ON X-COORDINATES OF PELL EQUATIONS WHICH ARE REPDIGITS

ON X-COORDINATES OF PELL EQUATIONS WHICH ARE REPDIGITS BERNADETTE FAYE AND FLORIAN LUCA Abstract. Let b ≥ 2 be a given integer. In this paper, we show that there only finitely many positive integers d which are not squares, such that the Pell equation X 2 − dY 2 = 1 has two positive integer solutions (X, Y ) with the property that their X-coordinates are base b-repdigits. Recall that a base b-repdigit is a positive integer all whose digits have the same value when written in base b. We also give an upper bound on the largest such d in terms of b. 1. Introduction Let d > 1 be a positive integer which is not a perfect square. It is well known that the Pell equation X 2 − dY 2 = 1 (1.1) has infinitely many positive integer solutions (X, Y ). Furthermore, putting (X1 , Y1 ) for the smallest solution, all solutions are of the form (Xn , Yn ) for some positive integer n, where √ √ Xn + dYn = (X1 + dY1 )n . There are many papers in the literature which treat Diophantine equations involving members of the sequences {Xn }n≥1 and/or {Yn }n≥1 , such as when are these numbers squares, or perfect powers of fixed or variable exponents of some others positive integers, or Fibonacci numbers, etc. (see, for example, [2], [4], [5], [8], [9]). Let b ≥ 2 be an integer. A natural number N is called a base b-repdigit if all of its base b-digits are equal. Setting a ∈ {1, 2, . . . , b − 1} for the common value of the repeating digit and m for the number of base b-digits of N , we have   m b −1 . (1.2) N =a b−1 When a = 1, such numbers are called base b-repunits. When b = 10, we omit mentioning the base and simply say that N is a repdigit. In [6], A. Dossavi-Yovo, F. Luca and A. Togbé proved that when d is fixed there is at most one n such that Xn is a repdigit except when d = 2 (for which both X1 = 3 and X3 = 99 are repdigits) or d = 3 (for which both X1 = 2 and X2 = 7 are repdigits). In this paper, we prove that the analogous result holds if we replace “repdigits” by “base b-repdigits”, namely that there is at most one n such that Xn is a base b-repdigit except for finitely many d, and give an explicit bound depending on b on the largest possible exceptional d. We thank the anonymous referee for a careful reading of the manuscript. B. F. thanks OWSD and Sida (Swedish International Development Cooperation Agency) for a scholarship during her Ph.D. studies at Wits. She also thanks Mouhamed M. Fall of AIMS-Senegal for useful conversations. Parts of this research work was done when F. L. visited the Max Planck Institute for Mathematics in Bonn, Germany in 2017. He wants to thank the institution for the invitation. He is also supported by grants CPRR160325161141 and an A-rated researcher award both from the NRF of South Africa and by grant no. 17-02804S of the Czech Granting Agency. MONTH YEAR 1 THE FIBONACCI QUARTERLY Of course, for every integer X ≥ 2, there is a unique square-free integer d ≥ 2 such that X 2 − 1 = dY 2 (1.3) for some positive integer Y . In particular, if we start with an N given as in (1.2), then N is the X-coordinate of the solution to the Pell equation corresponding to the number d obtained as in (1.3). Thus the problem becomes interesting only when we ask that the Pell