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PELL FACTORIANGULAR NUMBERS Florian Luca, Japhet Odjoumani

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PELL FACTORIANGULAR NUMBERS Florian Luca, Japhet Odjoumani PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 105(119) (2019), 93–100 DOI: https://doi.org/10.2298/PIM1919093L PELL FACTORIANGULAR NUMBERS Florian Luca, Japhet Odjoumani, and Alain Togbé Abstract. We show that the only Pell numbers which are factoriangular are 2, 5 and 12. 1. Introduction Recall that the Pell numbers Pm m> are given by { } 1 αm βm P = − , for all m > 1, m α β − where α =1+ √2 and β =1 √2. The first few Pell numbers are − 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025,... Castillo [3], dubbed a number of the form Ftn = n!+ n(n + 1)/2 for n > 1 a factoriangular number. The first few factoriangular numbers are 2, 5, 12, 34, 135, 741, 5068, 40356, 362925,... Luca and Gómez-Ruiz [8], proved that the only Fibonacci factoriangular numbers are 2, 5 and 34. This settled a conjecture of Castillo from [3]. In this paper, we prove the following related result. Theorem 1.1. The only Pell numbers which are factoriangular are 2, 5 and 12. Our method is similar to the one from [8]. Assuming Pm = Ftn for positive integers m and n, we use a linear forms in p-adic logarithms to find some bounds on m and n. The resulting bounds are large so we run a calculation to reduce the bounds. This computation is highly nontrivial and relates on reducing the diophantine equation Pm = Ftn modulo the primes from a carefully selected finite set of suitable primes. 2010 Mathematics Subject Classification: Primary 11D59, 11D09, 11D7; Secondary 11A55. Key words and phrases: factoriangular numbers, Pell numbers. Communicated by Žarko Mijajlović. 93 94 LUCA, ODJOUMANI, AND TOGBÉ 2. p-adic linear forms in logarithms Our main tool is an upper bound for a non-zero p-adic linear form in two logarithms of algebraic numbers due to Bugeaud and Laurent [1]. Let η be an algebraic number of degree d over Q with minimal primitive polynomial over the integers d f(x) := a (X η(i)) Z[X], 0 − ∈ i=1 Y (i) where the leading coefficient a0 is positive and the η , i =1,...,d are the conju- gates of η. The logarithmic height of η is given by d 1 h(η) := log a + log(max η(i) , 1 ) . d 0 {| | } i=1 X Let K be an algebraic number field of degree dK. Let η1, η2 Kr 0, 1 and b1 b1 ∈ { } b ,b positive integers. We put Λ = η η . For a prime ideal π of the ring K 1 2 1 − 2 O of algebraic integers in K and η K, we denote by ordπ(η) the order at which π ∈ appears in the prime factorization of the principal fractional ideal η K generated O by η in K. When η is an algebraic integer, η K is an ideal of K. When K = Q, O O π is just a prime number. Let eπ and fπ be the ramification index and the inertial degree of π, respectively, and let p Z be the only prime number such that π p. Then ∈ | k k eπ i fπi p K = π , K/π = p , dK = eπi fπi , O i |O | i=1 i=1 Y X where π1 := π,...,πk are prime ideals in K. We set D := dK/fπ. Let A1, A2 be positive real numbers such that O log p log A > max h(η , (i =1, 2). i i D Further, let n o ′ b b b := 1 + 2 . D log A2 D log A1 With the above notation, Bugeaud and Laurent proved the following result (see [1, Corollary 1 to Theorem 3]). Theorem 2.1. Assume that η1, η2 are algebraic integers which are multiplica- tively independent and that π does not divide η1η2. Then 24p(pfπ 1) ord (Λ) 6 − D5(log A )(log A ) π (p 1)(log p)4 1 2 − ′ 10 log p 2 max log b + log(log p)+0.4, , 10 . × D n o In the actual statement of [1], there is only a dependence of D4 in the right- hand side of the above inequality, but there all the valuations are normalized. Since we work with the actual order ordπ(Λ), we must multiply the upper bound of [1] by another factor of dK/fπ = D. PELL FACTORIANGULAR NUMBERS 95 3. Proof of Theorem 1.1 We study the Diophantine equation (3.1) Pm = Ftn. We may assume that n > 10, since the smaller values can be checked by hand. Before proving our main result, let us prove some preliminary results that are useful for the proof of Theorem 1.1. Lemma 3.1. The following inequalities n−2 n−1 (3.2) α 6 Pn 6 α hold for all n > 1. Proof. Follows immediately by induction on n. n n 6 6 n > Further (see [8]), the inequalities e Ftn n hold for all n 3. Taking logarithms, the above inequalities yield n(n + 1) (3.3) n(log n 1) < log n!