RAMANUJAN-NAGELL CUBICS

RAMANUJAN-NAGELL CUBICS MARK BAUER AND MICHAEL A. BENNETT ABSTRACT. A well-nown result of Beuers [3] on the generalized Ramanujan-Nagell equation has, at its heart, a lower bound on the quantity x 2 2 n. In this aer, we derive an inequality of the shae x 3 2 n x 4/3, valid rovided x 3 2 n and x, n 5, 7, and then discuss its imlications for a variety of Diohantine roblems. 1. Introduction. Surfing the internet one day, the second author came across a conversation on a hysics forum [10], in which a Diohantine roblem was roosed. The rooser wished to find a roof of his conjecture, to the effect that the Diohantine equation 1 x 3 x + 8 = 2 has no solutions in integers x > 8. Since it indeed has solutions with x 2, 1, 0, 1, 3, 5, 8, such a result would be, in some sense, best ossible. That equations such as 1 have at most finitely many solutions is an immediate consequence of a classical result of Siegel [12], and, in fact, if we denote by P m the greatest rime factor of a nonzero integer m, one may show see e.g. [7] that 2 P fx c 1 log log max x, 3. Here c 1 = c 1 f > 0 and f is, say, an irreducible olynomial with integer coefficients and degree at least two. Even more, if F x, y is an irreducible binary form i.e. homogeneous olynomial with integer coefficients and degree at least 3, wor of Mahler [8], as extended by 2000 AMS Mathematics subject classification. Primary: 11A63, Secondary: 11D61, 11J68. Keywords and hrases. Polynomial-exonential equations, digital roblems. Research suorted in art by grants from NSERC. Received by the editors January 2016. 1

2 MARK BAUER AND MICHAEL A. BENNETT Shorey et al [11], imlies that 3 P F x, y c 2 log log max x, y, 3, with c 2 = c 2 F > 0, so that, given rimes 1, 2,..., n, the Thue- Mahler equation 4 F x, y = α1 1 α2 2 αn n has at most finitely many solutions in corime integers x and y, and nonnegative integers α 1,..., α n. In articular, the equation x 3 xy 2 + 8y 3 = 2 can be satisfied by at most finitely many corime integers x and y, and nonnegative those with y = 1 rovide the solutions to 1. Statements 2 and 3 can both be made effective; the interested reader is directed to [13] for details. The motivation for studying such an equation in [10] was, aarently, to find a cubic analog of the Ramanujan-Nagell equation x 2 + 7 = 2, which is nown see e.g. [9] to have recisely the integer solutions corresonding to x = 1, 3, 5, 11 and 181. This is extremal in the sense that there exists no monic quadratic fx for which fx = 2 has more than ten solutions in integers x, via the following theorem of Beuers [3]. Theorem 1. Beuers If D is an odd integer, then the equation x 2 + D = 2 n has at most 5 solutions in ositive integers x. The only monic irreducible cubics f we now for which the equation fx = 2 has more solutions than the seven to 1 are those corresonding to the olynomial fx = x 3 13x + 20 and its translates, each with 8 solutions. The results of this aer mae it a routine matter to solve such an equation for any monic cubic and the machinery we emloy readily generalizes to certain non-monic cases, though we omit the details in the interest of eeing our exosition reasonably simle. An examle of what we rove is the following.