equation corresponding to some d (fixed or variable) has at least two positive integer solutions whose X-coordinates are base b-repdigits. Here, we apply the method from [6], together with an explicit estimate on the absolute value of the largest integer solution to an elliptic equation and prove the following result. Theorem 1.1. Let b ≥ 2 be fixed. Let d ≥ 2 be squarefree and let (Xn , Yn ) := (Xn (d), Yn (d)) be the nth positive integer solution of the Pell equation X 2 − dY 2 = 1. If the Diophantine equation   m b −1 with a ∈ {1, 2, . . . , b − 1} (1.4) Xn = a b−1 has two positive integer solutions (n, a, m), then   5 d ≤ exp (10b)10 . (1.5) The proof proceeds in two cases according to whether n is even or odd. If n is even, we reduce the problem to the study of integer points on some elliptic curves of a particular form. Here, we use an upper bound on the naive height of the largest such point due to Baker. When n is odd, we use lower bounds for linear forms in complex and p-adic logarithms. For a number field K, a nonzero algebraic number η ∈ K and a prime ideal π of OK , we use νπ (η) for the exact exponent of π in the factorization in prime ideals of the principal fractional ideal ηOK generated by η in K. When K = Q is the field of rational numbers and π is some prime number p, then νp (η) coincides with the exponent of p in the factorization of the rational number η. 2. Case when some n is even Assume that n satisfies (1.4). Put n = 2n1 . Then using known formulas for the solutions to Pell equations, (1.4) implies that  m  b −1 2 . (2.1) 2Xn1 − 1 = X2n1 = Xn = a b−1 Assume first that a = b − 1. Then (2.1) gives 2Xn21 = bm . (2.2) We first deduce that b is even. If m = 1, then d ≤ dYn21 = Xn21 − 1 < 2Xn21 = b, so d < b, which is a much better inequality than the one we aim for in general. From now on, we assume that m > 1. Then Xn21 = bm /2. Since m > 1, it follows that Xn1 is even. This shows that n1 is odd, for otherwise, if n1 = 2n2 is even, then Xn1 = X2n2 = 2Xn22 −1 is odd, a contradiction. Furthermore, the prime factors of Xn1 are exactly all the prime factors of b. Let us show that n is then unique. Indeed, assume that there exists n′ = 2n′1 such that ′ (n′ , b − 1, m′ ) is a solution of (1.4) and n′ 6= n. Then also Xn2′ = bm /2, so Xn1 and Xn′1 1 have the same set of prime factors. Since Xn1 = Y2n1 /(2Yn1 ) and Xn′1 = Y2n′1 /(2Yn′1 ), the conclusion that Xn1 and Xn′1 have the same set of prime factors is false if max{n1 , n′1 } ≥ 7 2 VOLUME, NUMBER ON X-COORDINATES OF PELL EQUATIONS WHICH ARE REPDIGITS by Carmichael’s Theorem on Primitive Divisors for the sequence {Ys }s≥1 (namely that Yk has a prime factor not dividing any Ys for any s < k provided that k ≥ 13, see [3]). Thus, n1 , n′1 ∈ {1, 3, 5} and one checks by hand that no two of X1 , X3 , X5 can have the same set of prime factors. Thus, n is unique and noting that m is odd (for example, because the exponent of 2 in the left–hand side of (2.2) is odd), we get that 2(m−1)/2 divides Xn1 . It is known from the theory of Pell equations that ν2 (Xn1 ) = ν2 (X1 ). Hence, X1 is a multiple of 2(m−1)/2 . Further, since we are assuming that equation (1.4) has two solutions (n, a, m), it follows that there exists another solution either with n even and a 6= b − 1, or with n odd. We now move on to analyze the case in which there exists a solution with n even and a 6= b − 1. Put m = 3m0 + r with r ∈ {0, 1, 2}. Here, m0 is a non-negative integer. Putting x := Xn1 and y := bm0 , equation (2.1) becomes   r 3 b y −1 . (2.3) 2x2 − 1 = a b−1 Equation (2.3) leads to X 2 = Y 3 + A0 , (2.4) where X := 4a(b − 1)2 br x, Y := 2a(b − 1)br y, A0 := 8a2 (b − 1)3 b2r ((b − 1) − a). If r = 0, then A0 < 2b6 ≤ 0.25b10 (here, we used the fact that a(b−1−a) ≤ ((b−1)/2)2 < b2 /4, a consequence of the AGM inequality). If r ∈ {1, 2}, then since one of b or b − 1 is even we get that X ′2 = Y ′3 + A′0 , (2.5) holds with integers (X ′ , Y ′ , A′0 ) = (X/23 , Y /22 , A0 /26 ) and A′0 = A0 /26 < 0.25b10 . Note that A0 A′0 6= 0. Let us now recall the following result of Baker (see [1]). Theorem 2.1. Let A0 6= 0. Then all integer solutions (X, Y ) of (2.4) satisfy 4 max{|X|, |Y |} < exp{(1010 |A0 |)10 }. We will apply the above theorem to equation (2.4) for r = 0 and to equation (2.5) for r ∈ {1, 2}. Note that since |A0 | < 0.25 · b10 when r = 0 and |A′0 | < 0.25b10 when r ∈ {1, 2}, we get that 4 4 5 (1010 |A0 |)10 ≤ (0.25 · 1010 b10 )10 < 0.25(10b)10 , and a similar inequality holds for A0 replaced by A′0 . Theorem 2.1 applied to equations (2.4), (2.5) tells us that in both cases 5 Since X ≥ Xn1 > √ X/23 < exp(0.25(10b)10 ). d, we get that 5 5 d < 26 exp(0.5(10b)10 ) < exp((10b)10 ), which is what we wanted. So, let us conclude this section by summarizing what we have proved. Lemma 2.2. Assume that there is a solution (n, a, m) to equation (1.4) with n even. Then one of the following holds: MONTH YEAR 3 THE FIBONACCI QUARTERLY (i) a < b − 1 and 5 d < exp((10b)10 ). (ii) a = b − 1, m = 1, and d < b. (iii) a = b − 1, m > 1. Then b is even, m is odd and n = 2n1 is unique with this property, Xn1 = bm /2 and 2(m−1)/2 divides X1 . 3. On the greatest divisor of repdigits From now on, we look at the case when equation (1.4) has solutions (n, a, m) with n odd. Say,   m b −1 . Xn = a b−1 If some other solution (n′ , a′ , m′ ) to equation (1.4) does not have n′ odd, then n′ must be even so Lemma 2.2 applies to it. If we are in one of the instances (i) or (ii) of Lemma 2.2, then we are done. So, let us assume that we are in instance (iii) of Lemma 2.2, so n′ is even, m ′ is odd and 2(m −1)/2 divides X1 . But since n is odd, X1 also divides Xn . Since b is even, ′ ′ ′ (bm − 1)/(b − 1) is odd, so 2(m −1)/2 divides a. Hence, 2m −1 ≤ 2a2 ≤ b3 . Since m′ ≤ 2m −1 , we get that m′ ≤ b3 . Writing n′ = 2n′1 , and using again (iii) of Lemma 2.2, we get that ′ ′ 3 d < X12 ≤ Xn2′ = bm /2 < bm < bb < exp(b4 ), 1 which is good enough for us. From now on, we assume that both solutions of equation (1.4) have an odd value for n. Let such indices be n1 6= n2 . Then  m1  m2   b −1 b −1 Xn1 = a1 , Xn 2 = a 2 , where a1 , a2 ∈ {1, 2, . . . , b − 1}. b−1 b−1 Let n3 := gcd(n1 , n2 ). Since n1 and n2 are odd, from known properties of solutions to the Pell equation, we get that Xn3 = gcd(Xn1 , Xn2 ). ′ We put a3 := gcd(a1 , a2 ), a1 := a1 /a3 , a′2 := a2 /a3 . We also put m3 := gcd(m1 , m2 ) and use the fact that gcd(bm1 − 1, bm2 − 1) = bm3 − 1. We then get, Xn 3 = gcd(Xn1 , Xn2 )   bm 1 − 1 bm 2 − 1 = gcd a1 , a2 b−1 b−1  m3    m2 − 1 m1 − 