Rev. Real Acad. Cienc. Exactas Fis. Nat. Ser. A-Mat. (2023) 117:89 https://doi.org/10.1007/s13398-023-01425-7 ORIG INAL PAPER The Diophantine equations Pxn + Pyn+1 = Pxm or Pyn + Pxn+1 = Pxm Bernadette Faye1 · Florian Luca2,3,4 · Salah Eddine Rihane5 · Alain Togbé6 Received: 15 September 2022 / Accepted: 16 March 2023 © The Author(s) under exclusive licence to The Royal Academy of Sciences, Madrid 2023 Abstract Here, we find all positive integer solutions of the Diophantine equations in the title, where (Pn)n≥0 is the Pell sequence P0 = 0, P1 = 1 and Pn+2 = 2Pn+1 + Pn for all n ≥ 0. Keywords Pell numbers · Linear form in logarithms · Reduction method Mathematics Subject Classification 11B39 · 11J86 1 Introduction We look at the title equation in positive integer variables (n, m, x, y), where (Pk)k≥0 is the Pell sequence given by P0 = 0, P1 = 1, and Pk+2 = 2Pk+1 + Pk for all k ≥ 0. We have the following theorem. B Salah Eddine Rihane salahrihane@hotmail.fr; salaheddine.rihane@nhsm.edu.dz Bernadette Faye bernadette.fayee@gmail.com Florian Luca Florian.Luca@wits.ac.za Alain Togbé atogbe@pnw.edu 1 UFR SAT, Section Maths Appliquées, University Gaston Berger of Saint-Louis (UBG), Saint-Louis, Senegal 2 School of Mathematics, University of the Witwatersrand, Private Bag 3, Wits, Johannesburg 2050, South Africa 3 Research Group in Algebraic Structures and Applications, King Abdulaziz University, Jeddah, Saudi Arabia 4 Centro de Ciencias Matemáticas UNAM, Morelia, Mexico 5 National Higher School of Mathematics, P.O. Box 75, Mahelma, Sidi Abdellah, 16093 Algiers, Algeria 6 Department of Mathematics and Statistics, Purdue University Northwest, 1401 S, U.S. 421, Westville, IN 46391, USA 0123456789().: V,-vol 123 http://crossmark.crossref.org/dialog/?doi=10.1007/s13398-023-01425-7&domain=pdf http://orcid.org/0000-0002-4145-7942 89 Page 2 of 13 B. Faye et al. Theorem 1.1 The only solution in positive integers (n, m, x, y) of either of the equations Px n + P y n+1 = Px m or P y n + Px n+1 = Px m (1.1) is (n, m, x, y) = (1, 3, 1, 2), for which 1 + 22 = 5 = P3. (1.2) Our work complements recent results on ternary equations ax + by = cz , where a, b, c are positive integers belonging to either the Fibonacci or the Pell sequence and x, y, z are positive integers [5, 7, 9, 10, 13]. Our proof uses Carmichael’s theorem on primitive divisors for Lucas sequences, lower bounds for linear forms in two or three logarithms, both complex and p- adic, as well as the LLL reduction algorithm and reductions based on continued fractions. We will recall such results when we need them. 2 The proof Since gcd(Pn, Pn+1) = 1, it follows that in Eq. 1.1 the two summands in the left-hand side are coprime. By Fermat’s last theorem proved by Wiles, gcd(x, y) ∈ {1, 2}. Either one of Eq. 1.1 implies Px m > Px n , so m > n. The case m = n + 1 is impossible since Pn and Pn+1 are coprime. Thus, m ≥ n + 2. If y ≤ x , then Px m ≤ Px n + Px n+1 ≤ (Pn + Pn+1) x < Px n+2, so m < n + 2, which is impossible. So, we have the first lemma. Lemma 2.1 Assume that (n, m, x, y) is a solution in positive integers of equation 1.1. Then: (i) m ≥ n + 2; (ii) y > x; (iii) gcd(x, y) ∈ {1, 2}. Let (Qk)k≥0 be the companion Pell numbers given by Q0 = 2, Q1 = 2 and Qk+2 = 2Qk+1 + Qk for all k ≥ 0. We have the following well-known property. Lemma 2.2 Assume m ≡ n (mod 2). Then, we have Pm − Pn = P(m−εn)/2Q(m+εn)/2 where ε = 1, − 1, respectively according to m − n ≡ 0, 2 (mod 4). Lemma 2.3 The only solution (n, m, x, y) in positive integers of Eq. 1.1 with n ∈ {1, 2} is (n, m, x, y) = (1, 3, 1, 2). Proof If n = 1, the equation is one of 1 + 2y = Px m or 1 + 2x = Px m . The second one is impossible for x > 1 since Px m − 2x = (Pm − 2)(Px−1 m + · · · + 2x−1) > 1 and also for x = 1 since 3 is not a Pell number. The first one is impossible for x > 1 by what is known about the Catalan equation solved by Mihăilescu [12]. So, x = 1. Since Pn ≡ n (mod 2), it follows that m is odd. Hence, we obtain 2y = Pm − 1 = Pm − P1 = P(m±1)/2Q(m∓1)/2. 123 The Diophantine equations Pxn + Pyn+1 = Pxm ... Page 3 of 13 89 The primitive divisor theorem for Lucas sequences with real roots (see [3]) implies that Pk has a primitive prime factor for all positive integers k, except eventually, k ∈ {1, 2, 3, 4, 6}. The primitive prime factor of Pk does not divide P� for any � ∈ [1, k − 1]. In the case of the Pell sequence one checks by inspecting the cases k ∈ {1, 2, 3, 4, 6} that Pk has a primitive divisor for all k ≥ 2. Since P2 = 2, we get that (m ± 1)/2 ≤ 2, which shows that m ≤ 5 and one checks that only m = 3 giving y = 2 is convenient. Assume next that n = 2. We then need to solve 2x + 5y = Px m or 2y + 5x = Px m . The second one implies that 2y = Px m − 5x = (Pm − 5) ( Px m − 5x Pm − 5 ) . Thus, Pm − 5 is a power of 2. Since 6 is not a Pell number, it follows that Pm − 5 is even so m is even. Thus, Pm − 5 = Pm − P3 = P(m±3)/2Q(m∓3)/2. Since this is a power of 2, again by the primitive divisors theorem, we get that (m ± 3)/2 ≤ 2, so m ≤ 7. One now checks Pm − 5 for m ∈ {5, 7} and one does not find a number of this form which is a power of 2. The first one implies 5y = Px m − 2x = (Pm − 2) ( Px m − 2x Pm − 2 ) . Thus, Pm − 2 is a power of 5. Since 3 is not a member of the Pell sequence, we get that Pm ≡ 2 (mod 5). The period of the Pell sequence modulo 5 is 12 and the only residues k (mod 12) such that Pk ≡ 2 (mod 5) are k ∈ {2, 4}. But then m ≡ 2, 4 (mod 12), so Pm is even, which is impossible. This shows that the only solutions with n ∈ {1, 2} is the one shown at Eq. 1.2. �� From now on, we assume n ≥ 3. Lemma 2.4 Assume that (n, m, x, y) is a solution of Eq. 1.1 with n ≥ 3. Then m ≥ 5 and (n + 1)y > (m − 2)x and (n − 2)y < (m − 1)x . (2.1) Proof The fact that m ≥ 5 follows from (i) of Lemma 2.1. For Eq. 2.1, we use that αk−2 ≤ Pk ≤ αk−1 where α = 1 + √ 2 for all k ≥ 1, where both inequalities are strict for k ≥ 3. We thus get α(m−2)x < Px m ≤ max{Px n + P y n+1, P y n + Px n+1} < (Pn + Pn+1) y < P y n+2 < α(n+1)y, and αx(m−1) > Px m ≥ min{Px n + P y n+1, P y n + Px n+1} > P y n > αy(n−2), from where we get the desired inequalities. �� Now note that since one of Pn, Pn+1 is even and the other is odd it follows that m is odd. Further, in the left–hand side exactly one of the two summands is even. Lemma 2.5 Assume that (n, m, x, y) is a solution in positive integers of Eq. 1.1 with n ≥ 3. Then the even number in the left-hand side has exponent x. 123 89 Page 4 of 13 B. Faye et al. Proof Say the equation is Px n + P y n+1 = Px m , and Pn+1 is even. Then n is odd and we get P y n+1 = Px m − Px n = (Pm − Pn) ( Px m − Px n Pm − Pn ) = P(m±n)/2Q(m∓n)/2 ( Px m − Px n Pm − Pn ) . In the left, both P(m±n)/2 and Q(m∓n)/2 have primitive divisors (by abuse we say that 1 is the primitive divisor of P1). Since these primitive divisors are prime factors of Pn+1 as well, we get that (m ± n)/2 and (m ∓ n)/2 both divide n + 1. These two numbers are coprime, since if d divides both of them, then d divides their sum m and their difference n, so Pd divides both Px n and Px m ; hence P y n+1, which is impossible unless d = 1. Thus, (m2 − n2)/4 | n + 1. Since m ≡ n (mod 2), we get m ≥ n + 2. If m ≥ n + 4, then (m2 − n2)/4 > n + 1, a contradiction. Thus, m = n + 2, and so we get P y n+1 = 2Pn+1 ( Px n+2 − Px n Pn+2 − Pn ) . The numbers uk := (ak − bk)/(a − b) with (a, b) = (Pn+2, Pn) form a Lucas sequence with integer roots whose discriminant is (a − b)2 = 4P2 n+1. The above equation shows that ux does not have primitive divisors, so x ∈ {1, 2, 3, 4, 6}. If x = 1, we get P y n+1 = 2Pn+1, so P y−1 n+1 = 2, showing that n = 2, y = 2, which contradicts n ≥ 3. If x is even, then P y n+1 = (P2 n+2 − P2 n ) ( Px n+2 − Px n P2 n+2 − P2 n ) = (Pn+2 − Pn)(Pn+2 + Pn) ( Px n+2 − Px n P2 n+2 − P2 n ) = 4Pn+1(Pn+1 + Pn) ( Px n+2 − Px n P2 n+2 − P2 n ) , and the third factor Pn+1+Pn in the right-hand side above is coprime to Pn+1, a contradiction. Finally, if x = 3, then P y n+1 = 2Pn+1(P2 n+2 + Pn+2Pn + P2 n ) = 2Pn+1(4P2 n+1 + 6Pn+1Pn + 3P2 n ). Thus, any prime p dividing 4P2 n+1 + 6Pn+1Pn + 3P2 n divides also Pn+1 so it can be only 3. Since 9 does not divide a number of the form a2 + ab + b2 for coprime a and b, it follows that (4P2 n+1 + 6Pn+1Pn + 3P2 n )/ gcd(3, Pn+1) > 1 is both an integer coprime to Pn+1 and a divisor of P y n+1, contradiction. Assume next that P y n is the even number in the left. We then havem ≡ n+1 ≡ 1 (mod 2) and P y n = Px m − Px n+1 = (Pm − Pn+1) ( Px m − Px n+1 Pm − Pn+1 ) = P(m±(n+1))/2Q(m∓(n+1))/2 ( Px m − Px n+1 Pm − Pn+1 ) . Asimilar argument as before based on the existence of primitive divisors shows that (m±(n+ 1))/2 and (m∓(n+1))/2 are coprime and divide n so (m2−(n+1)2)/4 | n. Sincem ≡ n+1 (mod 2), we get thatm ≥ n+3, so (m2−(n+1)2)/4 ≥ ((n+3)2−(n+1)2)/4 = n+2 > n, a contradiction. �� 123 The Diophantine equations Pxn + Pyn+1 = Pxm ... Page 5 of 13 89 From now on, the even number in the left-hand side of our equation has x in the exponent. Lemma 2.6 Assume (n, m, x, y) is a solution of Eq. 1.1 with n ≥ 3. Then m > n + 2. Proof Assume m = n + 2. Since m is odd, it follows that n + 1 is even, so the equation is Px n+1 + P y n = Px n+2. Thus, we obtain P y n = Px n+2 − Px n+1 = (Pn+2 − Pn+1) ( Px n+2 − Px n+1 Pn+2 − Pn+1 ) = (Pn+1 + Pn) ( Px n+2 − Px n+1 Pn+2 − Pn+1 ) , and we see that Pn+1 + Pn is coprime to Pn , so it cannot divide a power of it. �� Lemma 2.7 We have either x < 162mn ( log ( x n + y m ))2 or x n + y m < 25000. Proof For