Gaps between Consecutive Perfect Powers

Transcription

1 Iteratioal Mathematical Forum, Vol. 11, 016, o. 9, HIKARI Lt, Gaps betwee Cosecutive Perfect Powers Rafael Jakimczuk Divisió Matemática, Uiversia Nacioal e Lujá Bueos Aires, Argetia Copyright c 016 Rafael Jakimczuk. This article is istribute uer the Creative Commos Attributio Licese, which permits urestricte use, istributio, a reprouctio i ay meium, provie the origial work is properly cite. Abstract Let P be the -th perfect power a = P +1 P the ifferece betwee the two cosecutive perfect powers P a P +1. I a previous article of the author the followig cojecture was establishe,. I this article we prove that this cojecture is false, sice we prove that sup = 1, if = 0. Therefore there exist small gaps betwee cosecutive perfect powers. We also prove the stroger result if = 0, ()(/3)+ɛ where ɛ is a fixe but arbitrary positive real umber. Besies, usig the ieas of this article, we obtai a shorter proof of a theorem prove i aother article of the author. Mathematics Subject Classificatio: 11A99, 11B99 Keywors: Perfect powers, cosecutive perfect powers, gaps 1 Itrouctio A atural umber of the form m where m a are positive iteger is calle a perfect power. The first few terms of the iteger sequece of perfect powers are 1, 4, 8, 9, 16, 5, 7, 3, 36, 49, 64, 81, 100, 11, 15, 18,...

2 430 R. Jakimczuk Let A() be the umber of perfect powers i the ope iterval (( 1), ), where 1 is a positive iteger. It is well-kow that A() = 0 for almost all itervals (( 1), ), sice we have the theorem (see [3]) Theorem 1.1 Let us cosier the ope itervals (0, 1 ), (1, ),..., (( 1), ). Let S() be the umber of these ope itervals that cotai some perfect power. The followig it hols S() = 0 Therefore, if S 1 () is the umber of these ope itervals that o ot cotai perfect powers the the followig it hols sice S() + S 1 () =. Clearly S 1 () = 1 (1) A() 1 () ifiite times, sice there are ifiite perfect powers ot a square. Let P be the -th perfect power a = P +1 P the ifferece betwee the two cosecutive perfect powers P a P +1. We have the iequality (see [1]) = P +1 P < ( 3). (3) Let P be the -th perfect power. We have (see [4]) P (4) Therefore P +1 P Mai Results I a previous article of the author [] the followig cojecture was establishe Now, we give a proof that this cojecture is false. Theorem.1 The cojecture is false

3 Gaps betwee perfect powers 431 Proof. Let us cosier the perfect powers P such that k P < (k + 1) (5) The umber of perfect powers i this iterval is A(k + 1) + 1. Note that there is always a perfect power P that satisfies iequality (5), amely P = k. We eote the sum of the correspoig A(k + 1) + 1 iffereces i the form = (k + 1) k = k + 1 (6) Iequality (5) gives 1 P (k + 1) < (7) k k Therefore, sice both sies i (7) have it 1, we have Now, equatios (8) a (4) give P k = k() k where k() 1. Therefore equatio (9) gives a cosequetly P k = 1 (8) k = 1 Note that k + 1 = (k + 1) k (see (6)) Suppose that ( ) = k() = 1 (9) k k + 1 = 1 (10) Therefore (see (10) a (11)) = 1 (11) k + 1 = Cosequetly, from a certai value of we have k + 1 = 1.1 = 1 k + 1 > 3 (1)

4 43 R. Jakimczuk Sice A(k + 1) 1 ifiite times (see ()) we have that the umber of i the sum is at least ifiite times. Hece (see (6) a (1)) 1 = k + 1 k k + 1 > = 4 3 > 1 That is, a eviet cotraictio. The theorem is prove. Theorem. We have sup = sup k + 1 = 1 (13) Proof. We have (see (3)) 0 if = if k + 1 < 1 (14) Therefore 0 1 O the other ha, we have (see (6)) 0 if sup 1 (15) Therefore 0 k + 1 k + 1 = k + 1 k + 1 = 1 Now 0 if = k + 1 k + 1 k + 1 Equatios (15), (16) a (17) imply if sup k (16) k + 1 = k + 1 = if k + 1 (17) sup = sup k + 1

5 Gaps betwee perfect powers 433 There are ifiite values of k such that A(k + 1) = 0 (see (1)). Therefore i the iterval [k, (k + 1) ) there is a uique perfect power P, amely k, a cosequetly a uique ifferece, amely (k + 1) k = k + 1. Hece, for these ifiite values of k we have sup = sup Therefore (13) is prove. O the other ha, suppose that = k+1 k+1 k+1 k + 1 = 1 = 1. That is The if = if k + 1 = 1 = 1 This is impossible by Theorem.1. Cosequetly 0 if = if a (14) is prove. The theorem is prove. k + 1 < 1 The followig theorem was prove i [1, Theorem 3.1 a Corollaries 3. a 3.3]. We have obtaie the followig proof shorter usig the ieas of this article. Theorem.3 Let ɛ > 0 a arbitrary but fixe real umber. Let us cosier the first cosecutive iffereces 1 = (P P 1 ), = (P 3 P ),..., = (P +1 P ). Let v() be the umber of these iffereces such that ( ɛ)i < i < i. We have the followig it v() = 1. Proof. There are ifiite values of k such that A(k + 1) = 0 (see (1)). I this proof we work as this sequece of values of k. Therefore i the iterval [k, (k + 1) ) there is a uique perfect power P i, amely k, a cosequetly a uique ifferece i, amely, (k + 1) k = k + 1. Hece for these ifiite values of k we have i = k+1 k+1 k+1 k = 1, a therefore i k + 1 = k 1 = 1 (18)