1 b −1 ′ b ′ b = a3 ,a gcd a1 m3 b−1 b − 1 2 bm 3 − 1  m3  b −1 := a3 c . b−1 (3.1) The quantities inside the greatest common divisor denoted by c have the properties that a′1 , a′2 are coprime, (bm1 − 1)/(bm3 − 1) and (bm2 − 1)/(bm3 − 1) are also coprime. Thus,    m1  b −1 ′ bm 2 − 1 , a2 , gcd m3 c = gcd a′1 , m3 b −1 b −1 4 VOLUME, NUMBER ON X-COORDINATES OF PELL EQUATIONS WHICH ARE REPDIGITS which implies that c ≤ a′1 a′2 = (a1 a2 )/a23 . Hence, a3 c ≤ (a1 a2 )/a3 ≤ (b − 1)2 . Replacing a3 c by a3 , we retain the conclusion that   m3 b −1 , where a3 ∈ {1, 2, 3, . . . , (b − 1)2 }. Xn3 = a3 b−1 Since n1 6= n2 , we may assume that n1 < n2 , and then n3 < n2 and n3 is a proper divisor of n2 . Putting n := n2 /n3 , D := Xn23 − 1 ≥ d, m := m3 , ℓ := m2 /m3 and relabeling a2 and a3 as c and a, respectively, we can restate the problem now as follows: Problem 3.1. What can you say about D such that   m b −1 , with a ∈ {1, 2, 3, . . . , (b − 1)2 } X1 = a b−1   mℓ b −1 Xn = c , with c ∈ {1, 2, 3, . . . , b − 1}, b−1 (3.2) where n > 1 is odd and ℓ, m are positive integers. From now on, we work with the system (3.2). By abuse of notation, we continue to use d instead of D. 4. Bounding ℓ in terms of n We may assume that m ≥ 100 otherwise √ d < X1 < bm+1 ≤ b101 , so d < b202 , which is better than the inequality (1.5). We put q √ α := X1 + X12 − 1 = X1 + dY1 . On one hand, from the first relation of (3.2), we have that   m
b −1 1 −1 , α+α =a X1 = 2 b−1 so   m √
1 b −1 −1 α = X1 + dY1 > X1 = > bm−1 . α+α =a 2 b−1 (4.1) Hence, α > bm−1 implying m−1< log α . log b One the other hand, √  m  X1 + dY1 α 1 b −1 −1 = < (α + α ) = a ≤ a(bm − 1) < abm < bm+2 . 2 2 2 b−1 Hence, α < 2bm+2 ≤ bm+3 implying (4.2) log α < m + 3. log b Thus, m−1< MONTH YEAR log α < m + 3. log b (4.3) 5 THE FIBONACCI QUARTERLY We now exploit the second relation of (3.2). On one hand, we get that   mℓ b −1 n > bmℓ−1 , α > Xn = c b−1 therefore (4.4)   log α mℓ − 1 < n < n(m + 3), log b where the last inequality follows from (4.3). Since m is large (m ≥ 100), we certainly get that ℓ < 2n + 1. (4.5) On the other hand,
1 n αn < α + α−n = Xn < bmℓ < (bα)ℓ , 2 2 where the last inequality follows from (4.1). The above inequalities lead to αn < (2bα)ℓ < (α3/2 )ℓ , (4.6) where we used the fact that α1/2 > 2b, or α > 4b2 , which follows from the fact that α > bm−1 together with the fact that m ≥ 100. Inequality (4.6) yields ℓ > 2n/3, (in particular, ℓ > 1 since n ≥ 3), which together with (4.5) gives 2n/3 < ℓ < 2n + 1. (4.7) 5. Bounding m in terms of n Here, we use the Chebyshev polynomial Pn (X) ∈ Z[X] for which Pn (X1 ) = Xn . Recall that  p p 1 Pn (X) = (X + X 2 − 1)n + (X − X 2 − 1)n . 2 Using the second relation of (3.2), we have by Taylor’s formula: (c/(b − 1))bmℓ − c/(b − 1) = Xn = Pn (X1 ) = Pn ((a/(b − 1))bm − a/(b − 1)) = Pn (−a/(b − 1)) + Pn′ (−a/(b (5.1) − 1))(a/(b − 1))b m (mod b2m ) Here, 1/(b − 1) (mod bm ) is to be interpreted as the multiplicative inverse of b − 1 modulo bm , which exists since b − 1 and b are coprime. Case 1. Suppose that a = b − 1. Then X1 + 1 = bm . Put Y = X + 1 and denote n  n  p p 1  2 2 Qn (Y ) := Pn (Y − 1) = . Y − 1 + (Y − 1) − 1 + Y − 1 − (Y − 1) − 1 2 It was proved in [6] that Qn (0) = −1 and dQn (Y ) dY Y =0 = Q′n (0) = n2 . By