RAMANUJAN-NAGELL CUBICS 3 Proosition 2. Let b and c be integers, and suose that x and are integers for which 5 x 3 + bx + c = 2. Then either b = c = 0, or x = c/b, or x, b, c, = 5, 0, 3, 7, 5, 0, 3, 7, or we have 6 x max 2b 3, 2c 3/4. Alying this to equatio imlies that its solutions satisfy x 8, whereby a routine chec leads to the desired conclusion. Our real motivation in writing this aer is to emhasize the unnaturally large influence a single numerical flue can have uon certain results in exlicit Diohantine aroximation. Underlying Theorem 1 of Beuers is an inequality of the shae x 2 2 n 2 0.1n, valid for odd n and derived through Padé aroximation, through aeal to the identity 181 2 + 7 = 2 15 which imlies that 2 181 2 is small. 7 Proosition 2 follows from a rather similar aroach and deends fundamentally uon the relation 5 3 +3 = 2 7 which imlies that 3 2 5 2 2 is also small. The fact that we are able to rove an effective inequality of the shae 3 2 r 2 λ 2, for r 1, 2 and, crucially, λ < 2, is what enables us to derive results lie Proosition 2. It is worth observing that techniques based uon lower bounds for linear forms in logarithms lead to uer bounds for the heights of solutions to much more general equations than 5, but have the disadvantage of these bounds being significantly worse than exonential in the coefficients b and c. The outline of this aer is as follows. In Section 2, we begin by stating our results, exressed both in terms of exlicit rational aroximation to certain algebraic numbers by rational with restricted denominators, and also as results about corresonding Diohantine equations. In Sections 3 and 4, we collect the necessary technical lemmata about Padé aroximation to binomial functions at least in terms of Archimedean valuations. Section 5 contains the roof of our main result, Theorem 3, modulo an arithmetic result on the coefficients of our Padé aroximants, which we rovide in Section 6. Sections 7,

4 MARK BAUER AND MICHAEL A. BENNETT 8 and 9 consist of the roofs of Corollary 4, Theorem 5 and Theorem 6, resectively. Finally, in Sectio0, we mae a few remars about more general Thue-Mahler equations. 2. Statements of our results. Our results are of two closely related tyes. First, we have an exlicit lower bound for aroximation to 3 2 or 3 4 by rational numbers with denominators restricted to being essentially a ower of 2. From this, with what are basically elementary arguments, we are able to derive a number of results on both the heights and the number of solutions of equations of the shae F x = 2 n, for monic cubic olynomial F Z. We start with the following : Theorem 3. Suose that r,, s and are integers with r 1, 2, s 1, 3 and > 12. Then we have 7 2 r/3 > 2 1.62 s2. As noted earlier, this restricted irrationality measure to 3 2 and 3 4 has a number of straightforward consequences for certain Diohantine equations. We will begin by stating an almost immediate corollary that will rove a useful form for later alications to Diohantine equations. Corollary 4. If x and n are integers with x 3 27 2 n, then either x 4, 5, 8, 15, 19, 38, 121 or we have 8 x 3 27 2 n 3 5/3 x 4/3. From the standoint of exlicit solution of Diohantine equations, our main result is the following from which Proosition 2 is an immediate corollary, taing a = 0. Theorem 5. Let a, b and c be integers, and suose that x and n are integers for which we have 9 x 3 + ax 2 + bx + c = 2 n.

RAMANUJAN-NAGELL CUBICS 5 Then either a, b, c = 3t, 3t 2, t 3 for some integer t, or a, b, c, x, n = 3t, 3t 2, t 3 +3, 5 t, 7 or 3t, 3t 2, t 3 3, 5 t, 7, for some integer t, or x = a3 27c 27b 9a 2, or we have 10 x max 8 b a 2 /3 3 + a/3, 4a 3 /27 + 2c 2ab/3 3/4 + a/3. Finally, as an analog of Theorem 1, we have Theorem 6. Let D be an odd integer. Then the number of solutions to the equation 11 x 3 + D = 2 n in airs of integers x, n is at most three. This last result is shar as equatio1 with D = 3 has recisely three solutions x, n = 1, 1, 1, 2 and 5, 7. In fairness, we should oint out that this is not an analog of comarable generality to Theorem 1, in that the latter rovides an uer bound for the number of solutions to the equation fx = 2 n, for all monic quadratic olynomials f, while the same is not true of Theorem 6 for monic cubics. 3. Padé aroximants to 1 x ν. All our results in this aer have, at their heart, Padé aroximation to the algebraic function 1 x ν, where ν Q/Z. Recall that an [ /n 2 ]-Padé aroximant to a function fx is a rational function x/qx, where the numerator and denominator are olynomials with, say, integer coefficients, of degrees and n 2, resectively, such that fx and x/qx have the same MacLaurin series exansion u to degree + n 2, i.e. such that x/qx is a good aroximation to fx in a neighbourhood of x = 0.

