A new proof of Lucas' Theorem

0.1 MB·English

by  Alexandre Laugier

A NEW PROOF OF LUCAS’ THEOREM
ALEXANDRELAUGIERANDMANJILP.SAIKIA
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Abstract. Wegiveanew proofofLucas’Theoreminelementarynumbertheory.
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KeyWords:Lucas’theorem.
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2010Mathematical ReviewsClassificationNumbers: 11A07,11A41,11A51,11B50,11B65,11B75.
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1. Introduction

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T One of the most useful results in elementary number theory is the following result of E. Lucas
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Theorem 1.1 ([1],E. Lucas (1878)). Let p be a prime and m and n be two integers considered in the
.
h following way,
at m=akpk+ak−1pk−1+...+a1p+a0,
m
n=blpl+bl−1pl−1+...+b1p+b0,
[
where all a and b are non-negative integers less than p. Then,
i j
1
0v m = akpk+ak−1pk−1+...+a1p+a0 ≡max(k,l) ai (mod p).
5 (cid:18)n(cid:19) (cid:18) blpl+bl−1pl−1+...+b1p+b0 (cid:19) iY=0 (cid:18)bi(cid:19)
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4 Notice that the theorem is true if ai ≥bi for i=0,1,2,...,max(k,l)
.
1 There has been many different proofs of this result in the years that followed its first publication.
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We present here an alternate approachusing elementary number theoretic techniques.
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: 2. Proof of Theorem 1.1
v
i
X First of all, we state and prove a few lemmas
r Lemma 2.1. If
a
a +a X +a X2+...+a Xn ≡b +b X +b X2+...+b Xn (mod p)
0 1 2 n 0 1 2 n
then
a ≡b (mod p) ∀i∈[[0,n]].
i i
Proof. Indeed, if a +a X +a X2 +···+a Xn ≡ b +b X +b X2 +···+b Xn (mod p), then
0 1 2 n 0 1 2 n
there exits a polynomial k(X) = k + k X + k X2 + ··· + k Xn at most of degree n such that
0 1 2 n
a +a X+a X2+···+a Xn =b +b X+b X2+···+b Xn+p(k +k X+k X2+···+k Xn).
0 1 2 n 0 1 2 n 0 1 2 n
Thisgivesa +a X+a X2+···+a Xn =b +pk +(b +pk )X+(b +pk )X2+···+(b +pk )Xn.
0 1 2 n 0 0 1 1 2 2 n n
Hence we get a = b +pk , a = b +pk , a = b +pk ,..., a = b +pk . Or equivalently
0 0 0 1 1 1 2 2 2 n n n
a ≡b (mod p), a =b (mod p), a =b (mod p) ,···, a =b (mod p).
0 0 1 1 2 2 n n
The reciprocal implication is trivial. (cid:3)
ThesecondauthorissupportedbyDSTINSPIREScholarship422/2009fromDepartmentofScienceandTechnology,
GovernmentofIndia.
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Lemma 2.2. If the base p expansion of a positive integer n is,
n=a +a p+a p2+···+a pl
0 1 2 l
then we have
n!=qa !(a p)!(a p2)!···(a pl)!
0 1 2 l
with q a natural number.
Proof. Since the factorial of a natural number is a natural number, there exists a rational number q
such that
n!
q = .
a !(a p)!(a p2)!...(a pl)!
0 1 2 l
Let S be the set,
S ={x ,x ,...,x }.
1 2 l
We consider lists of elements of S where x is repeated a times, x is repeated a p times,..., x is
0 0 1 1 l
repeateda pl timessuchthat,0≤a ≤p−1withi∈[[0,l]]. Insuchalist,therearel+1unlikegroups
l i
of identical elements. For instance the selection x ,x ,...,x ,x ,x ,...,x ,......,x ,x ,...,x 
0 0 0 1 1 1 l l l
 a0 a1p alpl 
is such a list of n elements which contains l+1 unl|ike gr{ozups o}f i|dentic{azl elem}ents. | {z }
The number of these lists is given by n! . It proves that the rational number q is
a0!(a1p)!(a2p2)!...(alpl)!
a natural number. And, since the factorial of a natural number is non-zero (even if this number is 0
because 0!=1), we deduce that q is a non-zero natural number. (cid:3)
Theorem 2.3. Let n=ap+b=a +a p+a p2+...+apl such that 0≤b≤p−1and 0≤a ≤p−1
0 1 2 l i
with i∈[[0,l]]. Then q ≡1 (mod p).