Taylor’s formula again, we get Pn (X) = Qn (X + 1) = n2 (X + 1) − 1 6 (mod (X + 1)2 ). VOLUME, NUMBER ON X-COORDINATES OF PELL EQUATIONS WHICH ARE REPDIGITS Specializing at X = X1 and a = b − 1 and using the fact that ℓ > 1 (see (4.7)), equation (5.1) becomes −c/(b − 1) ≡ Xn (mod b2m ) ≡ −1 + n2 bm (mod b2m ). If c 6= b − 1, we get that bm | b − 1 − c. Since m ≥ 100 and c < b2 and c 6= b − 1, we get that b100 ≤ |b − 1 − c| < b2 , a contradiction. If c = b − 1, then −1 ≡ −1 + n2 bm (mod b2m ). The above congruence implies that bm | n2 . (5.2) Since n is odd, b is odd also. Hence, b ≥ 3. Thus, b m ≤n 2 therefore m≤2  log n log b  < 2 log n. (5.3) Case 2. Suppose that a 6= b − 1. We put Equation (5.1) gives β := −a/(b − 1) + p (−a/(b − 1))2 − 1. bm | Pn (−a/(b − 1)) + c/(b − 1), (5.4) bm where the above divisibility is to be interpreted that divides the numerator of the rational number Pn (−a/(b − 1)) + c/(b − 1) written in reduced form. We observe that Pn (−a/(b − 1)) + c/(b − 1) = (1/2)β −n (β n − γ)(β n − γ −1 ), p where γ := −c/(b − 1) + (−c/(b − 1))2 − 1. Hence, bm | (β n − γ)(β n − γ −1 ). (5.5) It could be however that the right–hand side of (5.5) is zero and then divisibility relation (5.5) is not useful. In that case, we return to (5.1), use the fact that Pn (−a/(b − 1)) + c/(b − 1) = 0, to infer that (5.6) bm | Pn′ (−a/(b − 1)). Calculating we get Pn′ (X) = √ Thus, so   p p n (X + X 2 − 1)n − (X − X 2 − 1)n . X2 − 1
n(b − 1) bm | p β n − β −n , 2 2 a − (b − 1) bm | n(β n − β −n ). (5.7) To continue, we need the following lemma. Lemma 5.1. The simultaneous system of equations β −n = β n = γ i for some i ∈ {±1} has no solutions. MONTH YEAR 7 THE FIBONACCI QUARTERLY Proof. If β n = β −n , then β 2n = 1. Hence, β is a root of unity of order dividing 2n. Since β is rational or quadratic and n is odd, we deduce that the order of β is 1, 2, 3, 6. If the order of β is 1, 2, we then get β = ±1. With x := −a/(b − 1), we get p x + x2 − 1 = ±1, whose solutions are x = ±1. This leads to a = ±(b − 1), which is false because 1 ≤ a < b − 1. Hence, the order of β is 3, 6. It follows that p √ x + x2 − 1 = ±(1/2 ± 3i/2). This gives x = √±1/2. Hence, −a/(b − 1) = ±1/2, giving that a = (b − 1)/2, b is odd and β =p−(1/2 ± i 3/2). Thus, β n = 1, showing that γ = 1. With y = −c/(b − 1), we get  y + y 2 − 1 = 1, giving y = 1. Thus, c = −(b − 1), a contradiction. We summarize what we did so far. Lemma 5.2. (i) If a = b − 1, then (ii) If a 6= b − 1, then bm | n2 . bm | (β n − γ)(β n − γ −1 ). (5.8) bm | n(β n − β −n ), (5.9) Further, the expression appearing in the right-hand side of divisibility relation (5.8) either is nonzero, or it is zero in which case we additionally have and the expression appearing in the right–hand side of (5.9) is nonzero. Thus, bm | n2 Λ, where either Λ = 1 or Λ= s Y i=1 (5.10) (δidi − ηiei ), for some s ∈ {1, 2}, (δi , ηi ) ∈ {(β, γ), (β, β)}, (di , ei ) ∈ {(n, 1), (n, −1), (n, n)} for 1 ≤ i ≤ s and furthermore Λ 6= 0. Let K = Q[β, γ] be of degree D. Note that D ≤ 4. Let p be any prime factor of b and let π be some prime ideal in K dividing p. Then (5.3) and (5.7) tell us that m ≤ 2 max{νπ (δidi − ηiei ) : 1 ≤ i ≤ s} + 2Dνp (n). Note further that both β and γ are invertible modulo any prime dividing b. Indeed this follows, for example, because λ2 λ1 and β −1 = , β= b−1 b−1 p where λ1,2 = −a ± a2 − (b − 1)2 are algebraic integers. Thus, λ1 λ2 = b − 1, showing that if π is any prime ideal such that one of νπ (λ1 ) or νπ (λ2 ) is nonzero, then π | b − 1. In particular, π ∤ b. A similar argument applies to γ. Now, we use a linear form in p-adic logarithm due to K. Yu [11], to get an upper bound for m in terms of n. We recall the statement of Yu’s result. 