#A2 INTEGERS 12A (2012): John Selfridge Memorial Issue ON A CONJECTURE REGARDING BALANCING WITH POWERS OF FIBONACCI NUMBERS

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1 #A2 INTEGERS 2A (202): John Selfridge Memorial Issue ON A CONJECTURE REGARDING BALANCING WITH POWERS OF FIBONACCI NUMBERS Saúl Díaz Alvarado Facultad de Ciencias, Universidad Autónoma del Estado de México, Toluca, México sda@uaemex.mx Andrej Dujella Department of Mathematics, University of Zagreb, Zagreb, Croatia duje@math.hr Florian Luca Instituto de Matemáticas, Universidad Nacional Autonoma de México, Morelia, Michoacán, México fluca@matmor.unam.mx Received: 2/28/0, Accepted: 9/7/, Published: 0/2/2 Abstract Here, we show that (k,, n, r) = (8, 2, 4, 3) is the only solution in positive integers of the Diophantine equation F k + F k F k n = F n+ + + F n+r, where F m is the mth Fibonacci number. To the memory of John Selfridge. Introduction Let (F n ) n 0 be the Fibonacci sequence given by F 0 = 0, F = and F n+2 = F n+ + F n for all n 0. () The following conjecture was proposed in []. Conjecture. The only quadruple (k,, n, r) of positive integers satisfying the Diophantine equation is (8, 2, 4, 3). F k + F k F k n = F n+ + + F n+r (2)

2 INTEGERS: 2A (202) 2 Conjecture is a version involving powers of Fibonacci numbers of the classical problem concerning balancing numbers, which are positive integers n such that the equality (n ) = (n + ) + + (n + r) holds with some positive integer r. Several variations of this problem have been previously considered in the literature (see [2], [0]). The authors of [] also show that every solution of equation (2) has < k and that there is no such solution with (k, ) = (2, ), (3, ), or (3, 2). In particular, all solutions of equation (2) have k 4. Observe also that n 4. Here, we confirm Conjecture. We record the result as follows. Theorem 2. Conjecture holds. Our method uses linear forms in logarithms, LLL, and some elementary considerations. We recall that the formula F n = n β n β holds for n 0, where (, β) := + 5, 5. (3) 2 2 In particular, the inequality n 2 F n n holds for all n. We will use this inequality several times throughout. 2. Preliminary Inequalities Observe that On the other hand, F k + F k F k n > F k n > ( n 3 ) k = k(n ) 2k, F k + F k F k n (F + + F n ) k = (F n+ ) k < F k n+ < ( n ) k = nk, where we used the known fact that the identity F + F F m = F m+2 holds for all m. Thus, k(n ) 2k < F k + F k F k n < k(n )+k. (4) In a similar way, we also get that (n+r) 2 < F n+ + + F n+r < (n+r)+. (5) Comparing bounds (4) and (5), we get k(n ) 2k < (n + r) + and (n + r) 2 < k(n ) + k,

3 INTEGERS: 2A (202) 3 so k(n ) (n + r) < max{2k +, k + 2} = 2k +. (6) We record this as a lemma. Lemma 3. Any positive integer solution (k,, n, r) of equation (2) satisfies inequality (6). 3. Initial bounds on k and Write N := Fn+ + + Fn+r Fn+ = Fn+r + F n+r Fn+2 F n+r + + Fn+r F n+r + =: F n+r( + S). (7) Observe that S 0. The inequality F m F m 2 3 holds for all m 3. (8) Indeed, the above inequality is equivalent to 2F m 3F m, or 2(F m + F m 2 ) 3F m, or 2F m 2 F m = F m 2 + F m 3, or F m 2 F m 3, which is indeed true for all m 3. Hence, F n+r 2 F n+r 3, F n+r 2 = F n+r 2 Fn+r F n+r F n+r F n+r Thus, S = Fn+ F n+r + + Fn+r F n+r < j because. In a similar way, we write j 2 = 3 2 2,..., 3 F n+ F n+r < r 2. 3 (9) 2 3 (2/3) 3.5, (0) N = F k + + Fn k k F + F2 k + + Fn 2 k + = F k n F n F n F n =: Fn ( k + S ). () The argument which proved inequality (9) based on inequality (8) shows that F n 2 F n 2 3, F n 3 F n = F n 3 F n 2 Fn 2 F n 2 3 2,..., F 2 F n < 2 3 n 3.

4 INTEGERS: 2A (202) 4 Furthermore, Thus, S = F F n F F n = F 2 F n k + + Fn 2 F n n k < 2 3 k(n 3) + j jk 2 3 k (2/3) k 3.5 k, (2) where for the last inequality we used the fact that k 4. Since N = F k n ( + S ) = F n+r( + S), we get that F k n F n+r = F k n S F n+rs < M max{s, S }, (3) where M := max{fn, k Fn+r}. Dividing both sides of the above inequality by M, we get Fn F εk n+r ε < 3.5, (4) where ε = or, according to whether M = Fn+r or Fn, k respectively. We shall use several times a result of Matveev (see [9], or Theorem 9.4 in [3]), which asserts that if, 2,..., K are positive real algebraic numbers in an algebraic number field K of degree D, b, b 2,..., b K are rational integers, and is not zero, then where and Λ := b b2 2 b K K Λ > exp.4 30 K+3 K 4.5 D 2 ( + log D)( + log B)A A 2 A K, (5) B max{ b, b 2,..., b K }, A i max{dh( i ), log i, 0.6}, for all i =, 2,..., K. (6) Here, for an algebraic number η we write h(η) for its logarithmic height whose formula is h(η) := d log a 0 + log max{ η (i), }, d i=

