A NEW PROOF OF LUCAS’ THEOREM ALEXANDRELAUGIERANDMANJILP.SAIKIA 3 1 0 Abstract. Wegiveanew proofofLucas’Theoreminelementarynumbertheory. 2 n KeyWords:Lucas’theorem. a J 2010Mathematical ReviewsClassificationNumbers: 11A07,11A41,11A51,11B50,11B65,11B75. 6 1 1. Introduction ] T One of the most useful results in elementary number theory is the following result of E. Lucas N 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) 2 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. 0 We present here an alternate approachusing elementary number theoretic techniques. 3 1 : 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. 1 2 ALEXANDRELAUGIERANDMANJILP.SAIKIA 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 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 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 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

A new proof of Lucas' Theorem

Similar A new proof of Lucas' Theorem