Iteratioal Mathematical Forum, Vol. 11, 016, o. 9, 49-437 HIKARI Lt, www.m-hikari.com http://x.oi.org/10.1988/imf.016.63 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,...
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
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)
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 + 1 + k + 1 > 3 + 3 = 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 + 1 1 (16) k + 1 = k + 1 = if k + 1 (17) sup = sup k + 1
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)
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)
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/ + 1 3 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)
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/ ( 1 + 1 4s ( = 8s 3 1 + 3 ( ) ) 1 (3/)((3/) 1) 1 1 + 4s (4s ) + g 1 4s (4s ) ( ( )) 3 1 1 = 8s 3 + 3s + 8 + g 4s s (9) Therefore (see (9) a (8)) 3/ = 8s 3 + 3s (s ) a ɛ() = ( ( )) 3 1 1 8 + 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
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 16 + 6 + 1 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, 815-819. http://x.oi.org/10.1988/ijcms.013.3790 [3] R. Jakimczuk, O the istributio of perfect powers, Joural of Iteger Sequeces, 14 (011), Article 11.8.5. [4] R. Jakimczuk, Asymptotic formulae for the -th perfect power, Joural of Iteger Sequeces, 15 (01), article 1.5.5. Receive: March 1, 016; Publishe: April 1, 016