6 MARK BAUER AND MICHAEL A. BENNETT Since the function x/qx is a rational function with rational coefficients, it taes rational values for rational choices of its argument. In this way, we will be able to tae a single suitably good aroximation to, in our case, a articular algebraic number, and generate an infinite sequence of good aroximations to the same number. In order to obtain shar estimates for the quality of the aroximations that are generated with these functions, we will use reresentations coming from contour integrals, as well as exlicit descritions of the Padé aroximants as olynomials. Define I,n 2 x = 1 2πi γ 1 zx 1 zx ν z +1 1 z +1 where and n 2 are ositive integers, γ is a closed, ositively oriented contour enclosing z = 0 and z = 1, and x < 1. A straightforward alication of Cauchy s theorem yields that 12 I,n 2 x = P,n 2 x 1 x ν Q,n 2 x where P,n 2 x and Q,n 2 x are olynomials with rational coefficients of degrees and n 2, resectively. In fact, examining the relevant residues, it is ossible to show that + ν + n 2 13 P,n 2 x = x =0 and n 2 ν + n 2 14 Q,n 2 x = x. =0 In articular, if we choose ν 1/3, 2/3, we have that P,n 2 x, Q,n 2 x Z[1/3][x]. Our goal in the following sections will be twofold. First, we will derive estimates for the size of I,n 2 x and P,n 2 x using contour integral reresentations. Secondly, we will find lower bounds for the size of the greatest common divisor of the coefficients involved in P,n 2 x and Q,n 2 x. 4. Bounding our aroximants. Here and henceforth, given a ositive integer, let us set n 2 = 4 δ, where δ 0, 1, and write n 2 dz

RAMANUJAN-NAGELL CUBICS 7 ν = r/3 where r 1, 2. Define P,δ = P,4 δ3/128, so that, from 12, 15 I,δ = P,δ We further define F z by 16 F z = Q,δ = Q,4 δ3/128 and I,δ = I,4 δ3/128, r 5 Q,δ. 2 7/3 1 3z/1284 z1 z 4. It is easy to show, via calculus, that F z attains its minimum for z 0, 1 at the value 17 τ = 1 631 5 15865 = 0.20304..., 6 where we have 243 75 15865 4 F := F τ = 34359738368 631 5 15865 125 15865 4 = 11.97804... Our argument will require uer bounds uon P,δ, Q,δ and I,δ ; from 15 it suffices to bound one of the first two of these, together with the last. Lemma 7. We have P,δ < 1.26 F and I,δ < 4 π 2 35 3 5 F, Proof. We begin by searating the integral defining I,δ3/128 into two ieces, one involving a closed contour containing only the ole at z = 0. Taing τ as i7, we may write P,δ = 1 2πi Γ 1 3z/128 4 δ+r/3 z +1 1 z 4 δ+1 where Γ is the closed, ositively oriented contour with z = τ. Using the transformation z = τe iθ, we have that P,δ 1 2π 1 3z/128 4 δ+r/3 2π z +1 1 z 4 δ+1 τ dθ 0 dz

8 MARK BAUER AND MICHAEL A. BENNETT and so whereby P,δ 1 τ P,δ max 0 θ 2π 1 3τ/128r/3 τ 1 τ 1 3τe iθ /128 4 δ+r/3 1 τe iθ 4 δ+1, 1 3τ/128 1 τ Since r 1, 2, the desired bound for P,δ follows. 4 δ. To bound I,δ, we argue as in [1] to arrive at the identity 18 I,δ = 3/1285 δ+1 π 1 We may thus rewrite the integrand here as v 4 1 v f δ,r v 1 31 v/128 4 where 0 v 4 δ+r/3 1 v r/3 dv 1 31 v/128 4 δ+1. 1 f δ,r v = v4+r/3 δ 1 v 1 r/3 1 31 v/128 5 δ, δ 0, 1 and r 1, 2. Changing variable via v = 1 z, we thus have I,δ 3/1285+1 δ π maxf δ,r v : v 0, 1 F 1. Since a little calculus reveals that, in all cases, maxf δ,r v : v 0, 1 < 1/3 and since we have F < 12, it follows that I,δ < 4 π 2 35 3 5 F, as desired. 5. Proof of a restricted irrationality measure : Theorem 3. Let us suose that r 1, 2, s 1, 3, and are integers, and write m = 3 + r. From 13, 14 and the fact that n ± r/3 3 [3j/2] Z j

