Bernadette Faye, Florian Luca, and Amadou Tall

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1 Bull Koran Math Soc 52 (205), No 2, pp ON THE EQUATION φ(5 m ) = 5 n Brnadtt Fay, Florian Luca, and Amadou Tall Abstract Hr, w show that th titl uation has no positiv intgr solutions (m,n), whr φ is th Eulr function Introduction Lt φ(m) b th Eulr function of th positiv intgr m Problm 0626 from th Amrican Mathmatical Monthly [6] asks to find all positiv intgr solutions (m, n) of th Diophantin uation () φ(5 m ) = 5 n To our knowldg, no solution was vr rcivd to this problm Hr, w prov th following rsult Thorm Euation () has no positiv intgr solution (m, n) Rsults in this spirit appar in [3], [4], [6], [7], [8], [9], [0], [] 2 Th proof of Thorm For th proof, w mak xplicit th argumnts from [9] togthr with som spcific faturs which w dduc from th factorizations of 5 k for small valus of k Writ (2) 5 m = 2 α p α pαr r Thus, (3) φ(5 m ) = 2 α p α (p ) pr αr (p r ) W achiv th proof of Thorm as a sunc of lmmas Th first on is known but w giv a proof of it for th convninc of th radr Lmma 2 In uation (), m and n ar not coprim Rcivd January 30, 204; Rvisd May 20, Mathmatics Subjct Classification D6 Ky words and phrass Diophantin uations, applications of primitiv divisors, applications of siv mthods 53 c 205 Koran Mathmatical Socity

2 54 B FAYE, F LUCA, AND A TALL Proof Supposthatgcd(m,n) = Assumfirstthatnisodd Thnord 2 (5 n ) = 2, whr for a prim p and a nonzro intgr k w writ ord p (k) for th xact xponnt of p in th factorization of k Applying th ord 2 function in both sids of (3) and comparing it with (), w gt r (α )+r (α )+ ord 2 (p i ) = 2 If m is vn, thn α 3, and th abov inuality shows that α = 3, r = 0, so 5 m = 8, fals Thus, m is odd, so α = 2 and r = If α 2, thn p α gcd(φ(5 m ),5 m ) = gcd(5 m,5 n ) = 5 gcd(m,n) = 4, i= a contradiction So, α =, 5 m = 4p, and 5 n = 2(p ) = 5m 2 2 = 5m 5, 2 which is impossibl Thus, n is vn and sinc gcd(m,n) =, it follows that m is odd so α = 2 Furthrmor, a prvious argumnt shows that in (2) w hav α = = ( α r ) = Sinc m is odd, w hav( that ) 5 (5 (m )/2 ) 2 (mod p i ), 5 a thrfor p i = for i =,,r Hr, p is th Lgndr symbol of a with rspct to th odd prim p Hnc, p i,4 (mod 5) If p i (mod 5), it follows that 5 φ(5 m ) = 5 n, a contradiction Hnc, p i 4 (mod 5) for i =,,r Rducing now rlation (2) modulo 5, w gt Rducing now uation modulo 5, w gt a contradiction 4 4 +r (mod 5), thrfor r 0 (mod 2) 2(p ) (p r ) = φ(5 m ) = 5 n 2 (3 r/2 ) 2 4 (mod 5), thrfor ( ) 2 =, 5 Lmma 3 If (m,n) satisfis uation (), thn m is not a multipl of any numbr d such that p 5 d for som prim p (mod 5) Proof This is clar, for if not, thn 5 (p ) φ(5 d ) φ(5 m ) = 5 n, fals Sinc is a prim dividing 5 32, it follows by Lmma 3 that ord 2 (m) 4 From th Cunnigham projct tabls [], w dducd that if 52 is an odd prim, thn 5 has a prim factor p (mod 5) xcpt for {7,4,7,03,223,257} So, if m is odd, thn (4) Q := {7,4,7,03,223,257} { > 52}

3 ON THE EQUATION φ(5 m ) = 5 n 55 Throughout th rst of th papr, w put (5) k = m n Not that k 2 bcaus m and n ar not coprim by Lmma 2 Th nxt lmma givs an uppr bound on k Lmma 4 Th following inuality holds: (6) log log Proof Writ m = 2 α0 s i= < 20 m >2 loglog αi i, i odd prim, i =,,s Rcall that α 0 4 Non of th valus m =,2,4,8,6 satisfis uation () for som n, so s Thn (7) 5 k < 5m 5 n = 5m φ(5 m ) = ( + ) p p 5 m For ach prim numbr p 5, w writ l p for th ordr of apparanc of p in th Lucas sunc of gnral trm 5 n That is, l p is th ordr of 5 modulo p Clarly, if p 5 m, thn l p = d for som divisor d of m Thus, w can rwrit inuality (7) as (8) 5 k < ( + ) p d ml p=d If p m and l p is a powr of 2, thn l p 6, thrfor p 5 6 Hnc, p P = {2,3,3,7,33,489} Thus, (9) l p 6 ( + ) ( + ) < 35 p p p P Insrting (9) into (8), w gt (0) 5 k 35 < d m l p=d P(d)>2 ( + ), p whr for a nonzro intgr l w writ P(l) for th largst prim factor of l with th convntion that P(±) = W tak logarithms in inuality (0)