+ <n log n, − 2 for all n > 3. Inequalities (3.2) imply (m 2)log α 6 log Pm 6 (m 1)log α. Using (3.1), we get that for positive integers m,−n satisfying (3.1) and n−> 10, we have n(n + 1) (m 2)log α 6 log n!+ 6 (m 1)log α. − 2 − Combining the last inequality above with (3.3), one has n(log n 1) < (m 1)log α and (m 2)log α<n log n. − − − Hence, n(log n 1) n log n (3.4) 1+ − <m< 2+ . log α log α If n 6 100, the above inequality implies that m 6 525. We listed all Pell numbers Pm with m 6 525 and all factoriangular numbers Ftn with n 6 100 and intersected these two lists. All solutions in this range of (3.1) are listed in the Theorem 1.1. We assume from now on that n> 100. Rewriting (3.1) as αm βm n(n + 1) − = n!+ , 2√2 2 we get, after some algebraic manipulations using the fact that β = α−1, that − −m 2m m 2√2 n!= α α n(n + 1)√2α + εm , − m+1 where εm := ( 1) 1 . We note that − ∈ {± } 2m m m m α n(n + 1)√2 α + εm = (α z )(α z ), − − 1 − 2 where n(n + 1)√2 2n2(n + 1)2 4ǫ (3.5) z1,2 := ± − . p2 96 LUCA, ODJOUMANI, AND TOGBÉ Thus, equation (3.1) is equivalent to − (2√2)n!= α m(αm z )(αm z ). − 1 − 2 Let K = Q(z1) and let π be a prime ideal lying above 2 in K. As α is unit and π 2, one has O | m m (3.6) ord (n!) 6 ordπ(2√2n!) 6 ordπ(α z )+ordπ(α z ). 2 − 1 − 2 m We use Theorem 2.1 to get an upper bound on ordπ(α zi), for i =1, 2. We fix i 1, 2 and put − ∈{ } m η := α, η := zi, b := m, b := 1, Λi := α zi. 1 2 1 2 − Note that z1z2 = ǫ. Then z1, z2 and α are all units so π cannot divide any of them. Further, these three numbers belong to K. Next we prove that α and zi are −1 multiplicatively independent for i =1, 2. Of course, since z2 = z1 , it suffices to show that this is so only for i = 1. Well, note first that since ±n > 100, it follows that ∆ > 0. Let d be that positive squarefree integer such that for some positive integer u we have ∆ = 2n2(n + 1)2 4ǫ = du2. Since the left–hand side above is a multiple of 4 and d is squarefree, it− follows that 2 u. Then, using (3.5), we have | z = A√2 B√d, 1 ± where (A, B) = (n(n + 1)/2,u/2) Z2. ∈ Hence, z2 = C + D√2d, where C, D are integers. However, since z2 Q(2√d) and 1 1 ∈ α Q(√2) and they are multiplicatively dependent, it follows that d 1, 2 . The case∈ d = 2, leads to ∈{ } u2 (n(n + 1))2 = 2ε, − − or (u n(n + 1))(u + n(n +1)) = 2ε, − − which is impossible since the left–hand side above is an integer multiple of the number u + n(n + 1) > 1002. The case d = 1 leads to u 2 n(n + 1) 2 2 = ε 1 . 2 − 2 − ∈ {± } Hence, (X, Y ) := (u/2,n (n+1)/2) is a positive integer solution of the Pell equation X2 2Y 2 = 1. − ± It is known that Y = Pk for some k > 1. Hence, Pk = n(n + 1)/2 is a triangular number. Luckily all Pell triangular numbers have been found by McDaniel [9]. Lemma 3.2 (Theorem [9]). If Pk is triangular then k =1. Since for us n> 100, it follows that the equation Pk = n(n+1)/2 is impossible. This proves that indeed z1 and α are multiplicatively dependent. Thus, K = Q(√2, √d) has dK = 4. Further, since the discriminant of K is even (because 2 ramifies in Q(√2) K), it follows that for our prime ideal π, we have ⊆ fπ > 2 and so D = dK/fπ 6 2. PELL FACTORIANGULAR NUMBERS 97 Next, we calculate upper bounds for the logarithmic heights of α and zi. The 2 1 minimal primitive polynomial of α over the integers is x 2x 1, so h(α)= 2 log α. 1 − − Since α > 2, one can take log A1 = 2 log α.
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