RAMANUJAN-NAGELL CUBICS 9 for all ositive integers n and j, it follows that 3 [/2] 2 7 P,δ and 3 2 δ 2 74 δ Q,δ are both integers. Here, is a ositive integer to be chosen later. We set Π,δ,r = gcd3 [/2] 2 7 P,δ, 3 2 δ 2 28 7δ Q,δ so that A = 3[/2] 2 7 Π,δ,r P,δ and B = 32 δ 2 28 7δ Q,δ Π,δ,r are integers. Equatio2 therefore imlies that Π 1 n I r 1,δ,r,δ = A 5 B 3 [/2] 2 7 2 7/3 3 2 δ 2 28 7δ. and If we define we have that Λ = s 2 m/3 Ω = 1. s 2 m/3 A 2 73 δ 3 2 [/2] δ, 5 r B 2 7/3 Λ < Ω + Π 1,δ,r A 1 3 [/2] 2 7 I,δ note that the nonvanishing of A is a consequence of the contour integral reresentation for P,n 2 x. From Lemma 4 of Beuers [3], for one of our two choices of δ 0, 1, we have Λ 0; we fix δ to be this value and choose [ ] m 7r + 21δ 19 = 1 +, 63 so that 7r/3 + 73 δ > m 3. Since s 3, we thus have Λ A 2 73 δ+r/3 3 2 [/2] δ 1.

10 MARK BAUER AND MICHAEL A. BENNETT Combining our uer and lower bounds for Λ, we find that 20 1 < ΩΛ 1 + Λ 2, where, uon substituting for A, and Λ 1 = Π 1,δ,r 32 δ 2 74 δ+r/3 P,δ Λ 2 = Π 1,δ,r 32 δ 2 74 δ+r/3 I,δ. and Alying Lemma 7, we thus have whereby, from 20, 21 Ω = Λ 1 < 1.26 Π 1,δ,r 214/3 3 2 2 28 F 1 Λ 2 < 4 3 7 π Π 1,δ,r 214/3 2 7, F 4 Π,δ,r 2 > 7 F. s 2 m/3 1.26 2 14/3 3 2 2 28 F π 214/3 In order for inequality 21 to be nontrivial, it remains therefore to show that 1 3 7 22 lim inf log Π,δ,r > log 2 7. F 3 7 In the next section, we will in fact rove the following result. Proosition 8. For r 1, 2, 497 and δ 0, 1, we have that Π,δ,r > 1.8. 3 Since we have that 7 2 7 F < 1.427, inequality 22 follows as desired. Assuming Proosition 8 and further that 11000, so that m 33000, 19 thus imlies that 497, We therefore have Π,δ,r 4 3 7 π 214/3 2 7 > 1.79 F

RAMANUJAN-NAGELL CUBICS 11 and hence, from 21, 1 > 16166467234. s 2 m/3 From 19, we have that m + 77/63 and so 1 > 16166467234 m+77/63 > 1.454 m, s 2 m/3 where we have aealed to the fact that m 33000. Since m = 3 + r, we thus have 2 r/3 > 2 r/3 s2 1.454 3 r > 2 1.62, for both r 1, 2,, Z and s 1, 3. For 100, a brute force search shows that the only aroximations in this range that fail to satisfy the desired bound 7 have r, s,, in the following list where we assume that gcd, s2 = 1 to avoid redundancy : 1, 1, 1, 0, 1, 1, 2, 0, 1, 1, 3, 1, 1, 1, 5, 2, 1, 3, 4, 0, 1, 3, 7, 1, 1, 3, 31, 3, 1, 3, 61, 4, 1, 3, 121, 5, 2, 1, 1, 0, 2, 1, 2, 0, 2, 1, 3, 1, 2, 1, 3251, 11, 2, 3, 5, 0, 2, 3, 19, 2, 2, 3, 305, 6. In articular, in every case we have < 12. All that remains is to verify the inequality for, say, 100 < < 11000. We do this by considering the binary exansions of 2 r/3 and 3 2 r/3, for r 1, 2, and searching for either unusually long strings of zeros or unusually long strings of ones each of which would corresond to a very good aroximation to 2 r/3 or 3 2 r/3 by a rational with denominator a ower of two. Such an argument is described in detail in [1] see, in articular, Lemma 9.1 and the remars following it. The fact that no such strings occur comletes the roof of Theorem 3 again, assuming Proosition 8. 6. Arithmetic roerties of the coefficients. We now turn our attention to roving Proosition 8. To do this, we require first a good understanding of the content of the olynomials P,n 2 x and Q,n 2 x.