Before we prove Theorem 2.3 we shall state and prove the following non-trivial lemmas.
Lemma 2.4. The integers q and p are relatively prime.
Proof. If 0<q <p, since p is prime, q and p are relatively prime.
If q ≥ p, let us assume that p and q are not relatively prime. It would imply that there exist
an integer x > 0 and a non-zero natural number q′ such that q = q′px with gcd(q′,p) = 1. Since
n! = qa0!(a1p)!...(al−1pl−1)!(alpl)! we get n! = q′pxa0!(a1p)!...(alpl)!. It follows that n! would
contain a factor a pl+x such that q′ =a q′′ with a ∈[[1,p−1]]. But, a pl+x >n.
l+x l+x l+x l+x
Indeed we know that 1+p+...+pl = pl+1−1. So pl+1 = 1+(p−1)(1+p+...+pl). Then
p−1
pl+1 >(p−1)+(p−1)p+...+(p−1)pl. Since0≤a ≤p−1withi∈[[0,l]],wehave0≤a pi ≤(p−1)pi
i i
with i∈[[0,l]]. Therefore, pl+1 >a +a p+...+a pl so pl+1 >n ⇒ a pl+x >n.
0 1 l l+x
Since n! doesn’t include terms like a pl+x >n with a ∈[[1,p−1]], we obtain a contradiction.
l+x l+x
It means that the assumption q = q′px with x ∈ N⋆ and gcd(q′,p) = 1 is not correct. So, q is not
divisible by a power of p. It results that q and p are relatively prime. (cid:3)
We know that n! = (ap + b)! = qa !(a p)!...(a pl)! with a = ⌊n⌋, 0 ≤ a ≤ p − 1 with i =
0 1 l p i
0,1,2,...,p−1 and b=a . Let q with 0≤i≤a ≤a be the natural number
0 a,l,1,i 1
(ap+b−ip)!
q = .
a,l,1,i a !((a −i)p)!(a p2)!...(a pl)!
0 1 2 l
In particular, we have q =q .
a,l,1,0
Lemma 2.5. q ≡q (mod p).
a,l,1,i+1 a,l,1,i
Proof. We have (0≤i<a )
1
ap+b−ip q (a −i)p
a,l,1,i 1
=
(cid:18) p (cid:19) q (cid:18) p (cid:19)
a,l,1,i+1
LUCAS’ THEOREM 3
Or equivalently
ap+b−ip (a −i)p
q =q 1 .
a,l,1,i+1(cid:18) p (cid:19) a,l,1,i(cid:18) p (cid:19)
Now
ap+b−ip (a−i)p+b
= ≡a−i≡a −i (mod p),
(cid:18) p (cid:19) (cid:18) p (cid:19) 1
and
(a −i)p
1 ≡a −i (mod p).
(cid:18) p (cid:19) 1
Therefore
q (a −i)≡q (a −i) (mod p).
a,l,1,i+1 1 a,l,1,i 1
Since a −i with i=0,1,2,...,a −1 and p are relatively prime, so q ≡q (mod p). (cid:3)
1 1 a,l,1,i+1 a,l,1,i
Noticethatq correspondstothecasewherea =0,(a +a p2+...+a pl)!=q a !(a p2)!...(a pl)!.
a,l,1,a1 1 0 2 l a,l,1,a1 0 2 l
Lemma 2.6. For n = ap+b = a pk +b with 0 ≤ b ≤ pk −1 and defining (1 ≤ k ≤ l and
(k) (k) (k)
0≤i≤a )
k
(a pk+b −ipk)!
q = (k) (k)
a,l,k,i a !(a p)!...((a −i)pk)!...(a pl)!