5 INTEGERS: 2A (202) 5 with d being the degree of η over Q and f(x) := a 0 d (X η (i) ) Z[X] i= being the minimal primitive polynomial over the integers having positive leading coefficient a 0 and η as a root. In a first application of Matveev s theorem, we take K := 2, := F n, 2 := F n+r. We also take b := εk, and b 2 := ε. Thus, Λ := F εk n F ε n+r (7) is the expression appearing under the absolute value in the left hand side of inequality (4). Let us check that Λ = 0. If Λ = 0, it follows that F k n = F n+r. Hence, F n and F n+k are multiplicatively dependent. However, Carmichael s Primitive Divisor Theorem (see [4]) asserts that that if n > 2, then F n has a primitive prime factor; that is, a prime factor p such that p does not divide F m for any m < n. In particular, if n + r > 2, then F n and F n+r are multiplicatively independent. A quick look at the remaining cases shows that the only instance in which F n and F n+r are multiplicatively dependent is when F n = 2 and F n+r = 8, so n = 4 and r = 2. But in this case, equation (2) is k + k + 2 k = 5 + 8, which has no solutions anyway since its left hand side is even and its right hand side is odd. Hence, indeed Λ = 0. Since < k, it follows that B = k. Since and 2 are rational numbers, it follows that we can take D :=. Next, since the inequality F m < m holds for all positive integers m, we can take A := (n ) log and A 2 := (n + r) log, and then inequalities (6) hold for both i =, 2. Now Matveev s theorem tells us that Λ > exp ( C (n ) log (n + r) log ( + log k)), (8) where C := < Taking logarithms in inequality (4) and comparing the resulting inequality with (8), we get C (log ) 2 (n )(n + r)( + log k) < log Λ < log(.5) + log 3, so log 3 log(.5) < C (log ) 2 (n )(n + r)( + log k), (9) log(.5)

6 INTEGERS: 2A (202) 6 which leads to < n(n + r)( + log k) < 0 9 n(n + r) log k, (20) because log k log 4 >. Recall now that, by Lemma 3 and the fact that n 4, we have Thus, 3k (n )k (n + r) + 2k +, therefore k (n + r + ). (2) < 0 9 n(n + r) log((n + r + )). (22) If n + r, then we have an inequality which is better than inequality (22). Otherwise, n + r +, therefore n(n + r) log, so log < 2 09 n(n + r). (23) It is well-known and easy to prove that if A 3 and x/ log x < A, then x < 2A log A (see, for example, [7]). Thus, taking A := n(n + r), inequality (23) gives us < 2(2 0 9 n(n + r)) log(2 0 9 n(n + r)) < n(n + r)(log(2 0 9 ) + 2 log(n + r)) < n(n + r)( log(n + r)) < n(n + r)(6 log(n + r)) < n(n + r) log(n + r). In the above chain of inequalities, we used that fact that n + r 5, which implies that log(n + r) log 5 = > 22/4 = Thus, From estimate (2), we also deduce that < n(n + r) log(n + r). (24) k < (n + r + ) < n(n + r)(n + r + ) log(n + r) < n(n + r) 2 log(n + r), (25) where we used the fact that n + r 5, which implies that (n + r + )/(n + r) 6/5. We record what we have just proved. Lemma 4. If (k,, n, r) is a solution in positive integers of equation (2), then both inequalities hold. < n(n + r) log(n + r), k (n + r + ) < n(n + r) 2 log(n + r)

7 INTEGERS: 2A (202) 7 4. The Case of Small n and r Here, we assume that n 3000, r Thus, by Lemma 4, we have < log 6000 <. 0 9, k (n + r + ) < Now put Γ := k log F n log F n+r. Inequality (4) tells us that e Γ = Λ < 3.5. Assuming that 5, we then have that 3/.5 < /2, so that e Γ < /2. This leads to e Γ < 2, therefore Γ < e Γ e Γ < 6.5. Dividing the last inequality above by log F n, we get that log F n+r k log F n < 6 (log F n ).5 6 (log 2).5. The left hand side above is < /(2 2 ) for all 4. Thus, by a criterion of Legendre, it follows that if 4, then k/ is a convergent of the continued fraction of the number γ := (log F n+r )/(log F n ). Hence, k/ = p i /q i, where p i /q i is the ith convergent of γ and furthermore q i < This gives a certain number of possibilities for the ratio k/ once n and r are fixed. Fixing the ratio k/ = κ/λ with coprime positive integers κ and λ, we can write k = κd and = λd for some positive integer d, which is the greatest common divisor of k and. Then, again with n, r fixed and κ and λ fixed also, inequality (4) gives (Fn F εκ n+r ελ ) d 3 < (.5 λ ) d, which gives a few possibilities for d. Hence, when n and r are fixed, we get a certain number of possibilities for the pair (k, ) = (κd, λd). All this was when 4, but if 3 and n and r are fixed, then we have only a few possibilities for k as well. Then we test all such possible quadruples (k,, n, r) and check whether equation (2) is satisfied. This computation took some 20 hours and revealed no additional solutions (k,, n, r) to equation (2) aside from (8, 2, 4, 3). We record what we have obtained as follows. Lemma 5. If (k,, n, r) is a positive integer solution to equation (2) other than (8, 2, 4, 3), then max{n, r} 300.

