−1
X
k=1
αp−k
(mod p).
Ap−k (p − k)2
Therefore the second desired congruence follows.
37 / 42
Proof of
Pp−1
k=1 Lk /k
2
≡ 0 (mod
p)
P
p−1
Let p > 5 be a prime. We prove k=1
Lk /k 2 ≡ 0 (mod p).
Let α and β be the two roots of the equation x 2 − x − 1 = 0.
Applying the Proposition with A = 1 and B = −1, we get
p−1 k
p−1 2k
X
X
Lp − 1 2
α
α
2p
≡−2
− 2β
(mod p),
(1)
p
k
k2
k=1
k=1
p−1 k
p−1
X
X
Lp − 1 2
α
(−α)k
2p
≡ − 2αp
−
2β
(mod p). (2)
p
k2
k2
k=1
k=1
2p−1
p−1
j
X
X
2α2k
j α
=
(1
+
(−1)
)
(2k)2
j2
k=1
j=1
p−1 k
X
α + (−α)k
αp+k + (−α)p+k
=
+
k2
(p + k)2
k=1
p
≡(1 + α )
p−1 k
X
α
k=1
k2
p
+ (1 − α )
p−1
X
(−α)k
k=1
k2
(mod p),
38 / 42
Proof of
Pp−1
k=1 Lk /k
2
≡ 0 (mod p)
so (1) can be rewritten as
Lp − 1
p
2
p
2p
≡ − 2(1 + 2(1 + α )β )
p−1 k
X
α
k=1
p
− 4(1 − α )β
2p
p−1
X
(−α)k
k=1
k2
k2
(mod p).
(3)
Multiplying (2) by 2(1 − αp ) and then subtracting it from (3) we
obtain
p
(2α − 1)
Lp − 1
p
2
p−1 k
X
α
≡ 4α (1 − α ) − 2 − 4(1 + α )β
k2
p
p
p
2p
k=1
=(4Lp − 4L2p − 2)
p−1
X
k=1
αk
k2
(mod p).
39 / 42
Proof of
Pp−1
k=1 Lk /k
2
≡ 0 (mod p)
Now that Lp ≡ 1 (mod p) and
L2p = α2p +β 2p ≡ (α2 +β 2 )p = (α + β)2 − 2αβ
p
= 3p ≡ 3 (mod p),
we have
p
(2α −1)
Lp − 1
p
2
≡ (4−4×3−2)
p−1 k
X
α
k=1
k2
= −10
p−1 k
X
α
k=1
k2
(mod p).
Similarly,
p
(2β − 1)
Lp − 1
p
2
≡ −10
p−1 k
X
β
k=1
k2
(mod p).
As 2αp − 1 + (2β p − 1) = 2Lp − 2 ≡ 0 (mod p), we finally obtain
p−1
X
Lk
k=1
k2
=
p−1 k
X
α + βk
≡ 0 (mod p).
k2
k=1
40 / 42
More conjectures on congruences
For more conjectures of mine on congruences, see
Z. W. Sun, Open Conjectures on Congruences, arXiv:0911.5665
which contains 100 unsolved conjectures raised by me.
You are welcome to solve my
conjectures!
41 / 42
Thank you!
42 / 42