12 MARK BAUER AND MICHAEL A. BENNETT Lemma 9. Let be a ositive integer, n 2 = 4 δ for δ 0, 1 and r 1, 2. Suose that is rime, with 23 > max 3n 2 + 2, 5. Assume that 24 if r mod 3, or 25 2 3, 3 5 4 6, 1, 5 12, 1 2 2 3, 3 11 4 12, 1, if r mod 3. Then we have + r/3 + n 2 ord 1 for 0 and n 2 r/3 + n 2 ord 1 for 0 n 2. Proof. We begin by considering the case when [ r mod 3. 2 3 + r 3, 3 4, Observe that, from 24, we have that either whereby 4 δ 26 = [ n or that 1 5 6 + 4r 3 6,, 1 whence 27 4 δ = We conclude, in either case, that 28 + 7 20r 21 + 6 6 [ 2 3 + 4r 3 3, 1, [ 1 3 + 8r 9 3, 1. 7 6 1 6 1 + 1, where the last inequality follows from the assumtion that 7. We roceed to show that satisfying the hyotheses of the lemma have

RAMANUJAN-NAGELL CUBICS 13 ositive valuation for the desired binomial coefficients. If = 0, then + r/3 + n 2 r/3 + n 2 + n 2 = = n 2 and, since < n 2, our assumtion that 2 > 3n 2 + 2 > 3n 2 imlies that + n 2 + n 2 ord = +. It follows that ord +n 2 1 if and only if n + 2 1, whereby, from 28, we conclude as desired. Similarly, if = 1, then the fact that + r/3 + n 2 ord 1 is a consequence of the inequalities 1 + while follows from ord r/3 + 1 n 2 + 1 1, + n 2 1 + 1 1. Let us next suose that 2. From Lemma 4.5 of Chudnovsy [4], if n N and 2 > 3n + r, we have [ ] [ ] [ ] n + r/3 n q n q ord = where q = r/3. It follows that + r/3 q 29 ord = and 30 ord + n 2 n 2 = + + q + n 2.

14 MARK BAUER AND MICHAEL A. BENNETT Let us suose that ord +r/3 contradiction. From 29, we have and 27, imlies that q 31 = Similarly, from 30, we have 32 + < 1, i.e. n If 1 <, since +n 2 n 2 q = 0 and see to derive a which, with 26 q = r 3 +. 1 1. 2 3 + 1 3, we have from 31, that > 2 3 + 1 3 + r 3 = 1 + 1 r 3 1 1 3, an immediate contradiction. It follows that = We therefore have from 31 and 32 that + 1 1 +. 1 1 + and hence r 3, whence 2 3 + r < 2 3 3, contradicting 24. Similarly, we may write r/3 q 33 ord = and 34 ord + n 2 = + + q + n 2

RAMANUJAN-NAGELL CUBICS 15 where now q = 2 + r/3. Let us suose that r/3 + n 2 35 ord = 0 so that, in articular, we have q 36 = q = It follows that we necessarily have < 1 2 + r 3 whereby + = + 2 + r 3 = 1 3 r 3 Arguing as before, we find that < 1 1 2 + r 3 again contradicting 26 and 27., < 1 3 1,. 1 1. The argument for r mod 3 is essentially similar. Relation 25 imlies that we have one of 37 38 or 39 [ 5 12 + 9 4r 12, 1 2 [ 2 3 + 3 r 3, 3 4 [ 11 12 + 9 4r 12, 1 = = = 4 δ 4 δ 4 δ [ 2 3 + 6 4r 3, 1, [ 2 3 + 9 4r 3, 1 [ 2 3 + 6 4r 3, 1. In each case, we may thus conclude that 40 + 13 33 20r + 13 12 12 12 7 12 1 + 1.