8 INTEGERS: 2A (202) 8 5. A Bound for r in Terms of n From now on, we assume that max{n, r} 300. We look at the right hand side of (2) more closely. Recall that N := F n+ + + F n+r (see (7)). We let t := max{n, (n + r)/2} and put N := F n+ + + F t. Observe that N = 0 if r n +. If N = 0, then r n + 2, and therefore we have that t = (n + r)/2 (n + r)/2. Thus, Fn+ N = Ft Fn+2 Ft < F t F t F t F t 3F t < 3(F 2t) /2 3(F n+r) /2 3N /2. (26) In the above estimates, we invoked the argument used at (0) to bound S, as well as the known fact that the inequality F 2m Fm 2 holds for all positive integers m. Thus, the inequality N 3 N holds regardless of whether N is zero or not. Before moving further, observe that the inequality N < N (n+r)/2 (27) holds, because this inequality is equivalent to N n+r 2, which holds since Hence, using (26) and (27), we get We next look at Let j [t +, n + r]. Write N F n+r F n+r > n+r 2. N < 3 N < 3N (n+r)/2 < N 2 := N N = F t+ + + F n+r. 5 /2 5 /2 5N. (28) (n+r)/2 Fj j β j = = j ( )j. (29) Observe that, by Lemma 4, we have that 2j 2j < 7 00 (n + r) 2 log(n + r) n+r <. (n+r)/2

9 INTEGERS: 2A (202) 9 The last inequality holds whenever n + r 53, which is the case for us. Since n + r 3002, the right hand side above is < 500 < If j is odd, then < ( )j 2j = < exp + 2j = exp log + 2j < exp 2j 2 < +, (30) (n+r)/2 (n+r)/2 because the argument inside the exponential is < Similarly, if j is even, then > ( )j 2j = 2j = exp log 2j (3) > exp j > exp > (n+r)/2. (n+r)/2 Formula (29), together with bounds (30) and (3), gives F j j < j ( )j j < Put 5 /2 5 /2 2j x := (n+r)/2. 5 /2 2 (n+r)/2. (32) Since x < 0 300, inequality (32) certainly implies that j /5 /2 <.5Fj, therefore F j j j < 2x < 3xFj. (33) 5 /2 The above inequality applied for j = t +,..., n + r, gives immediately that if we put n+r j N 3 := 5, /2 then the inequality j=t+ 5 /2 N 2 N 3 < 3x Ft+ + + Fn+r = 3xN2 3xN = 3N (34) (n+r)/2 holds. It remains to estimate N 3. Observe first that N 3 = (n+r+) (t+). (35) 5 /2 ( ) In particular, N (t+) ( (n+r t) ) 5 /2 ( ) = N N 3 N + N 2 N 3 < 8N, (36) (n+r)/2

10 INTEGERS: 2A (202) 0 by estimates (28) and (34). Now, by estimate (33) for j = t +, we have (t+) < 2F Fn+r 5 /2 t+ 2 F n+r t 2N F n+r t 2N (n+r t 2), where we used the fact that for positive integers a and b the inequality F a+b F a F b+ holds (with a := n + r t and b := t). Hence, (t+) 5 /2 ( ) < 2N ( ) < 6N, (37) (n+r t 2) (n+r t ) where we used the fact that the inequality /( ) < 2.62 < 3 holds for all positive integers. Thus, from formula (35) and estimates (36) and (37), we get that N (n+r+) 5 /2 ( ) N N 3 + (t+) 5 /2 ( ) < 8N + (n+r)/2 by (36). The above inequality (38) implies that N( ) (n+r+) 5 /2 < 8N( ) + (n+r)/2 < 8N +. (n+r)/2 (n+r t 2) Hence, so that (n+r t ) (n+r t ) N (n+r+) 5 /2 < N , (n+r)/2 (n+r t 2) N (n+r) 5 /2 < 8N + + (n+r)/2 (n+r t ), (38). (39) Let s take a break and see what we have done so far. Let us look at the last term on the right hand sides both in (38) as well as in (39) above. If t = (n + r)/2, then r n, and (n + r t ) (n + r (n + r)/2 ) ((n + r)/2 ), so the third term on the right hand side of (39) is majorized by the second term because n + r If t = n, then (n + r t ) = (r ) r/2 if r 2. However, if r =, we then get n + r t = 0, so the third term on the right hand side of both inequalities (38) and (39) is too large to be useful. So, let us take a closer look at the case r =. In this case, we simply have N = F n+.

11 INTEGERS: 2A (202) Hence, N (n+) 5 /2 < 3xF n+ = 3xN = 3N, (40) (n+r)/2 by estimate (33) with j = n + = n + r. Furthermore, the above estimate (40) implies that N (n+2) 3N 5 /2 ( ) < < 8N (n+)/2. (n+)/2 Hence, N (n+2) N 5 /2 ( ) < + 8N < 8N (n+)/2 + (n+)/2 = 8N +. (4) (n+r)/2 (n+r t)/2 Comparing estimates (38) and (4), as well as (39) and (40), we get that estimates (38) and (39) hold also when r = with the exponent of on the last term in the right hand side replaced by (n + r t)/2 (instead of n + r t ). Since n + r t (n + r t)/2 holds whenever r >, we can record our conclusion as follows: Lemma 6. Let (k,, n, r) be a solution other than (8, 2, 4, 3) of equation (2). Putting N := F n+ + + F n+r, and t := max{n, (n + r)/2}, then all three inequalities N (n+r) 5 /2 < 8N + + (n+r)/2 N (n+r+) 5 /2 ( ) < 8N + (n+r)/2 and hold. Now we return to (n+r t)/2 (n+r t)/2, (42), (43) N (t+) ( (n+r t) ) 5 /2 ( ) < 8N (44) (n+r)/2 N = F k + + F k n = F k n ( + S ), where 0 < S < 3/.5 k (see formula () and estimate (2)). Since k 4, we get that S < 3/5, and in particular 5N/8 < F k n < N. Hence, N F k n < F k n S < 3N.5 k. (45)