4 56 B FAYE, F LUCA, AND A TALL abov and us th inuality log(+x) < x valid for all ral numbrs x to gt ( ) 5 k log < 35 p d m l p=d P(d)>2 If l p = d, thn p (mod d) If P(d) > 2, thn sinc d m, w gt that vry odd prim factor of d is in Q In particular, it is at last 7 Thus, p > 34 Hnc, ( ) 5 k log < 35 d m l p=d P(d)>2 p + p 37 W thus gt that ( ) 5 k () log whr (2) S d := p(p ) < d m < d m P(d)>2 l p=d W nd to bound S d For this, w first tak p P d = {p : l p = d} S d, l p=d P(d)>2 Put ω d := #P d Sinc p (mod d) for all p P d, w hav that (3) (d+) ω d p p P d p < 5 d < 5 d, thrfor ω d < dlog5 log(d+) W us th Brun-Titchmarsh thorm in th vrsion du to Montgomry and Vaughan [2] which assrts that (4) π(x;d,) < 2x φ(d) log(x/d) for all x > d 2, whr π(x;d,) stands for th numbr of prims p x with p (mod d) Put Q d := {p < 4d : p (mod d)} Clarly, Q d {d+,2d+,3d+} and sinc d m and 3 Q, it follows that d is not a multipl of 3 In particular, on of d + and 2d + is a multipl of 3, so that at most on of ths two numbrs can b a prim W now split S d as follows: (5) S d p<4d p (mod d) p + 4d p d 2 p (mod d) l p=d p + p>d 2 l p=d p := T +T 2 +T 3

5 ON THE EQUATION φ(5 m ) = 5 n 57 Clarly, (6) T = p Q d p For S 2, w us stimat (4) and Abl s summation formula to gt T 2 π(x;d,) x d 2 d2 + x=4d 4d t 2 d 2 2d 2 d 2 φ(d)logd + 2 φ(d) π(t;d,) dt 4d dt tlog(t/d) 2 φ(d)logd φ(d) loglog(t/d) d = 2loglogd φ(d) + 2 φ(d) t=4d ( logd loglog4 Th xprssion /logd loglog4 is ngativ for d 34, so (7) T 2 < 2loglogd φ(d) for all d 34 Inuality (7) holds for d = 7 as wll, sinc thr T 2 < S 7 = ) 2loglog7 < 0003 < 03 < φ(7) Hnc, inuality (7) holds for all divisors d of m with P(d) > 2 As for T 3, w hav by (3), (8) T 3 < ω d d 2 < log5 dlog(d+) Hnc, collcting (6), (7) and (8), w gt that (9) S d < p + 2loglogd log5 + φ(d) dlog(d+) p Q d W now show that (20) S d < 3loglogd φ(d) Sinc φ(d) < d and at most on of d+ and 2d+ is prim, w gt, via (9), that S d < d+ + < φ(d) 3d+ + 2loglogd log5 + φ(d) dlog(d+) ( ) 4 log5 +2loglogd+ 3 log(d+)

6 58 B FAYE, F LUCA, AND A TALL So, in ordr to prov (20), it suffics that log5 < 2loglogd, which holds for all d > 200 log(d+) Th only possibl divisors d of m with P(d) > 2 (so, whos odd prim factors ar in Q), and with d 200 ar (2) R := {7,34,4,68,7,82,03,36,42,64} W chckd individually that for ach of th valus of d in R givn by (2), inuality (20) holds Now w writ d = 2 α d d, whr α d {0,,2,3,4} and d is odd Sinc d 7 > 2 α d, w hav that d < d 2 Hnc, kping d fixd and summing ovr α d, w hav that (22) 4 S 2 α dd < 3 α d =0 4 α d =0 log(2logd ) φ(d ) < 87log(2logd ) φ(d ) ( ) 8 Insrting inualitis (20) and (22) into (), w gt that ( ) 5 k (23) log < 87log(2logd ) φ(d ) d m d > d odd Th function a 87log(2loga) is sub multiplicativ whn rstrictd to th st A = {a 7} That is, th inuality 87log(2log(ab)) 87log(2loga) 87log(2logb) holds if min{a,b} 7 Indd, to s why this is tru, assum say that a b Thn logab 2logb, so it is nough to show that 87log2+87log(2logb) 87log(2loga) 87log(2logb) which is uivalnt to 87log(2logb)(87log(2loga) ) > 87log2, which is clar for min{a,b} 7 It thus follows that 87log(2logd ) < + φ(d ) m i d m d > d odd 87log(2log i ) φ( i )