16 MARK BAUER AND MICHAEL A. BENNETT Arguing as reviously, 40 imlies the desired conclusion almost immediately in case = 0 or 1. Let us assume that 2. From Lemma 4.5 of Chudnovsy [4], we once again have 29 and 33, only this time with q = 2 r/3 and q = + r/3, resectively. If we suose that ord +r/3 = 0, then n 2 2 3 + r 3, whereby, from 37, 38 and 39, 41 2 3 r 3 +, n whence < 1/3 < 1. If also ord +n 2 = 0, we again have 32, and so, from 41, 2 3 r 3 + + n 2 1 1, a contradiction. If, on the other hand, we assume 35, then both +r 3 and + 1 1. The first of these imlies that whence 1 1 1 3 + r 3 + 2 3 r + 3 3 and so, from 37, 38 and 39, 1 1 + = + whereby 2 3 r + 3 3., 1 3 + r 3 + The resulting contradiction to 37, 38 and 39 comletes the roof.,

RAMANUJAN-NAGELL CUBICS 17 To aly this result, we observe that if 1 c < d are integers, then we have / > c/d recisely when + 1,. + c/d Let us define =0 θx, q, r = x r mod q log, where the sum is over rimes. Fixing r 1, 2, it follows from Lemma 9 that log Π,δ,r is bounded below by 0 42 L r, = T 1, + T 2, + T 3, + T 4, + T 5,, where and =0 T 1, = θ + 2/3, 3, r θ + 3/4, 3, r, T 2, = θ + 5/6, 3, r θ + 1, 3, r, T 3, = θ + 5/12, 3, r θ T 4, = θ, 3, r + 2/3 T 5, = θ, 3, r + 11/12, 3, r, + 1/2 θ, 3, r + 3/4 θ, 3, r, + 1 for 0 = [ ] 12 2 we can actually, in most cases, use a slightly larger value for 0 ; it is simly chosen so that inequality 23 is satisfied. To estimate L r,, for large values of, we will aeal to recent bounds on θx, 3, r due to the second author, Martin, O Bryant and Rechnitzer [2]. In articular we use that 43 θx, 3, r x 1.798158 x if x 10 13 < 2 0.00144 x log x if x > 10 13.

18 MARK BAUER AND MICHAEL A. BENNETT Note that together these inequalities imly that we have θx, 3, r x < 0.00144 x 2 log x for all x > 641239201. We will show that L r, > 0.58779, which immediately imlies Proosition 8. Let us assume first that > 2 10 9. Then T 1,0 > n 1 12 0.00144 3 2 log3 /2 + 4 > 0.083145, 3 log4 /3 T 2,0 > n 1 10 0.00144 6 5 log6 /5 + n 1 > 0.099852, log T 3,0 > 5 0.00144 T 4,0 > 12 0.00144 T 5,0 > 11 0.00144 T 1,1 > 70 0.00144 and T 3,1 > 51 0.00144 12 5 log12 /5 + 2 log2 > 0.199714, 3 2 log3 /2 + 4 > 0.083145, 3 log4 /3 12 11 log12 /11 + n 1 > 0.090768, log 3 5 log3 /5 + 4 7 log4 /7 12 17 log12 /17 + 2 3 log2 /3 > 0.014204, > 0.019513, whereby L r, > 0.58779 as desired. Next suose that 2 10 9. Then, for each 0, we have T 1, > 23 + 24 + 3 1.798158 + 2/3 +, + 3/4 T 2, > 26 + 5 + 1 1.798158 + 5/6 +, + 1 T 3, > 2 + 112 + 5 1.798158 + 5/12 +, + 1/2 T 4, > 23 + 24 + 3 1.798158 + 2/3 + + 3/4