12 INTEGERS: 2A (202) 2 Comparing estimates (45), (42) and (43), and using also the fact that >.5, we get F n k (n+r) 5 /2 < 8N k +.5 +, (46) (n+r)/2.5 (n+r t)/2 and F n k (n+r+) 5 /2 ( ) < 8N.5 k (n+r)/2.5 (n+r t)/2. (47) We divide both sides of equations (46) and (47) by Fn k and keep in mind that N/Fn k < 8/5, to get Fn k (n+r) 5 /2 < k +.5 +, (48) (n+r)/2.5 (n+r t)/2 and Fn k (n+r+) 5 /2 ( ) < 3.5 k +.5 (n+r)/2 +.5 (n+r t)/2 (49) Recall that our goal in this section is to bound r. We distinguish several cases. Case. r n. In this case, by Lemma 4, we get that n(2n) log(2n) =.4 0 n 2 (log 2 + log n).4 0 n 2 (3/2) log n < 2. 0 n 2 log n, (50) where we used the fact that n 4 2 2, so that log n 2 log 2. Furthermore, n + r k < (n + r + ) = (n + r) (2n) < 4.3 n 3 log n, (5) n + r 3002 where we used the fact that n + r We also record that for future referencing.. r n (52) From now on, we assume that r > n. In particular, t = (n+r)/2, and therefore (n + r t)/2 (n + r)/4. (53) We shall apply Matveev s theorem to bound from below the left hand sides of (48) and (49). Let us check that they are not zero. If the left hand side of (48) is zero, then 2(n+r) = F 2k n 5 Z, which is impossible since no power of of

13 INTEGERS: 2A (202) 3 positive integer exponent is an integer. If the left hand side of (49) is zero, we then get that (n+r+) = F k n 5 /2. Conjugating the above relation in Q( 5) and multiplying the two resulting relations we get ( ) (n+r+) ( )(β ) = ( ) Fn 5 2k. The right hand side above is an integer of absolute value larger than, while the left hand side above is the reciprocal of an integer. This is a contradiction. Hence, the left hand sides of (48) and (49) are non-zero. We start with a lower bound on the left hand side of inequality (48). For this, we take K := 3, = F n, 2 :=, 3 := 5. We also take b := k, b 2 := (n + r), b 3 :=. Hence, Λ 2 := b b2 2 b3 3 = F k n (n+r) 5 /2 is the expression which appears under the absolute value in the left hand side of inequality (48). We have already checked that Λ 2 = 0. Observe that, 2, 3 are all real and belong to the field K := Q( 5), so we can take D := 2. Next, since F n < n, it follows that we can take A := 2n log > D log F n = Dh( ). Next, since h( 2 ) = (log )/2 = , it follows that we can take A 2 := 0.5 > Dh( 2 ). Since h( 3 ) = (log 5)/2 = , it follows that we can take A 3 :=.6 > Dh( 3 ). Finally, Lemma 4 tells us that we can take B = r 4 > (2r) 4 > (n + r) 4 Matveev s theorem tells us that > n(n + r) 2 log(n + r) (n + r + ) max{k, (n + r), } = max{ b, b 2, b 3 }. Λ 2 > exp( C 2 ( + log B)A A 2 A 3 ), (54) where Thus, C 2 := ( + log 2) < 0 2. C 2 ( + log B)A A 2 A 3 < 0 2 (2 log ) ( + log( ) + 4 log r)n < 8 0 ( log r)n < 8 0 (8 log r)n < n log r, (55)

14 INTEGERS: 2A (202) 4 where we used the fact that log r < 8 log r, because r 300 (otherwise, that is if r 3000, then we get that n < r 3000 also, which is a case already treated). Now inequalities (5), (48), (53), and (54) show that if we put λ := min{, k, (n + r)/2, (n + r)/4}, (56) then the inequality exp n log r < 60.5 λ holds. This in term yields λ < log 60 log (log.5) n log r < n log r. (57) We already know that < k. We distinguish the following cases. Case 2. λ {(n + r)/2, (n + r)/4}. Inequality (57) gives in this case that Hence, r/4 < λ < n log r, therefore r < n log r. r < 2 (8 0 3 )n log(8 0 3 n) = n(log(8 0 3 ) + log n) < n(32. + log n) < n(33 log n) < n log n, (58) where we used the fact that n 4. With Lemma 4, we get that Also, n(n + r) log(n + r) < n(2r)(log 2r) <.4 0 (nr)(log 2 + log r) <.4 0 (6 0 5 )(n 2 log n)(log 2 + log(6 0 5 ) + 2 log n) < (n 2 log n)( log n) < 0 27 (n 2 log n)(40 log n) < n 2 (log n) 2. (59) k < (n + r + ) (2r) < (n 2 (log n) 2 )(n log n) < n 3 (log n) 3. (60) Now we assume that λ {(n + r)/2, (n + r)/4}. Since < k, we get that λ =. Hence, inequality (57) gives < n log r. (6)

15 INTEGERS: 2A (202) 5 We next apply Matveev s theorem to get a lower bound on the expression appearing in the left hand side of (49). We take K := 4, := F n, 2 :=, 3 := 5, 4 :=. We also take b := k, b 2 := (n + r + ), b 3 :=, b 4 :=. Thus, Λ 3 := F k n (n+r+) 5 /2 ( ) is the expression appearing under the absolute value in the left hand side of inequality (49). We have already checked that Λ 3 = 0. As for parameters, we have again D := 2 and we can take A := 2n log, A 2 := 0.5, A 3 :=.6 and B := r 4 as in the previous application of Matveev s theorem. As for A 4 observe that Dh( 4 ) = 2h( 4 ) log( ) + max{0, log( β )} log + log 2 < (log 2)( + ) < (2 log 2) <.4, so we can take A 4 :=.4. Thus, the left hand side of (49) is where Λ 3 > exp( C 3 ( + log B)A A 2 A 3 A 4 ), (62) C 3 := ( + log 2) < Thus, using part of the calculation from (55), we get that the expression under the exponential in (62) is bounded as C 3 ( + log B)A A 2 A 3 A 4 <. 0 4 ( + log( r 4 )) (2n log ) Using also inequality (6), we get that (.4) < (n)( log r) < (n)(8 log r) < 0 5 (n) log r. (63) C 3 ( + log B)A A 2 A 3 A 4 < n 2 (log r) 2 < n 2 (log r) 2. (64) We now compare bound (62) with bound (49) and use also inequality (53), to get that exp( C 3 ( + log B)A A 2 A 3 A 4 ) < 40.5 λ2, where λ 2 := min{k, (n + r)/2, (n + r)/4}. Hence, λ 2 < We now distinguish the following cases. Case 3. λ 2 {(n + r)/2, (n + r)/4}. log 40 log.5 + C 3( + log B)A A 2 A 3 A 4 (log.5). (65)

