in the group Zp for any Lucas number Lk. Similar argument and using the identity : Lkn. LkL!nL%"k - $!kL!nL&"k we prove the following theorem : Theorem3 : If p ...
Primes And Some Lucas Identities Issam Kaddoura , Ghadir Makke Lebanese International University School Of Art And Sciences Abstract In this Paper , We prove Some identities including Lucas numbers and primes . Lucas Identities And Trinomial Coe¢ cients X Using the Trinomial expansion (x+y +z)n =
n! i j k i!j!k! x y z
we prove
i+j+k=n
the following lemma Lemma 1 : If n is any prime , then (x + y + z)n xn y n z n 0 mod n(x + y)(x + y)(y + z) 8x; y; z 2 Z Using the well known identity (L2k+1 ) (Lk+1 Lk ) the previous Lemma we prove: Theorem1 : If n is a prime number ,then (L2k+1 )p
(Lk+1 Lk )p
k+1
(L2k ) = 5 ( 1)
k+1 p
(L2k )p
) mod pL2k+2 Lk+1 L2k
(5 ( 1)
1 Lk
and
:
for any Lucas number Lk . Theorem2 : If p is a prime number ,then (Lk+1 )p (Lk )p (Lk Lk+1 Lk Lk 1
p 1)
=0
in the group Zp for any Lucas number Lk . Similar argument and using the identity : Lkn Lk L(n we prove the following theorem : Theorem3 : If p is a prime number ,then (Lkn )p
(Lk L(n
p 1)k )
( 1)k (L(n
p 2)k )
mod pLkn L(n
1)k
= ( 1)k L(n
1)k
L(n
2)k
2)k Lk
where Lk is any Lucas number . Next we utilize the well known identities s
L2s = 5(Fs )2 + 4 ( 1) ; Ln =
Ln
m Lm
+ 5Fm Fn 2
m
to prove the main theorem in this paper. Theorem 4. p is an odd prime factor of the integer m if and only if Lm = L mp K((L mp )2 ) where Ln is the Lucas number and K is a polynomial with the following properties : 1 .K has integeral coe¢ cients with degree (p-1)/2 and jK(0)j = p: 2. K is an irreducible polynomial over the …eld of rational numbers .
1
3. All the coe¢ cients of K are divisible by the prime p except the coe¢ cient of the higest power (p-1)/2 term is equal to 1:As a consequence of the previous theorem we prove : Corrolary : If p is any prime divides m , then Lm
(L mp )p Lp
(L mp ) mod pL mp
(1)
1 mod p :
(2)
The following Lucas identities illustrate the previous theorem: s L2s =L2s 2 ( 1) (1) s 2 L3s = Ls 3 ( 1) Ls (2) s 2 4 L4s =Ls 4Ls ( 1) +2 (3) s L5s = L4s 5L2s ( 1) +5 Ls (4) 4 s s 2 2 L6s =(Ls 4Ls ( 1) +1) Ls 2 ( 1) (5) s s 4 2 6 L7s = Ls 7Ls ( 1) +14Ls 7 ( 1) Ls (6) s s 6 2 4 8 L8s =Ls 8Ls ( 1) 16Ls ( 1) +20Ls +2 (7) s s s L2s 3 ( 1) Ls (8) L9s = L6s 6L4s ( 1) +9L2s 3 ( 1) s s s 8 (9) L10s =(Ls 8L6s ( 1) +19L4s 12L2s ( 1) +1) L2s 2 ( 1) s s s 10 2 6 4 8 (10) L11s =(Ls 11Ls ( 1) 77Ls ( 1) +44Ls +55Ls 11 ( 1) )Ls s s s 2 4 6 8 (11) 12L10 L12s =L12 s ( 1) +54Ls 112Ls ( 1) +105Ls 36Ls ( 1) +2 s s s s 12 2 4 6 8 (12) L13s =(Ls 13L10 s ( 1) +65Ls 156Ls ( 1) +182Ls 91Ls ( 1) +13)Ls F or p = 13; m = 13s 12
where K(x) = x 13x5 ( 1) +65x4 156x ( 1) +182x2 91x ( 1) +13 is irreducible polynomila and all coe¢ cients except the …rst one of K(x) satisfy 13 65 156 182 91 13 0 mod 13 .