a nonzero rational number r let ν2(r) be the exponent of 2 in the factorization of r . Let c ∈ {0, 1} be such that Px n+c is even. We then have Px n+c = Px m − P y n+1−c = Px m(1 − P−x m P y n+1−c), so x ≤ ν2(Px n+c) = ν2(1 − P−x m P y n+1−c). We will apply a linear form in two 2-adic logarithms. A convenient one for us is Corollary 1 of Bugeaud-Laurent [1]. We take α1 = Pn+1−c, α2 = Pm . In the notation of [1], we have D = 1, p = 2, g = 1. Since Pk < αk−1 holds for all positive integers k, we can take log A1 = n logα and log A2 = m logα. Then x n logα + y m logα < 1.14 ( x n + y m ) =: b′. We get x ≤ 24 · 2 (2 − 1)(log 2)4 ( max { log b′ + log log 2 + 0.14, 10 })2 mn(logα)2, so x < 162mnM2, where M is the max. If M = 10, then log b′ + log log 2 + 0.14 ≤ 10, which implies x n + y m < e9.86 1.14 × log 2 < 25000. Otherwise, since 1.14 × log 2 × e0.14 < 0.91 < 1, we get x < 162mn ( log ( x n + y m ))2 . �� 123 89 Page 6 of 13 B. Faye et al. Remark Observe that by Lemma 2.4, we have x n < ( n + 1 n ) y m ( m m − 2 ) ≤ ( 20 9 ) y m < 3y m , and y m < ( m − 1 m ) x n ( n n − 2 ) < 3x n . the following lemma, which is Lemma 7 in [6], is used often in the sequel. Lemma 2.8 If s ≥ 1 is an integer and T > (4s2)s and T > x/(log x)s , then x < 2s T (log T )s . Lemma 2.9 We have y < 66000m2(logm)2. Proof We use Lemma 2.7. If we are in the second case, then y < 25000m, a much better inequality. So, let us treat the first case. Then, one can see that x n < 162m ( log ( x n + y m ))2 < 162 ( log ( 4y m ))2 (see the Remark after Lemma 2.7). Thus, we obtain y m < 3x n < 3 × 162m(log(4y/m))2 < 500(log(4y/m))2. Hence, with z := 4y/m, we get z < 2000m(log z)2. We apply Lemma 2.8 with s = 2 and T := 2000m and get z < 4 × 2000m(log(2000m))2 = 8000m(logm)2 ( log(2000) log 5 + 1 )2 < 262000m(logm)2, which yields y < 66000m2(logm)2. �� Lemma 2.10 Let k, � be positive integers. We have (i) Pk+� ≥ 2� Pk; (ii) P2 k+� ≡ ±P� (mod Pk). Proof (i) For � = 1, we get Pk+1 = 2Pk + Pk−1 ≥ 2Pk . The statement now follows by induction on � ≥ 1. 123 The Diophantine equations Pxn + Pyn+1 = Pxm ... Page 7 of 13 89 (ii) In the formula 2Pk+� = Pk Q� + P�Qk, we divide both sides by 2 (note that Qk is always even) and square to get P2 k+� = (Pk(Q�/2) + P�(Qk/2)) 2 ≡ P2 � (Qk/2) 2 ≡ ±P2 � (mod Pk), where we used the fact that (X , Y ) = (Qk/2, Pk) are solutions of the Pell equation X2 − 2Y 2 = ±1. �� Lemma 2.11 Let c ∈ {0, 1} such that Px n+c is even. We then have |x log Pm − y log Pn+1−c| < 2 2(m−(n+1))x . Additionally, the right–hand side is less that 1. Proof The right-hand side is less than 1 by Lemma 2.6. Next, the equation Px n+c + P y n+1−c = Px m, leads to 1 − P y n+1−c P−x m = ( Pn+c Pm )x ≤ ( Pn+1 Pm )x ≤ 1 2(m−(n+1))x , (2.2) where the last inequality follows from Lemma 2.10. The right-hand side is at most 1/4. Putting � := y log Pn+1−c − x log Pm, we get |e� − 1| < 1 2(m−(n+1))x . This leads to |�| < 2 2(m−(n+1))x , which is what we wanted. �� Remark Note that from 2.2 we get that � < 0. Lemma 2.12 There are no solutions with n ≥ 3 and m ≤ 2500. Proof Assume m ≤ 2500. By Lemma 2.9, we get y < 66000m2(logm)2 < 66000(2500)2(log(2500))2 < 3 × 1013. Lemma 