16 INTEGERS: 2A (202) 6 In this case, we work with inequality (64) to get that inequality (65) implies that giving r/4 λ 2 log 40 log (log.5) n 2 (log r) 2 < n 2 (log r) 2, r < n 2 (log r) 2. One checks easily that if A > 00, then the inequality x (log x) 2 < A implies x < 4A(log A) 2. Indeed, for if not, since the function x/(log x) 2 is increasing for x > e 2, it follows that 4A(log A) 2 (log(4a(log A) 2 )) 2 x (log x) 2 < A, giving 2 log A < log(4a(log A) 2 ), or A 2 < 4A log(a) 2, or A < 4(log A) 2, which gives A < 75. Applying this with A := n 2, we get that r < 4 ( )n 2 (log( n 2 )) 2 With inequality (6), we get < n 2 (log( ) + 2 log n) 2 < n 2 ( log n) 2 < n 2 (70 log n) 2 < n 2 (log n) 2. (66) < n log( n 4 ) < ( log n) < n(82 log n) < n log n. (67) Finally, using inequalities (66), (67) and (2), we also get k (n + r + ) 2nr n 3 (log n) 3 = n 3 (log n) 3. (68) Now we move on to the last case, namely: Case 4. λ 2 = k. Then inequality (65) together with inequality (63) gives k < log 40 log (log.5) n log r < n log r. From Lemma 3, we deduce that r < (n + r) < k(n + 2) < n(n + 2) log r n(.5n) log r < n 2 log r,

17 INTEGERS: 2A (202) 7 giving Hence, r < n 2 log r. r < n 2 log(4 0 5 n 2 ) < n 2 ( log n) < (26 log n) < n 2 log n. (69) With inequality (6), we get that < n log( n 3 ) < n( log n) < n(32 log n) < n log n, (70) while by (2), (69) and (70), we get k (n + r + ) 2r < (n log n)(n 2 log n) < n 3 (log n) 2. (7) Let us summarize what we have done. The following lemma follows by picking up the worse upper bounds for r, and k from estimates (52), (58), (66), (69) (for r), (50), (59), (67), (70) (for ) and (5), (60), (68) and (7) (for k), respectively. Lemma 7. If (k,, n, r) is a solution of equation (2) with n + r 3002, then the estimates hold. r 0 34 n 2 (log n) 2 ; (72) 0 29 n 2 (log n) 2 ; (73) k 0 49 n 3 (log n) 3 (74) 6. The Case of the Small n Here, we treat the case when n Then, by Lemmas 3 and 7, we have (n + r) k(n + 2) 0 49 n 3 (log n) 3 (n + 2) < for n (75) From inequality (48), we infer that where Fn k (n+r) 5 /2 < 60.5, λ3 λ 3 := min{, k, (n + r)/2, (n + r t)/2}. (76)

18 INTEGERS: 2A (202) 8 Put Λ 4 := F k n (n+r) 5 /2, Γ 4 := k log F n (n + r) log log 5. (77) Assume that λ 3 3. We then have that which gives that e Γ4 < 3/2. Hence, e Γ4 = Λ 4 < 60.5 λ3 < 3, Γ 4 < e Γ4 e Γ4 <.5 Λ 4 < 00.5 λ3. Observe that Γ 4 is an expression of the form x log F n + y log + z log 5, (78) where x := k, y := (n + r), z := are integers with max{ x, y, z } (see (75)). For each n [4, 3000], we used the LLL algorithm to compute a lower bound for the smallest nonzero number of the form (78) with integer coefficients x, y, z not exceeding in absolute value. We followed the method described in [5, Section 2.3.5], which provides such bound using the approximation for the shortest vector in the corresponding lattice obtained by LLL algorithm. In these computations, we used the PARI/GP function qflll. The minimal such value is > 00/.5 750, which gives that λ Observe that since n 3000 and we already covered the range when both n and r were in [, 3000], it follows that r > In particular, (n+r)/2 > 500 and r > n, therefore t = (n+r)/2. Thus, (n + r t)/2 (n + r)/4 >. Since also k >, we learn from this computation that = λ 3 and 750. Next we move to inequality (49) and rewrite it as Fn k (n+r) 5 /2 ( ) < 40.5, λ4 where We now put λ 4 := min{k, (n + r)/2, (n + r t)/2}. (79) Λ 5 := F k n (n+r) 5 /2 ( ), Γ 5 := k log F n + (n + r) log + log(5 /2 ( ). (80) Assume that λ 4 2. We then have that e Γ5 = Λ 5 < 40.5 λ4 < 3,