RAMANUJAN-NAGELL CUBICS 19 and T 5, > 212 + 11 + 1 1.798158 + 11/12 +. + 1 If we suose that 10 6 2 10 9, then it is readily checed that the inequalities here are nontrivial i.e. that the right-hand-sides are ositive for, in each case, 0 4, whereby we find that, once again, L r, 4 T 1, + T 2, + T 3, + T 4, + T 5, > 0.58779. =0 It remains to treat values of < 10 6. a = 1, = 496 and n 2 = 4 1, then We note that if we have Π 1/,δ,r = 1.79954218... and hence we cannot exect to extend Proosition 8 to smaller values of. By direct if slow comutation of Π,δ,r, we find that the inequality of Proosition 8 is satisfied for each r 1, 2, 497 1000 and n 2 = 4 δ with δ 0, 1. For larger values of, instead of relying uon the definition of Π,δ,r, we aeal to the bound log Π,δ,r L r,, where L r, is as defined in 42. For 1000 < 10000 and r 1, 2, we chec that, in each case, exl r, / > 1.8. This taes roughly 20 minutes in Male on an elderly Macboo Air. We find that the largest value of in this range for which we have exl r, / < 1.9 corresonds to exl 2,3319 /3319 = 1.89773. This is unsurrising since, from 43, we have that L := lim L r, / is equal to =0 1 3 + 24 + 3 + 1 26 + 5 + 1 + 1 2 + 112 + 5 + 1 212 + 11 + 1. Defining ψx = d dx ln Γx = Γ x Γx,

20 MARK BAUER AND MICHAEL A. BENNETT we have that whence ψx = γ + n=0 1 n + 1 1, for x 0, 1, 2,..., n + x L = ψ1+ψ3/4+ψ1/2/2 ψ2/3 ψ5/6/2 ψ5/12/2 ψ11/12/2. Using nown identities for ψ, we thus have L = γ 3 l+ π 2 5π 3 ln 3 +9 12 and so 4 ψ5/12 2 ψ11/12 2 lim ex L r, / = 2.019084. = 0.70264, To finish the comutation verifying the inequality exl r, / > 1.8 for 10000 < < 10 6, we emloy the bootstraing rocedure described in detail in Section 7 of [1], which exloits that lim ex L r, / greatly exceeds 1.8 by way of examle ex L 1,10000 /10000 > 2.006523 together with the fact that the difference between L r, and L r,+ is small, rovided is much larger than. This enables us to significantly reduce the number of times we actually comute L r,. Full details are available from the authors on request. This comletes the roof of Proosition 8.. 7. Proof of Corollary 4. To go from Theorem 3 to Corollary 4 is straightforward. Suose that x and n are integers with x 3 27 2 n and x 4, 5, 8, 15, 19, 38, 121. We may suose further that x is ositive since our desired conclusion is trivial otherwise. If n 0 mod 3, say n = 3n 0, we have x 3 27 2 n x 3 x 1 3 > 3x 3 3x 2 3 5/3 x 4/3, where the last inequality holds for x 3 and, for x = 1 or 2, the desired result follows immediately. We may thus suose that n = 3n 0 + r for r 1, 2, and so x 3 27 2 n = 3 2 n 0 2 r/3 x 3 2 n0 x 2 + 3 x2 n0+r/3 + 9 2 2n0+2r/3.

RAMANUJAN-NAGELL CUBICS 21 If x 3 2 n0+r/3 > 1, then, once again, we have x 3 27 2 n > 3x 3 3x 2 3 5/3 x 4/3. Otherwise, alying Theorem 3, we thus have x 3 27 2 n > 3 2 0.62n 0 x 2 + xx 1 + xx 1 2, at least rovided n 0 > 12. Since 0.62 x + 1 2 0.62n0, 3 we obtain inequality 8, after a little wor. The values n 0 12 corresond to n 38. For these, we readily chec that 8 is satisfied, excet for x 4, 5, 8, 15, 19, 38, 121. 8. Proof of Theorem 5. We next roceed with the roof of Theorem 5. Let us suose that a, b and c are given integers and that there exist integers x and n such that Writing u = x + a/3, we find that x 3 + ax 2 + bx + c = ±2 n. 44 u 3 + b a 2 /3u + 2a 3 /27 + c ab/3 = 1 δ 2 n, where now either u or 3u is an integer and δ 0, 1. If u 3 = 1 δ 2 n then, from 44, so that b a 2 /3x + a/3 = ab/3 c 2a 3 /27 b a 2 /3x = c + a 3 /27. If b = a 2 /3 then we necessarily have that a = 3t for some integer t, whereby b = 3t 2 and c = t 3, so that x 3 + ax 2 + bx + c = x + t 3. Otherwise, we conclude that x = a3 27c 27b 9a 2. Next, to treat the cases where 3 a and x = u a/3 for u 4, 5, 8, 15, 19, 38, 121,