0 1 k l
with a ≥ 1, (where it is understood that when a and l appears together as the two first labels of
k
one q····, it implies that a is given by a = a(1) = (a1a2...al)p = a1 +a2p+...+alpl−1) we have
(0≤i<a )
k
a pk+b −ipk q (a −i)pk
(k) (k) a,l,k,i k
=
(cid:18) pk (cid:19) q (cid:18) pk (cid:19)
a,l,k,i+1
Additionally, q ≡q (mod p).
a,l,k,i a,l,k,i+1
In particular for k =0, we have (0≤i<a ),
0
q
ap+b= a,l,0,i (a −i)
q 0
a,l,0,i+1
with a ≥1.
0
We can provethis lemma by the following a similar reasoningas earlier andhence we omit it here.
Notice that q =q . And q corresponds to the case where a =0. Also
a,l,k,0 a,l,k,ak k
qa−i,l,k,0 =qa,l,1,i ≡qa,l,1,i+1 (mod p),
with 0 ≤ i < a and a ≥ 1. So, since q ≡ q (mod p) with 0 ≤ j < a , we have
1 1 a,l,k,j a,l,k,j+1 k
qa−i,l,k,j ≡ qa−i,l,k,0 ≡ qa,l,1,i ≡ qa,l,1,0 (mod p) and qa−i,l,k,0 = qa−i,l,l,0 ≡ qa−i,l,l,j ≡ qa−i,l,l,al ≡
qa−i,l−1,k,0 (mod p).
So qa−i,l,k,0 ≡qa−i,l−1,k,0 ≡···≡qa−i,l,k,0 ≡qa−i,1,1,0 ≡qa,1,1,i ≡qa,1,1,0 (mod p). Or qa−i,l,k,0 =
q ≡q ≡q (mod p).
a,l,1,i a,l,1,0 a,l,k,0
Finally we have q =q ≡q (mod p).
a,l,k,0 a,1,1,0
Lemma 2.7. We have also the congruence (a pi)!≡a !pai(1+p+...+pi−1) (mod p).
i i
Proof. We proceed by induction on i.
Indeed, we have (a1p)!≡1p·2p···(a1−1)p·a1p (mod p). So (a1p)!≡a1!pa1 (mod p).
Itfollowsthat(a2p2)!=((a2p)p)!≡(a2p)!pa2p ≡a2!pa2pa2p(modp). So(a2p2)!≡a2!pa2(1+p)(modp).
Let us assume that (a pi)!≡a !pai(1+p+...+pi−1) (mod p).
i i
We have (ai+1pi+1)! = ((ai+1pi)p)! ≡ (ai+1pi)!pai+1pi ≡ ai+1!pai+1(1+p+...+pi−1)pai+1pi (mod p).
Thus (ai+1pi+1)!≡ai+1!pai+1(1+p+...+pi−1+pi) (mod p).
Hence the result follows. (cid:3)
We now prove Theorem 2.3.
4 ALEXANDRELAUGIERANDMANJILP.SAIKIA
Proof. If a = 0 for i ∈ [[1,l]], then n = a and we have n! = qa ! with q = 1. So, if a = 0 for
i 0 0 i
i∈[[1,l]], q ≡1 (mod p).
Let consider the case where a = 0 for all i > 1, then n = a +a p and we have n! = qa !(a p)!.
i 0 1 0 1
Then
n! a0−1
qa != =(a +a p)(a +a p−1)...(a p+1)= (a p+a −r).
0 (a p)! 0 1 0 1 1 1 0
1 rY=0
Since 0<a −r ≤a for r ∈[[0,a −1]], we obtain
0 0 0
a0−1 a0
qa !≡ (a −r)≡ r ≡a ! (mod p).
0 0 0
rY=0 rY=1
Since a ! and p are relatively prime, we have q ≡1 (mod p).
0
Since q =q ≡q (mod p), we conclude that q ≡1 (mod p) whatever n is. (cid:3)
a,l,k,0 a,l,1,0
We come back to the proof of Theorem 1.1.