19 INTEGERS: 2A (202) 9 which gives that e Γ5 < 3/2. Hence, Γ 5 < e Γ5 e Γ5 <.5 Λ 5 < 60.5 λ4. Observe that Γ 5 is an expression of the form x log + y log 2 + log 3, (8) where := F n, 2 :=, 3 = 5 /2 ( ), and x := k, y := (n + r). Since n 3000, by Lemma 7, we have that max{ x, y } 0 49 (n log n) 3 < (82) For each n [4, 3000] and each [, 750], we performed the LLL algorithm to find a lower bound on the smallest number of the form (8) whose coefficients x, y are integers satisfying (82). In each case, we got that this lower bound is > 60/.5 800, which gives that λ Again, we have (n + r)/2 > 500; hence, λ 4 = k or λ 4 = (n + r t)/2. From what we have seen, (n + r t)/2 (n + r)/4 > 750 and this last number is 500 > 800 unless =. Thus, we always have k 800, unless =, and then (n + r)/4 800, which gives n + r We first deal with the second possibility. Fix n 3000 and r such that n + r Let =. Then N = F n+ + + F n+r is known. Furthermore, by inequality () and estimate (2), we get that N = Fn ( k + S), where 0 < S < 3/.5 k < 2/3 for k 4. Thus, (3/5)N < Fn k < N, therefore log N log(5/3) < k < log N. (83) log F n log F n For n and r fixed, there is at most one k satisfying inequalities (83). When this exists, we tested whether with this value of k the quadruple (k,, n, r) does indeed satisfy equation (2). No new solution turned up. So, from now on, we assume that k 800. Thus, (n + r) k(n + 2) < We now fix again n and apply again the LLL algorithm to get a lower bound on the minimum absolute value of the nonzero numbers of the form (78) where now x, y, z are integer coefficients of absolute values In all cases, we got a lower bound of 00/.5 00, which gives that λ Hence, 00. We now moved on to the number of the form (8), and for all values of n [4, 3000] and each [, 00], we applied the LLL algorithm to find a lower bound for the absolute value of the nonzero numbers of the form (8) when x, y are integer coefficients

20 INTEGERS: 2A (202) 20 not exceeding in absolute values. In all cases, we got a lower bound larger than 60/.5 30, showing that k 30. Now we covered the rest by brute force. That is, suppose that n [4, 3000] and < k are fixed in [, 00] and [4, 30], respectively. Then N = F k + + F k n is fixed and (3/5)N < F n+r < N. Thus, (3N/5) / < F n+r < N /. (84) For each fixed triple (k,, n), the above inequality gives some range for r. For each of these candidates r such that r 300, we checked whether the quadruple (k,, n, r) does indeed satisfy (2). As expected, no new solutions turned up. This completes the analysis of the case when n We record our conclusion as follows. Lemma 8. If (k,, n, r) is a solution of (2) other than (8, 2, 4, 3), then n 300. Furthermore, if =, then n + r Three Linear Forms in Logarithms Now we start working on the left hand side of equation (2) and do to it what we did to the right hand side of it in Section 3. Write m := n/2, and put N 4 := j m F k j. Then N 4 < F j j m k < (F m+2 ) k < Fn F n m 2 k < N N (n m 4)k < N (n/2 4)k n, because n 300 and k 4. Assume now that j [m +, n ]. Formula (29) gives us that By Lemma 7, we have that Fj k j β j k k = = jk ( )j. 5 /2 5 k/2 2j k 2j 049 n 3 (log n) 3 n <. n/2 The last inequality holds whenever n 572, which is the case for us. Set y := / n/2. Since n 300, it follows that y < 500 < The argument used (85)

21 INTEGERS: 2A (202) 2 to prove inequality (32), based on the inequalities (30) and (3), yields that k ( )j 2j < 2k < 2y, 2j and therefore F j k jk k = jk ( )j jk < 2y 5 k/2 5 k/2 Since y is small, we get, as in (33), that the inequality above implies that jk /5 k/2 <.5Fj k, and therefore the inequality F j k jk < 3yF j k (86) 5 k/2 holds for all j [m +, n ]. Now we sum up the above inequalities over all j getting that if we put then the inequality N N 4 N 5 = < 3y N 5 := n j=m+ n j=m+ holds. We now estimate N 5. Clearly, Note that (m+)k 5 k/2 ( k ) F k j 2j n j=m+ jk 5 k/2, n kj 5 k/2 j=m+ n j=m+ 5 k/2. F j k jk 5 k/2 F k j < 3yN (87) N 5 = nk (m+)k 5 k/2 ( k )..5F m+ k k 4 < F m+ k Fn < F n m < N N (n m 3)k < N (n 6)k/2 n. (88) From inequalities (85), (87) and (88), we get N nk 5 k/2 ( k ) N 4 + N N 4 N 5 + (m+)k 5 k/2 ( k ) < N n n/2 n < 4N. (89) n/2

22 INTEGERS: 2A (202) 22 Multiplying both sides above by k, we also get that N(k ) nk 5 k/2 < 4(k )N < 4N n/2, n/2 k and therefore Hence, Let N k nk 5 k/2 < N 4 +. n/2 k N (n )k < N k + 4 A := 5 k/2 nk 5 k/2 ( k ) n/2 and < 4 k +. (90) n/2 B := (n )k 5 k/2. From inequalities (89) and (90) together with the fact that k 4 and n 300, which implies that 4 k + < 0.6, n/2 we infer that both inequalities N < 2.5A and N < 2.5B hold. Now we put together the two inequalities (89) and (90) involving N together with the three inequalities (42), (43) and (44) involving also N, and get the following six inequalities: (n )k (n+r) 5 k/2 5 /2 < 8N (n )k k + n/2 + + (n+r)/2 + (n+r t)/2 < 6N + + n/2 (n )k (n+r+) 5 k/2 5 /2 ( ) < 8N k + + n/2 (n+r)/2 + (n+r t)/2 ; (n+r t)/2 < 6N k + + n/2 (t+) ( (n+r t) ) 5 k/2 5 /2 ( ) < 8N k + + n/2 (n+r)/2 < 6N k + ; n/2 (n+r t)/2 ;