7 ON THE EQUATION φ(5 m ) = 5 n 59 Insrting th abov inuality into (23), taking logarithms and using th fact that log(+x) < x for all ral numbrs x, w gt (24) log log < 87log(2log i ) φ( i ) m i Nxt w show that (25) i 87log(2log( i )) φ( i ) < 20loglog for Q W chck that it holds for = 7 So, from now on, 4 Sinc log(2log i ) = log(2i)+loglog < (+logi)+loglog i+loglog, w hav that log(2log( i )) φ( i < i ) i ( ) +loglog i ( ) i= i i 2 ( ) = ( ) 3 +(loglog) ( ) 2 ( ) 2 < (loglog) ( ) 3 + ( ) 2 ( 2 2 ) = (loglog) ( ) 3 bcaus loglog > Thus, it suffics that ( 2 2 ) 87 ( ) 3 < 20, which holds for 4 Hnc, (25) holds, thrfor (24) implis (26) log log < 20 m >2 loglog, which is xactly (6) This finishs th proof of th lmma Lmma 5 If < 0 4 and m, thn n Proof This is clar for = 2, sinc thn m, thrfor 8 = φ(24) φ(5 m ) = 5 n, so n is vn Lt now b odd Look at th numbr 5 (27) = r β 4 rβ l l Assum that l 2 Sinc r i (mod ) for i =,,l, w hav that 2 (r ) (r l ) φ(5 m ) = 5 n Sinc 5 for all odd < 0 4, w gt that, n, as dsird So, it rmains to show that l 2 in

8 520 B FAYE, F LUCA, AND A TALL (27) W do this by contradiction Suppos that l = Sinc r 4 (mod 5), rducing uation (27) modulo 5 w gt that 4 β (mod 5), so β is vn Hnc, 5 n 5 = Howvr, th uation x n x = for intgrs x > and n > 2 has bn solvd by Ljunggrn [5] who showd that th only possibilitis ar (x,n) = (3,5), (7,4) This contradiction shows that l 2 and finishs th proof of this lmma Rmark Apart from Ljunggrn s rsult, th abov proof was basd on th computational fact that if < 0 4 is an odd prim, thn 5 In fact, th first prim failing this tst is = 2077 Lmma 6 W hav k = 2 Proof W split th odd prim factors p of m in two substs U = { n} and V = { n} By Lmma 4, w hav (28) log log 20 U loglog + p V loglog := 20(T +T 2 ) W first bound T 2 By Lmma 5, if V, thn > 0 4 In particular, > 52 Lt t 9, and put I t = [2 t,2 t+ ) V Suppos that r,,r u ar all th mmbrs of I t By th Primitiv Divisor Thorm (s [2]), 5 dru has a primitiv prim factor for all divisors d of r r u, and this prim is congrunt to modulo r u Sinc th numbr r r u has 2 u divisors, w gt that 2 u ord ru (φ(5 m )) = ord ru (5 n ) Sinc r u n, w gt that ord ru (5 n ) = ord ru (5 ru ), so 2 u ord ru (5 ru ) < log5ru logr u = r ulog5 logr u < 2t+ log5 (t+)log2 Th abov inuality implis that u t, for if not, thn u t, and w would gt that 2 t 2t+ log5, or 4log5 (t+)log2 0log2, (t+)log2

9 ON THE EQUATION φ(5 m ) = 5 n 52 a contradiction This shows that #I t t for all t 9 Hnc, Hnc, w gt that (29) log 20T 2 t 9 log 20(t )loglog2 t 2 t < 4 < 20 gcd(m,k) >2 loglog +4 W us (29) to bound k by bttr and bttr bounds W start with log log < 20(loglogk) +4, k >2 which is implid by (29) Assum k 3 W hav < d = σ(k) < k 25 < 79loglogk + k φ(k) loglogk, k d k whr th last inuality abov holds for all k 3 (s inuality (34) in [3]) W thus gt that logk < log(klog5+ log()) < 20 79(loglogk) 2 +54, which givs logk < 263 Sinc it follows that Hnc, T = gcd(m,k) > loglog log > 266 > logk, < loglog < 48 log log < ; so k < By (4), th first fw possibl odd prim factors of m ar 7, 4, 7, 03, 223, 257 and all othrs ar > 52 Sinc it follows that > 0 4 > k, T loglog7 + loglog4 + loglog loglog257 < loglog loglog