2.11 gives us that∣∣∣∣ log Pn+1−c log Pm − x y ∣∣∣∣ < 2 y(log Pm)2(m−(n+1))x < 0.6 y2(m−(n+1))x . (2.3) There are two scenarios. The first one is when 0.6 2(m−(n+1))x < 1 2(3 × 1013) ( < 1 2y ) . (2.4) 123 89 Page 8 of 13 B. Faye et al. In this case, the right–hand side of Eq. 2.3 is less than 1/(2y2), so by Legendre’s criterion x/y is a convergent of log Pn+1−c/ log Pm . Here is a way to generate the potential candidates (x, y). Loop over all odd n′ := n + 1 − c ∈ [3, 2497] and odd m with n′ + 1 < m ≤ 2499. For each such pair, generate convergents pk/qk with qk < 3 × 1013 of log Pn′/ log Pm . Then (x, y) ∈ {(pk, qk), (2pk , 2qk)}. These are candidates for solutions to our equations in this range. Then simply check, for example, whether P2qk n′ ≡ 0 (mod Pm − Pn′±1). In Mathematica, we used the PowerMod command. This computation took a couple of hours on a desk computer and no solution was found. Another case is when Eq. 2.4 fails. Then 2(m−(n+1))x < 0.6 × 6 × 1012, so (m − (n + 1))x < 46. This shows that x ≤ 45 and m ≤ n + 1 + 45/x . This suggests to fix x ≤ 45 and n ≤ 2497. Then take odd m ∈ [n + 3, n + 1 + 45/x). Then check whether y arising from one of the two equations P y n+1 = Px m − Px n (so y = �log(Px m − Px n )/ log Pn+1�) verifies P y n+1 ≡ 0 (mod Pm − Pn), or P y n = Px m − Px n+1 (so, y = �log(Px m − Px n+1)/ log Pn�) verifies P y n ≡ 0 (mod Pm − Pn+1). Again we used the PowerMod function in Mathematica. This computation took less than a minute and no solution was found. �� From now on, we assume that m > 2500. In this case, we have log Pm = log(αm − βm) − log(2 √ 2) = m logα − log(2 √ 2) + log ( 1 ± 1 α2m ) = m logα − log(2 √ 2) + ζm, where |ζm | < 1 α2m . Inserting the above calculation into the inequality of Lemma 2.11, we get∣∣∣y log Pn+1−c − xm logα − x log(2 √ 2) ∣∣∣ < 2 2(m−(n+1))x + x α2m . (2.5) Since x < y < 66000m2(logm)2 and m > 2500, it follows that x < αm . We record Eq. 2.5 for future use. Lemma 2.13 Assume (n, m, x, y) is a positive integer solution of equation 1.1 with n ≥ 3 and m > 2500. Then, we obtain∣∣∣y log Pn+1−c − xm logα − x log(2 √ 2) ∣∣∣ < 2 2(m−(n+1))x + 1 αm . (2.6) Lemma 2.14 Assume (n, m, x, y) is a solution of Eq. 1.1 with n ≥ 3 and m > 2500. Then, we get y < 1040n2(log n)4. Proof If y ≤ 2x , then Eq. 2.1 shows that (m − 2)x < (n + 1)y ≤ 2(n + 1)x, so m < 2n + 5 < 4n. Then, Lemma 2.9 gives y < 66000(4n)2 log(4n)2 < (66000) × 16n2 log(n3)2 < 107n2(log n)2, 123 The Diophantine equations Pxn + Pyn+1 = Pxm ... Page 9 of 13 89 a much better inequality than the one aimed for. So, in the rest of this proof we may assume that y > 2x . Lemma 2.4 shows that (m − 1)x > (n − 2)y > 2(n − 2)x so m > 2n − 3. Since m is odd, we get m ≥ 2n − 1, so n ≤ (m + 1)/2. Thus, m − (n + 1) ≥ m − ((m + 1)/2 + 1) ≥ (m − 3)/2 ≥ 0.49m, where this last inequality holds since m > 2500. Invoking Lemma 2.13, we get ∣∣∣y log Pn+1−c − xm logα − x log(2 √ 2) ∣∣∣ < 2 2(m−(n+1))x + 1 αm < 3 20.49m . The right–hand side is a linear form in three logarithms. It is nonzero, since Pn+1−c, α, 2 √ 2 are multiplicatively independent. Indeed, assume they were. Since α is a unit and P2 n+1−c, (2 √ 2)2 are integers, it