23 INTEGERS: 2A (202) 23 nk 5 k/2 ( k ) (n+r) 5 /2 < 8N n/2 + + (n+r)/2 + (n+r t)/2 < 6N + + n/2 nk 5 k/2 ( k ) (n+r+) 5 /2 ( ) < 8N + n/2 + (n+r)/2 < 6N + ; n/2 (n+r t)/2 nk 5 k/2 ( k ) (t+) ( (n+r t) ) 5 /2 ( ) < 8N + n/2 (n+r)/2 < 6N. n/2 (n+r t)/2 ; (n+r t)/2 We will actually not use the third, fourth or sixth inequality above; we will only use the first, second and fifth. Since each one of them involves either A or B (but not both), it follows that dividing both sides of the respective inequality by its A or B term and using the fact that N < 2.5 max{a, B}, we get the following three inequalities: (n+r) (n )k 5 (k )/2 < ; (9) n/2 (n+r t)/2 k + n/2 + (n+r+) (n )k 5 (k )/2 ( ) < 40 (n+r+) nk 5 (k )/2 k < 40 + n/2 (n+r t)/2 (n+r t)/2 ; (92). (93) We next comment on the size of (n + r t). Assume first that r n. Then t = (n + r)/2, and therefore (n + r t) (n + r (n + r)/2) (n + r)/2 n/2. Otherwise, we have t = n, and therefore (n + r t) = r. But note that Lemma 3 tells us that the inequality holds. This can be rewritten as k(n ) (n + r) + 2k + k(n 3) n + r + = (n 3) + r + 4 (n 3) + 5r, giving that (n 3)(k ) 5r.

24 INTEGERS: 2A (202) 24 Hence, (n + r t) = r (n 3)(k )/5 holds when r < n. To summarize, we always have (n + r t)/2 (n 3)(k )/0. (94) In conclusion, on the right-hand side of inequalities (9) (93), the exponent of in the last term which is (n + r t)/2 is always of comparable size, at least, with the exponent of in the previous term which is n/2. We record the conclusions of this section as follows. Lemma 9. If (k,, n, r) is a positive integer solution of equation (2) other than (8, 2, 4, 3), then inequalities (9) (93) hold. Moreover, (n + r t)/2 (n 3)(k )/0. 8. Logarithmic Bounds for and k Our next goal is to bound and k as logarithmic functions in n. This will be achieved by applying Matveev s theorem to bound from below the left hand sides of (9) (93). Let us see whether there are instances in which the left hand side of one of these three inequalities can be zero. If the left hand side of (9) is zero, we then get that 2(n )k 2(n+r) = 5 k. Since no power of of nonzero integer exponent can be an integer, it follows that (n )k = (n + r) and k =, but this is impossible. If the left hand side of (92) is zero, we then get that (n+r+) (n )k 5 (k )/2 =. (95) Conjugating the above relation in Q( 5) and multiplying the two resulting equations we get ( ) (n+r+) (n )k+(k ) 5 k = ( )(β ) = β + + ( ). (96) It is well known that m + β m = L m, where (L m ) m 0 is the Lucas sequence given by L 0 = 0, L = and L m+2 = L m+ + L m for all m 0. Hence, equation (96) above is L ( ) = ±5 k. If is odd, we then get L = ±5 k, which is impossible since no member of the Lucas sequence is a multiple of 5. If is even, then L L 2 = 3, so that L ( ) = L 2 is positive. If /2 is odd, then 5 k = L 2 = + β 2 = + β + 2(β) /2 = ( /2 + β /2 ) 2 = L 2 /2, which is also impossible since L /2 cannot be a multiple of 5. multiple of 4, we then get that Finally, if is a 5 k = L 2 = ( /2 β /2 ) 2 = 5F 2 /2,

25 INTEGERS: 2A (202) 25 so F/2 2 = 5k. By the Primitive Divisor Theorem, the only Fibonacci numbers which are powers of 5 are = F = F 2 and 5 = F 5. Thus, /2 = 2, therefore = 4 and k =, so k = 5. Since 4 = 5 2, equation (95) also gives 4(n + r + ) 5(n ) = 2, therefore n = 4r + 7. The conclusion is that the left hand side of inequality (92) is nonzero except when (k,, n, r) = (5, 4, 4r + 7, r). However, later we shall use inequality (9) to get some bound for, and then inequality (92) to get some bound for k. If have already been bounded (like it is the case when = 4 and k = 5), then we will move on to inequality (93). Let us now check that the left hand side of inequality (93) is nonzero. Assuming that it is, we get the equation (n+r+) nk k = 5. (k )/2 Conjugating the above relation in Q( 5) and multiplying the two resulting relations, we get L k ( ) k L ( ) = ( k )(β k ) ( )(β ) = 5 k 5. Hence, since k 4, so L k 7, we get that 5L k 5 L k ( ) k L ( ) L + 2 < L k + 2, giving 4L k < 2, which is impossible. Hence, the left hand side of inequality (93) cannot be zero. We now apply Matveev s theorem to the left hand sides of inequalities (9) (93). We start with bounding by applying Matveev s theorem to inequality (9). We already checked that this is nonzero. Let λ 5 := min{, n/2, (n+r t)/2}, and note that (n + r t)/2 min{n/2, (n 3)(k )/0} (see Lemma 9). From inequality (9), we obtain (n+r) (n )k 5 (k )/2 < 20. (97) λ5 We apply Matveev s theorem to the left hand side of the above inequality with K := 2, :=, 2 := 5, b := (n + r) (n )k, b 2 := k l, D := 2. We take as in prior applications of this theorem A := log, A 2 := log 5. Furthermore, by (6), we have that max{ b, b 2 } 2k +, and by (73) and (74), we may take B := n 3 (log n) 3. Now, Matveev s theorem and inequality (97) give us that log 20 λ 5 < log ( + log( n 3 (log n) 3 ).