8 VOLUME, NUMBER ON X-COORDINATES OF PELL EQUATIONS WHICH ARE REPDIGITS Theorem 5.3. Let α1 , . . . , αt be algebraic numbers in the field K of degree D, and b1 , . . . , bt be nonzero integers. Put Λ = α1b1 . . . αtbt − 1 and B ≥ max{|b1 |, . . . , |bt |}. Let π be a prime ideal of K sitting above the rational prime p of ramification eπ and Hi ≥ max{h(αi ), log p} for i = 1, . . . , t, where h(η) is the Weil height of η. If Λ 6= 0, then √ νπ (Λ) ≤ 19(20 t + 1D)2(t+1) · et−1 π pf π log(e5 tD)H1 · · · Ht log B. (fπ log p)2 (5.11) Here fπ is the inertia degree of π, namely that positive integer such that the finite field OK /π has cardinality pfπ . In our application, we take t = 2, fix i ∈ {1, . . . , s} and put (α1 , α2 ) = (δi , ηi ) ∈ {(β, γ), (β, β)}, (b1 , b2 ) = (di , ei ) ∈ {(n, 1), (n, −1), (n, n)}, respectively according to which Λ = α1b1 α2−b2 − 1 is nonzero. Thus, B = n and α1 , α2 ∈ {β, γ}. Hence, we get √ pf π log(8e5 )H 2 log n, (5.12) νπ (Λ) ≤ 19(20 3 · 4)6 · eπ (fπ log 2)2 where H ≥ max{h(β), h(γ), log p}. Since β, γ are roots of the polynomials (b − 1)2 X 2 + (2a)X + 1 and (b − 1)2 X 2 + (2c)X + 1, are of degree at most 2 and both these numbers as well as their conjugates are in absolute values at most p (b − 1)2 /(b − 1) + ((b − 1)2 /(b − 1)) − 1 < 2b, we conclude that max{h(β), h(γ)} < 2 log(2b) ≤ 4 log b. Since p ≤ b, we can take H = 4 log b. Furthermore, since eπ ≤ 4 and fπ ≤ 4, and Dνp (n) ≤ 4(log n)/(log 2) < 8 log n, inequalities (5.10) and (5.12) yield m ≤ 1.3 × 1017 b4 (log b)2 log n + 16 log n < 2 × 1017 b6 log n. (5.13) We record this as a lemma. Lemma 5.4. If system (3.2) has a solution, then m < 2 × 1017 b6 log n. MONTH YEAR 9 THE FIBONACCI QUARTERLY 6. Bounding n in terms of b The second equation of (3.2) gives αn + α−n = 2Xn = (2c/(b − 1))bmℓ − (2c/(b − 1)). Since the above leads to (2c/(b − 1))bmℓ − αn = α−n + (2c/(b − 1)), (6.1) 1 3 < n−2 . (6.2) αn α The left hand side of (6.2) is non-zero by the equation (6.1). We find a lower bound on it using a lower bound for a nonzero linear form in logarithms of Matveev [10] which we now state. 0 < (2c/(b − 1))bmℓ α−n − 1 < Theorem 6.1. In the notation of Theorem 5.3, assume in addition that K is real and Hi ≥ max{Dh(δi ), | log δi |, 0.16} If Λ 6= 0, then for i = 1, . . . , t. log |Λ| ≥ −1.4 · 30t+3 · t4.5 · D2 (1 + log D)(1 + log B)H1 · · · Ht . (6.3) We take t = 3, δ1 = 2c/(b − 1), δ2 = b, δ3 = α, b1 = 1, b2 = mℓ, b3 = −n. Since ℓ ≤ 2n (see (4.7)), we can take B = 2mn. Now the algebraic numbers δ1 , δ2 , δ3 belong to L = Q[α], a field of degree D = 2. Since h(δ1 ) ≤ log(2b) ≤ 2 log b, we can take H1 = 4 log b and H2 = 2 log b . Furthermore, since α is a quadratic unit, we can take H3 = log α. Thus, we get, by (6.2), that (n − 2) log α ≤ − log Λ ≤ 1.4 · 306 · 34.5 · 22 · (1 + log 2)(1 + log 2mn)(8(log b)2 ) log α, giving n ≤ 1013 (1 + log(2mn))(log b)2 . Inserting (5.13) into the above inequality we get n ≤ 1013 (1 + log(4 × 1017 b6 n log n))(log b)2 < 1013 (1 + log(4 × 1017 ) + 6 log b + 2 log n)(log b)2 < 1013 · 43 · (6 log b)(2 log n)(log b)2 < 6 × 1015 (log b)3 log n. (6.4) In the above and in what follows we use the fact that if x1 , . . . , xk are real numbers > 2, then x1 + · · · + xk ≤ x1 · · · xk . Lemma 1 in [7] says that if T > 3 and n < T, log n then n < (2T ) log T. Taking T := 6 × 1015 (log b)3 in the above implication and using (6.4), we get that n < 12 × 1015 (log b)3 log(6 × 1015 (log b)3 ) < 12 × 1015 (log b)2 (log(6 × 1015 ) + 3 log b) < 12 × 1015 (log b)3 × 37 × (3 log b) < 2 × 1018 (log b)4 . 10 VOLUME, NUMBER ON X-COORDINATES OF PELL EQUATIONS WHICH ARE REPDIGITS Inserting this back into the inequality of Lemma 5.4, we get m ≤ 2 × 1017 b6 log(2 × 1018 (log b)4 ) = 2 × 1017 b6 (log(2 × 1018 ) + 4 log b) < 2 × 1017 b6 × 43 × 4 log b < 1020 b7 . Since ℓ ≤ 2n (see (4.7)), we conclude the following result. Lemma 6.2. Under the hypothesis of Problem 3.1, we have that n < 1018 b4 , ℓ < 2 × 1018 b4 , m < 1020 b7 . Finally, from (3.2) we have that d < X12 < b2m , therefore d < b2×10 20 b7 < exp(1020 b10 ), which together with Lemma 2.2 implies the desired conclusion of Theorem 1.1. Acknowledgements We thank the referee for spotting a wrong calculation in a previous version of this manuscript. References [1] A. Baker On the representation of integers by binary forms.. Philos. Trans. A 263 (1968), 173-208. [2] Y. Bugeaud, M. Mignotte and S. Siksek, Classical and modular approaches to exponential Diophantine equations. I. Fibonacci and Lucas perfect powers, Ann. of Math. (2) 163 (2006), 969–1018. [3] R. D. Carmichael,On the numerical factors of arithmetic forms αn ± β n , Ann. Math. (2) 15 (1913), 30–70. [4] J. H. E. Cohn, The Diophantine equation x4 + 1 = Dy 2 , Math. Comp. 66 (1997), 1347–1351. [5] J. H. E. Cohn, The Diophantine equation x4 ?Dy 2 = 1.II, Acta Arith. 78 (1997), 401–403. [6] A. Dossavi-Yovo, F. Luca, A. Togbé On the x-coordinate of Pell Equation which are rep-digits, Publ. Math. Debrecen, 88 (2016), 381–399. [7] S. Gúzman and F. Luca, Linear combinations of factorials and S-units in a binary recurrence sequence, Ann. Math. Québec 38 (2014), 169–188. [8] F. Luca A note on the Pell equation. Indian J. Math. 39 (1997), 99–105. [9] F. Luca and A. Togbé, On the x-coordinates of Pell equations which are Fibonacci numbers, Math. Scand., to appear. [10] E.M. Matveev, An explicit lower bound for a homogenous rational linear form in logarithms of algebraic numbers, II, Izv. Ross. Akad. Nauk Ser. Mat. 64 (2000), 125-180. [11] K.Yu, p-adic logarithm forms and groups varieties II, Act.Arith. 89 (1999), 337-378. Ecole Doctorale de Mathématiques et d’Informatique Université Cheikh Anta Diop de Dakar BP 5005, Dakar Fann, Senegal School of Mathematics, University of the Witwatersrand Private Bag X3, Wits 2050, South Africa E-mail address: [email protected] School of Mathematics, University of the Witwatersrand Private Bag X3, Wits 2050, South Africa and Department of Mathematics, Faculty of Sciences University of Ostrava 30 dubna 22, 701 03 Ostrava 1, Czech Republic E-mail address: [email protected] MONTH YEAR 11