6 434 R. Jakimczuk Now (see (18) a (10)) i k i = i k + 1 k k + 1 i = 1.1 = 1 (19) Let ɛ > 0 a fixe but arbitrary real umber. There exists k ɛ such that if k k ɛ + 1 a A(k + 1) = 0 we have (see (19) a (3)) 1 ɛ < i i < 1 That is ( ɛ)i < i < i The iequality s P +1 has the solutios s = 1,,..., P+1 a cosequetly P+1 solutios. Here (as usual). is the iteger part fuctio. The umber of k such that A(k + 1) = 0 a (k + 1) P +1 will be (see (1) a (4)) ( ) ( ) S 1 P +1 = a() P +1 = a() P +1 α() = a()(b() α()) = c() where a() 1, b() 1, c() 1 a 0 α() < 1. Now c() q ɛ = S 1 ( P +1 ) q ɛ v() where q ɛ is the umber of k such that k k ɛ a A(k + 1) = 0. Cosequetly v() The theorem is prove. I the followig theorem we prove that there exist small gaps i the sequece of perfect powers. Theorem.4 We have if = if Proof. We shall ee the followig Taylor s formulae. 1 1 x k + 1 = 0 (0) = 1 + x + f(x)x (1)

7 Gaps betwee perfect powers 435 where x 0 f(x) = 0. This is the Taylor s formula of the geometric power series. (1 + x) α = 1 + αx + α(α 1) x + g(x)x () where x 0 g(x) = 0. This is the Taylor s formula of the biomial power series. Suppose that is ot a square. Cosequetly 3/ < 3/ < 3/ + 1 (3) That is 3/ < 3 < ( 3/ + 1 ) (4) Therefore the perfect power 3 is i the iterval [ k, (k + 1) ) [ 3/ = (, 3/ + 1 ) ) (5) Let P i be the first perfect power greater tha 3/. Hece we have P i 3. The first ifferece i i iterval (5) will be i = P i 3/ 3 3/ (6) Equatios (6) a (1) give That is i 3/ / 3/ 3/ 0 < = 3 3/ ( 3 3/ = ( 3/ ɛ()) ɛ() ) = 3/ 1 3/ 1 ɛ() + ɛ() ( 3/ = 3/ 1 + ɛ() ( ) ) ɛ() ɛ() + f 3/ 3/ 3/ 3/ + ɛ() = ɛ() ( ) ɛ() ɛ() + f = ɛ() + o(1) ( ) 3/ 0 < i 3/ + 1 where ɛ() is the fractioal part ɛ() + o(1) ( ) (7) ɛ() = 3/ 3/ (8)

8 436 R. Jakimczuk a therefore (see (3)) 0 < ɛ() < 1 Suppose that is of the form = 4s + 1. Cosequetly (see equatio ()) ) 3/ 3/ = ( 4s + 1 ) 3/ ( ) = 4s 3/ ( s ( = 8s ( ) ) 1 (3/)((3/) 1) s (4s ) + g 1 4s (4s ) ( ( )) = 8s 3 + 3s g 4s s (9) Therefore (see (9) a (8)) 3/ = 8s 3 + 3s (s ) a ɛ() = ( ( )) g 4s s = o(1) (s ) (30) Therefore (see (7) a (30)) if = 4s + 1 the s Equatio (31) implies (0). The theorem is prove. i 3/ + 1 = 0 (31) From the proof of this theorem we ca euce the followig stroger result. Theorem.5 Let ɛ be a fixe but arbitrary positive real umber. We have if (k + 1) = if = 0 (3) (/3)+ɛ ()(/3)+ɛ Proof. We have (see the proof of theorem.1) Therefore (see (10)) k = 1 () (/3)+ɛ (k + 1) (/3)+ɛ = 1

9 Gaps betwee perfect powers 437 a cosequetly (see the proof of theorem.) if (k + 1) (/3)+ɛ = if () (/3)+ɛ If we cosier the umbers of the form 4s + 1, from the proof of theorem.4 we obtai the iequality (8s 3 + 3s) < (4s + 1) 3 < (8s 3 + 3s + 1) Cosequetly (see the proof of theorem.4) = i 0 < ((8s 3 + 3s) + 1) (4s + 1) 3 (8s 3 + 3s) (/3)+ɛ ((8s 3 + 3s) + 1) (/3)+ɛ 3s + 1 ( ) (/3)+ɛ 0 (s ) s +3ɛ s s 3 a hece s Limit (33) implies (3). The theorem is prove. i = 0 (33) ((8s 3 + 3s) + 1) (/3)+ɛ Ackowlegemets. The author is very grateful to Uiversia Nacioal e Lujá. Refereces [1] R. Jakimczuk, Some results o the ifferece betwee cosecutive perfect powers, Gulf Joural of Mathematics, 3 (015), o. 3, 9-3. [] R. Jakimczuk, A cojecture o the ifferece betwee cosecutive perfect powers, Iteratioal Joural of Cotemporary Mathematical Scieces, 8 (013), o. 17, [3] R. Jakimczuk, O the istributio of perfect powers, Joural of Iteger Sequeces, 14 (011), Article [4] R. Jakimczuk, Asymptotic formulae for the -th perfect power, Joural of Iteger Sequeces, 15 (01), article Receive: March 1, 016; Publishe: April 1, 016