26 INTEGERS: 2A (202) 26 Since we know that n 300, we get that λ 5 <.59 0 log n. Assume that λ 5 =. Then λ 5 (n 3)(k )/0. But then n < log n, which gives that n < Now we apply continued fractions to a variant of the inequality (97), which is log 5 (n )k (n + r) log k < 240 λ5 (k ) log < 2(k ) 2. (98) The above inequality holds because λ 5 (n 3)(k )/0 299(k ). Hence, we get that ((n )k (n + r))/(k ) = p i /q i for some convergent p i /q i of γ := log 5/ log. We computed all convergents p i /q i of γ satisfying q i < , which is an upper bound for k when n < by Lemma 7. We find that all convergents in that range satisfy q i log 5 p i log > Hence, we conclude that λ Since λ 5 (n 3)/2, we get that n 938, which contradicts previously established result that n 300. Thus, we have shown that λ 5 =, and so <.59 0 log n. (99) We will now establish a similar logarithmic bound for k using inequality (92). The cases = and = 2 will be treated separately because in these cases is a power of. From (92), for = we get while for = 2 we get n+r+2 (n )k 5 (k )/2 < 20, λ6 2n+2r+ (n )k 5 (k )/2 < 20, λ6 where λ 6 := min{k, n/2, (n + r t)/2}. The left hand sides of the two inequalities above are nonzero by the argument used to prove that the left hand side of inequality (9) is nonzero, since assuming that it were zero, we would get that k =, which is not allowed. Thus, we may apply again Matveev s theorem as previously with :=, 2 := 5. After some calculation, we get the same bound for λ 6 as the bound obtained previously for λ 5. Hence, λ 6 <.59 0 log n in this case. The

27 INTEGERS: 2A (202) 27 assumption that λ 6 = k leads to a contradiction as before, and thus we obtain that for {, 2} we have k <.59 0 log n. (00) Assume now that 3 but (, k) = (4, 5). From inequality (92), we get (n+r+) (n )k 5 (k )/2 ( ) < 20. (0) λ6 We apply Matveev s theorem to inequality (0). We have K := 3, :=, 2 := 5, 3 = ( ), b := (n + r + ) (n )k, b 2 := k l, b 3 :=, D := 2. We take A := log, A 2 := log 5, A 3 :=. Here we use that 3. Note that, by (99), we have A 3 <.59 0 log n. Furthermore, by inequality (6), we have max{ b, b 2, b 3 } 2k + 2, and by inequalities (73) and (73) we may take B := n 3 (log n) 3. Now, Matveev s theorem and inequality (0) give us that log 20 λ 6 < log ( + log( n 3 (log n) 3 ) log n. Since we know that n 300, we get that Assume that λ 6 = k. Then λ 6 (n 3)/0. But then λ 6 < (log n) 2. (02) n < (log n) 2, which gives that n < The application of the above described continued fraction method for inequality (98) in this new range for n gives that = λ Now we apply the LLL algorithm, as explained in Section 6, to find a lower bound for the smallest nonzero value of a number of form x log + y log 5 ± log( ), (03) with max{ x, y } < , which is the bound for (n + r + ) (n )k 2k + 2 when n < , by Lemma 7. The computation shows that this minimal value is > 240/ 400, which gives that λ If λ 6 = n/2, we get n 2880, contradicting the bound n If λ 6 = (n + r t)/2, then from (n 3)(k )/0 λ and n 3000, we get that k 4 and k <.6 0 log n. (04) It remains to treat the case when λ 6 = k. Then, by (02), we have k < (log n) 2. (05) We summarize the results of this section in the following lemma. Lemma 0. Let (k,, n, r) be a solution of equation (2). Then <.59 0 log n and k < (log n) 2. If {, 2}, then k <.59 0 log n.

28 INTEGERS: 2A (202) Absolute Upper Bound for n and the End of the Proof Now that we have upper bounds for and k as logarithmic functions of n (see Lemma 0), we may apply Matveev s theorem to the left hand side of inequality (93), in order to obtain an absolute upper bound for n. We have already checked that this expression is nonzero. We take K := 3, :=, 2 := 5, 3 := ( k )/( ), b := (n + r + ) nk, b 2 := k l, b 3 :=, D := 2. We also take A := log, A 2 := log 5, A 3 := k. By Lemma 0, we have A 3 < (log n) 2 and by (6), max{ b, b 2, b 3 } 3k + 2, so by (73) and (73) we may take B := (log n) 2. Now, Matveev s theorem and inequality (93) give us that min{n/2, (n + r t)/2} < (log 80) log ( + log( (log n) 2 ))(log n) 2. Since, (n + r t)/2 (n 3)/0, we get an absolute upper bound for n, namely n < (06) Inserting the bound for n given by (06) in the bound for k from Lemma 0, we get k < Applying the continued fraction method to inequality (98) for k < gives that = λ Applying the LLL algorithm to the numbers of the form (03) with the bounds k < and < 55 gives that λ 6 = k 30. Now we consider inequality (93). For k 30 and min{55, k }, we compute the smallest value of x 5 k ( k )/( ), for an integer x. We get that this value is always > 80/ 30. From inequality (93), we obtain that (n 3)/0 30, i.e. n 303, a contradiction. Hence, Theorem 2 is proved. Then we had a beer. Acknowledgements. We thank the referee for pointing out some relevant references. This work was done while A. D. visited the Mathematical Institute of the UNAM in Morelia in December 200. He thanks the people of this institution for their hospitality and support. S. D. A. and F. L. were supported in part by the joint project France Mexico 2469 Linear recurrences, arithmetic functions and additive combinatorics, and F. L. was also supported in part by Grant SEP-CONACyT

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