follows that Pn+1−c and 2 should be multiplicatively depen- dent, which would then make Pn+1−c a power of 2, which is not the case by the primitive divisor theorem and the fact that n ≥ 3.We apply a lower bound for linear forms in logarithms due to Matveev [11]. Here, we use the version due to Bugeaud–Mignotte–Siksek [2]. We get that |P y n+1−cα mx (2 √ 2)x − 1| = |e − 1| < 2|�| < 6 20.49m , where is the linear form in logarithms appearing in the left-hand side of Eq. 2.5. We take α1 = Pn+1−c, α2 = α, α3 = 2 √ 2. We have D = 2, We take A1 = 2n logα (> D log(Pn+1−c)), A2 = logα (= log(Dh(α2)), and A3 = 4 log(2 √ 2) = 2 log(8) (= Dh(α3)). We take b1 = y, b2 = xm, b3 = x . Since n ≥ 3, Lemma 2.4 gives mx > (m − 1)x > y(n − 2) ≥ y, so we can take B = mx . Applying Matveev’s inequality, we get 0.49m log 2 − log 6 < 1.4 × 306 × 34.5 × 22(1 + log 2)(1 + log(mx))(2n logα)(logα)(2 log 8) < 6.3 × 1012n log(emx), and then m < 7 × 1012n log(emx) 0.49 logα < 1.63 × 1013n log(emx). By Lemma 2.9, we have ex < ey/2 < 33000em2(logm)2 < 100000m2(logm)2 < m5, where the last inequality holds since m > 2500. We get m < 1.63 × 1013n log(emx) < 1.63 × 1013n log(m6) < 1014n logm. Thus, Lemma 2.8 with s = 1 and T := 1014n gives m < 2 × 1014n log(1014n) = 2 × 1014n(log n) ( log(1014) log n + 1 ) < 7 × 1015n log n. 123 89 Page 10 of 13 B. Faye et al. This together with Lemma 2.9 gives y < 66000m2(logm)2 < 66000 ( 7 × 1015n log n )2 ( log(7 × 1015n log n) )2 < 3.3 × 1036n2(log n)2(log n)2 ( log(7 × 1015) log n + 2 )2 < 1040n2(log n)4, which is what we wanted to prove. �� We now write log Pn+1−c = log(αn+1−c − βn+1−c) − log(2 √ 2) = (n + 1 − c) logα − log(2 √ 2) + log ( 1 ± 1 α2n ) = (n + 1 − c) logα − log(2 √ 2) + ζn, where |ζn | < 1 α2n . We get y log(Pn+1−c) = y(n + 1 − c) logα − y log(2 √ 2) + yζn, where |yζn | < y α2n < 1040n2(log n)4 α2n . Assume next that n > 400. Then 1040n2(log n)4 < αn , so the above fraction is smaller than 1/αn . Inserting this into Lemma 2.13, we get the following result. Lemma 2.15 Assume that (n, m, x, y) is a solution of Eq. 1.1 with n > 400 and m > 2500. Then ∣∣∣((n + 1 − c)y − mx) logα − (y − x) log(2 √ 2) ∣∣∣ < 2 2(m−(n+1))x + 2 αn . To continue, we need the following fact. Lemma 2.16 We have (m − (n + 1)) ≥ (n − 2)/2. Proof We put � = m − n, so Px n+c + P y n+1−c = Px n+�. We square both sides and reduce them modulo Pn getting P2z n+1 ≡ P2x n+� (mod Pn), for z ∈ {x, y}. Now Lemma 2.10, (ii) gives ±1 ≡ ±P2x � (mod Pn), so Pn | P2x � ± 1. The above expression is not zero since � ≥ 3. Hence, we obtain αn−2 < Pn ≤ Pcx � + 1 < α2(�−1)x + 1 < 2α2(�−1)x < α2(�−1)x+1, 123 The Diophantine equations Pxn + Pyn+1 = Pxm ... Page 11 of 13 89 and then n − 2 < 2(� − 1)x + 1. This shows that (m − (n + 1))x = (� − 1)x ≥ (n − 2)/2, which is what we wanted. �� Using Lemma 2.16 into the right–hand side of Lemma 2.15, we get the following. Lemma 2.17 Assume (n, m, x, y) is a solution of Eq. 1.1 in positive integers with n > 400 and m > 2500. Then, we get ∣∣∣((n + 1 − c)y − mx) logα − (y − x) log(2 √ 2) ∣∣∣ < 5 2n/2 . (2.7) now we are ready to find an upper bound for n. Lemma 2.18 We have n < 107. Proof We may assume that n > 6000. The