Proof. Let m,n be two positive integers whose base p expansion with p a prime, are
m=a +a p+...+a pk,
0 1 k
and
n=b +b p+...+b pl,
0 1 l
such that m ≥ n. We assume that a ≥ b with i = 0,1,2,...,max(k,l). We denote a = ⌊m⌋ and
i i p
b=⌊n⌋. Since a ≥b with i=0,1,2,...,max(k,l), we have a≥b. We define
p i i
0 if k <l
a =
max(k,l) (cid:26) ak if k ≥l
and
b if k ≤l
b = l
max(k,l) (cid:26) 0 if k >l
In particular if max(k,l) = k, b = 0 for i > l and if max(k,l) = l, a = 0 for i > k. Since m ≥ n,
i i
max(k,l)=k. So, we have a =a and b =b with b =0 when l<k.
max(k,l) k max(k,l) k k
Using these
m!=q a !(a p)!...(a pk)!,
a,k,1,0 0 1 k
n!=q b !(b p)!...(b pl)!,
b,l,1,0 0 1 l
and
(m−n)!=qa−b,k,1,0(a0−b0)!((a1−b1)p)!...((ak−bk)pk)!.
We have
m q a a p a pmax(k,l)
= a,k,1,0 0 1 ... max(k,l) .
(cid:18)n(cid:19) qb,l,1,0qa−b,k,1,0(cid:18)b0(cid:19)(cid:18)b1p(cid:19) (cid:18)bmax(k,l)pmax(k,l)(cid:19)
Rearranging
m a a p a pmax(k,l)
qb,l,1,0qa−b,k,1,0(cid:18)n(cid:19)=qa,k,1,0(cid:18)b0(cid:19)(cid:18)b1p(cid:19)...(cid:18)bmax(k,l)pmax(k,l)(cid:19).
0 1 max(k,l)
Since qa,k,1,0 ≡qb,l,1,0 ≡qa−b,k,1,0 ≡1 (mod p), we get
m a a p a pmax(k,l)
≡ 0 1 ... max(k,l) (mod p).
(cid:18)n(cid:19) (cid:18)b (cid:19)(cid:18)b p(cid:19) (cid:18)b pmax(k,l)(cid:19)
0 1 max(k,l)
Notice that if for some i, a =0, then b =0 since we assume that a ≥b and b ≥0. In such a case
i i i i i
aipi = ai =1.
bipi bi
(cid:0) W(cid:1)e as(cid:0)sum(cid:1)e that a ≥ 1. For k ∈ [[1,a pi −1]] and for i ∈ [[0,max(k,l)]] with 1 ≤ a ≤ pi −1,
i i i
aipi ≡0 (mod p)
k
(cid:0) (cid:1)
LUCAS’ THEOREM 5
Therefore for a =1 we have
i
(1+x)pi = pi pi xk ≡1+xpi (mod p),
(cid:18)k(cid:19)
kX=0
and for any a ∈[[1,pi−1]]
i
(1+x)aipi =((1+x)pi)ai ≡(1+xpi)ai (mod p).
Now comparing
(1+x)aipi =aipi aipi xk,
(cid:18) k (cid:19)
Xk=0
and
(1+xpi)ai = ai ai xlpi
(cid:18)l(cid:19)
Xl=0
we get by taking k =b pi and l=b ,
i i
a pi a
i ≡ i (mod p).
(cid:18)b pi(cid:19) (cid:18)b (cid:19)
i i
Finally we have
m a a a
≡ 0 1 ... max(k,l) (mod p).
(cid:18)n(cid:19) (cid:18)b (cid:19)(cid:18)b (cid:19) (cid:18)b (cid:19)
0 1 max(k,l)
(cid:3)
References
[1] E. Lucas, The´orie des Fonctions Nume´riques Simplement Pe´riodiques, Amer. J. Math., 1 (2), 184–196; 1 (3),
197–240; 1(4),289–321 (1878).
Lyc´eeprofessionnelhotelierLaCloserie,10ruePierreLoti-BP4,22410Saint-Quay-Portrieux,France
E-mail address: [email protected]
Departmentof MathematicalSciences, Tezpur University,Napaam,Sonitpur,Assam,Pin-784028,India
E-mail address: [email protected]