10 522 B FAYE, F LUCA, AND A TALL Hnc, log log < ; so k 44 If follows that gcd(k,m) can hav at most on odd prim factor, so thrfor T loglog7 7 < 007, log log < = 28; so k Thus, in fact k has no odd prim factor, giving that T = 0, so log log < 4, thrfor k 2 Sinc by Lmma 2, m and n ar not coprim, it follows that in fact k 2, so k = 2 Lmma 7 k > 2 Proof Lt b th smallst odd prim factor of m which xists for if not m 6, which is not possibl Lt,, s b all th prim factors of m For ach divisor d of 2 s, th numbr 5 d has a primitiv divisor which is congrunt to modulo Sinc thr ar 2 s divisors of 2 s, w gt that 2 s ord (φ(5 m )) = 5 n Sinc dos not divid n (othrwis it would divid k = 2), w gt that ord (5 n ) = ord (5 ), and Hnc, 2 s ord ( 5 ) < log5 log = log5 log < Lmma 4 now shows that ( ( 5 2 )) log log s < + log log2 < 20 m < 20 >2 loglog < 20sloglog ( + log ) loglog log2 This givs < 300, so by Lmma 5, w hav k, which finishs th proof of this lmma

11 ON THE EQUATION φ(5 m ) = 5 n 523 Obviously, Lmmas 6 and 7 contradict ach othr, which complts th proof of th thorm Acknowldgmnts W thank th rfr for a carful rading of th manuscript and for pointing out som computational rrors in a prvious vrsion of th manuscript W also thank Sam Wagstaff for snding us an updatd vrsion of som of th tabls in [] This papr was writtn during a visit of F L at th AIMS-Sngal in January of 204 H thanks th popl of this institution for thir hospitality During th prparation of this papr, F L was also supportd in part by Projct PAPIIT IN0452 (UNAM) Rfrncs [] J Brillhart, D H Lhmr, J L Slfridg, B Tuckrman, and S S Wagstaff, Jr, Factorizations of b n ±, b = 2,3,5,6,7,0,,2 up to high powrs, Amrican Mathmatical Socity, Providnc, RI, 988 [2] R D Carmichal, On th numrical factors of th arithmtic forms α n ± β n, Ann Math (2) 5 (93), no -4, [3] J Cillrulo and F Luca, Rpunit Lhmr numbrs, Proc Edinb Math Soc 54 (20), no, [4] M T Damir, B Fay, F Luca, and A Tall, Mmbrs of Lucas suncs whos Eulr function is a powr of 2, Fibonacci Quart 52 (204), no, 3 9 [5] W Ljunggrn, Som thorms on indtrminat uations of th form xn /x = y, Norsk Mat Tidsskr 25 (943), 7 20 [6] F Luca, Problm 0626, Amrican Math Monthly 04 (997), 87 [7], Euations involving arithmtic functions of Fibonacci and Lucas numbrs, Fibonacci Quart 38 (2000), no, [8], Eulr indicators of binary rcurrnc suncs, Collct Math 53 (2002), no 2, [9], On th Eulr function of rpdigits, Czchoslovak Math J 58 (2008), no, 5 59 [0] F Luca and M Mignott, φ(f ) = 88, Divulg Mat 4 (2006), no 2, 0 06 [] F Luca and P Stănică, Euations with arithmtic functions of Pll numbrs, Bull Soc Mat Roumani, to appar [2] H L Montgomry and R C Vaughan, Th larg siv, Mathmatika 20 (973), 9 34 [3] J B Rossr and L Schonfld, Approximat formulas for som functions of prim numbrs, Illinois J Math 6 (962), Brnadtt Fay AIMS-Sénégal Km 2 rout d Joal (Cntr IRD Mbour) BP: Dakar-Fann, Sénégal addrss: brnadtt@aims-sngalorg Florian Luca School of Mathmatics Univrsity of th Witwatrsrand P O Box Wits 2050, South Africa addrss: florianluca@witsacza

12 524 B FAYE, F LUCA, AND A TALL Amadou Tall AIMS-Sénégal Km 2 rout d Joal (Cntr IRD Mbour) BP: Dakar-Fann, Sénégal addrss: amadoutall@aims-sngalorg

BINOMIAL COEFFICIENTS INVOLVING INFINITE POWERS OF PRIMES. 1. Statement of results

BINOMIAL COEFFICIENTS INVOLVING INFINITE POWERS OF PRIMES DONALD M. DAVIS Abstract. If p is a prim and n a positiv intgr, lt ν p (n dnot th xponnt of p in n, and u p (n n/p νp(n th unit part of n. If α