left-hand side of Eq. 2.7 is nonzero since α and 2 √ 2 are multiplicatively independent. To get a lower bound on it we use Corollaire 2 in Laurent–Mignotte–Nesterenko [8]. We have α1 = α, α2 = 2 √ 2, D = 2. We take log A1 = h(α)/2 = (1/2) logα (= h(α1)) and log A2 = log(2 √ 2) (= h(α2)). Since n is large, the left-hand side of Eq. 2.7 is smaller than 0.01, so (n + 1 − c)y − mx < (y − x) log(2 √ 2) + 0.01 logα < 1.18(y − x) + 0.02 < 1.2y. (2.8) We put b′ = ((n + 1 − c)y − mx log A2 + y − x log A1 < ( 1.2 log(2 √ 2) + 1 logα ) y < 2.3y. Then Corollaire 2 in [8] together with inequality Eq. 2.7 gives (n/2) log 2 − log 5 < 24.34 × 24 (max{log(2.3y) + 0.14, 10.5})2 × (1/2)(logα) log(2 √ 2) < 180M2, where M is the above max. This gives n < 181M2 (log 2)/2 < 523M2. If M = 10.5, we then get n < 60000. Otherwise, we get n < 523(log(2.13y) + 0.14))2 < 523 log(2.5y)2. Using Lemma 2.14, we get n < 523 ( log ( 2.5 × 1040n2(log n)4 ))2 which gives n < 107. �� We can now show that there are no solutions with n ≥ 420. Lemma 2.19 We have n < 420. 123 89 Page 12 of 13 B. Faye et al. Proof Assume n ≥ 420. We know that n < 107. By Lemma 2.14, we get that y < 1040n2(log n)4 < 7 × 1058, and inequality Eq. 2.8 shows that (n + 1 − c)y − mx < 1.23y < 1059. So, putting a := (n + 1 − c)y − mx, b := y − x , we have∣∣∣∣ logα log 2 √ 2 − b a ∣∣∣∣ < 5 a log(2 √ 2)2n/2 < 5 a · 2n/2 . (2.9) We know that max{a, b} < 1059. Since F300 > 1060, it suffices to compute the first 300 partial quotients of τ := logα/ log 2 √ 2. Writing the continued fraction of τ as τ = [a0, a, . . . , a300], we have that max{ai : 0 ≤ i ≤ 300} = 1136. Thus, we get that the left–hand side of Eq. 2.9 is bounded below by 1/(1138a2). So, we get 1 1138a2 < 5 a2n/2 , which yields 2n/2 ≤ 1138 × 5 × a < 1063, showing that n < 2 × 63(log 10)/ log 2 < 420, a contradiction. �� So, we have n ≤ 420 and m > 2500. If y ≤ 2x , then as in the beginning of the proof of Lemma 2.14 using Eq. 2.1, we get (m − 2)x < (n + 1)y ≤ 2x(n + 1), so m ≤ 2n + 5 ≤ 945, which is not the case since m > 2500. So, we have y > 2x . Then, as in the proof of Lemma 2.14, or using again inequalities Eq. 2.1, we get (m − 1)x > (n − 2)y > 2(n − 2)x, so m > 2n − 3 and since m is odd we get m ≥ 2n − 1. Thus, n ≤ (m + 1)/2, so that m − (n + 1) ≥ m − ((m + 1)/2 + 1) ≥ (m − 3)/2 > 0.49m since m > 2500. Inequality Eq. 2.6 gives ∣∣∣y log Pn+1−c − mx logα − x log(2 √ 2) ∣∣∣ < 2 2(m−3)x/2 + 1 αm < 3 20.49m . Now y < mx . Lemma 2.14 gives y < 1040n2(log n)4 ≤ 1040(420)2(log 420)4 < 2.35 × 1049. and Eq. 2.1 gives 0.99mx < (m − 2)x < (n + 1)y < 2.35 × 421 × 1049 < 0.99 × 1053, showing that mx < 1053. So, we take α1 = Pn+1−c for some n +1−c ≤ 420, α2 = α, α3 = 2 √ 2. We use LLL (see Proposition 2.3.20 in [4, Section 2.3.5]) to compute a lower bound on |x1 logα1 + x2 logα2 + x3 logα3| > 1 T , (2.10) 123 The Diophantine equations Pxn + Pyn+1 = Pxm ... Page 13 of 13 89 for x = (x1, x2, x2) ∈ Z 3 with max{|xi | : 1 ≤ i ≤ 3} ≤ 1062 := X . Here, n + 1 − c is an odd number in [3, 419]. Taking n + 1 − c = 3, C = (10X)3 was enough and we got√ d2 − 3X2 − (1 + 3X)/2 > 6 · 1062, so 2.10 holds with T = 10127. But this is only for n + 1 − c = 3. As n + 1 − c became larger in [3, 219], larger and larger values of C were needed. At the end, C = X3 ×10220 worked for all odd values of n +1− c in [3, 419]. The smallest √ d2 − 3X2 − (1 + 3X)/2 computed was greater than 8.3 × 1069 and was obtained for n + 1 − c = 219. At any rate, this shows that one can take T = X3 × 10220/1069 = 10337 in Eq. 2.10. Thus, we get 1 10337 < 3 20.49m , so m < 2300, which is a contradiction and finishes the proof of Theorem 1.1. Acknowledgements We thank the referee for comments which improved the quality of our manuscript. The project started in June 2022, during the CIMPA school at IMSP, Institut de Mathématiques et de Sci- ences Physiques de l’Université d’Abomey-Calavi. FL and AT thank the institute for the excellent working environment. References 1. Bugeaud, Y., Laurent, M.: Minoration effective de la distance p-adique entre puissances de nombres algébriques. J. Number Theory 61, 311–342 (1996) 2. Bugeaud, Y., Mignotte, M., Siksek, S.: Classical and modular approaches to exponential Diophantine equations I. Fibonacci and Lucas perfect powers. Ann. Math. 163, 969–1018 (2006) 3. Carmichael, R.D.: On the numerical factors of the arithmetic forms αn ± βn . Ann. Math. (2) 15, 30–70 (1913) 4. Cohen, H.: Number Theory Volume I: Tools and Diophantine Equations. Springer, New York (2007) 5. Faye, B., Gómez, C.A., Luca, F., Rihane, S.E., Togbé, A.: Complete solutions of the exponential Diophantine equation Px n + Px n+1 = P y m . Math. Commun. 27, 163–185 (2022) 6. Guzmán Sanchez, S., Luca, F.: Linear combinations of factorials and S-units in a binary recurrence sequence. Ann. Math. Qué. 38, 169–188 (2014) 7. Hirata Kohno, N., Luca, F.: On the Diophantine equation Fx n + Fx n+1 = F y m . Rocky Mt. J. Math. 45(2), 509–538 (2015) 8. Laurent, M., Mignotte, M., Nesterenko, Yu.: Formes linéaires en deux logarithmes et déterminants d’interpolation. J. Number Theory 55, 285–321 (1995) 9. Luca, F., Oyono, R.: An exponential Diophantine equation related to powers of two consecutive Fibonacci numbers. Proc. Japan Acad. Ser. A 87, 45–50 (2011) 10. Marques, D., Togbé, A.: On the sum of powers of two consecutive Fibonacci numbers. Proc. Japan Acad. Ser. A 86, 174–176 (2010) 11. Matveev, E.M.: An explicit lower bound for a homogeneous rational linear form in the logarithms of algebraic numbers, II. Izv. Math. 64(6), 1217–1269 (2000) 12. Mihăilescu, P.: Primary cyclotomic units and a proof of Catalan’s conjecture. J. Reine Angew. Math. 572, 167–195 (2004) 13. Rihane, S.E., Luca, F., Faye, B., Togbé, A.: On the exponential Diophantine equation Px n + Px n+1 = Pm . Turk. J. Math. 43, 1640–1649 (2019) Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Springer Nature or its licensor (e.g. a society or other partner) holds exclusive rights to this article under a publishing agreement with the author(s) or other rightsholder(s); author self-archiving of the accepted manuscript version of this article is solely governed by the terms of such publishing agreement and applicable law. 123 The Diophantine equations Pnx+Pn+1y=Pmx or Pny+Pn+1x=Pmx Abstract 1 Introduction 2 The proof Acknowledgements References