22 MARK BAUER AND MICHAEL A. BENNETT let us suose that for one of these choices of x we have 45 x > max 8 b a 2 /3 3 + a/3, 4a 3 /27 + 2c 2ab/3 3/4 + a/3. Since x a/3 + u we thus have 46 max 8 b a 2 /3 3, 4a 3 /27 + 2c 2ab/3 3/4 < u. It follows that b = a 2 /3 + b 0 for some integer b 0 with b 0 < 1 2 u 1/3. Since we also have c ab 0 3 a3 27 < 1 2 u 4/3, we may write c = ab 0 3 + a3 27 + c 0, where c 0 is an integer with c 0 < 1 2 u 4/3. Writing a = 3t, for t an integer, we thus have whence u t 3 + 3tu t 2 + 3t 2 + b 0 u t + t 3 + tb 0 + c 0 = ±2 n, 47 u 3 + b 0 u + c 0 = ±2 n, where u 4, 5, 8, 15, 19, 38, 121, b 0 < 1 2 u 1/3 and c 0 < 1 2 u 4/3. A short comutation reveals that 47 is satisfied only for u, b 0, c 0 ±4, 0, 0, ±5, 0, 3, ±8, 0, 0, the first and last of which corresond to x 3 + ax 2 + bx + c = x + t 3. The case u, b 0, c 0 = ±5, 0, 3 leads to a, b, c, x = 3t, 3t 2, t 3 +3, 5 t and 3t, 3t 2, t 3 3, 5 t. If we suose finally that u 3 1 δ 2 n and x u a/3 for any u satisfying u 4, 5, 8, 15, 19, 38, 121, from the fact that u 3 1 δ 2 n = b a 2 /3u + 2a 3 /27 + c ab/3 > 0 we thus have, in all cases, alying Corollary 4 to 3u 3 27 2 n, that

RAMANUJAN-NAGELL CUBICS 23 u 4/3 b a 2 /3u + 2a 3 /27 + c ab/3 2 max b a 2 /3 u, 2a 3 /27 + c ab/3 and so u max 8 b a 2 /3 3, 4a 3 /27 + 2c 2ab/3 3/4, whereby x max 8 b a 2 /3 3 + a/3, 4a 3 /27 + 2c 2ab/3 3/4 + a/3. This comletes the roof of Theorem 5. 9. Proof of Theorem 6. In this section, we will rove Theorem 6. Let us begin by suosing that D is an odd integer and that we have with x 3 i + D = 2 i, i 1, 2, 3, 4, x 1 < x 2 < x 3 < x 4 and 1 < 2 < 3 < 4. Then, for each i 1, 2, 3, x i+1 x i x 2 i+1 + x i+1 x i + x 2 i = x 3 i+1 x 3 i = 2 i 2 i+1 i 1, whereby we may write x i+1 = x i + a i 2 i with a i a ositive integer. Substituting this into 2 i+1 = x 3 i+1 + D, we find that 48 2 i+1 i = 1 + 3a i x 2 i + 3a 2 i 2 i x i + a 3 i 2 2i. Let us suose first that D is ositive and write x i = D 1/3 + y i for y i a ositive real number. From 48, we have 2 i+1 i = 1+a i 3D 2/3 3a i 2 i D 1/3 + a 2 i 2 +3a 2i i y i y i + a i 2 i 2D 1/3. Alying the arithmetic-geometric mean inequality to the first braceted term on the right-hand-side of this equation, we thus have that 49 2 i+1 i 1 + 2 3 3 a 2 i D 1/3 2 i + 3a i y i y i + a i 2 i 2D 1/3.

24 MARK BAUER AND MICHAEL A. BENNETT Notice that and hence 2 i = x 3 i + D = 3y i D 2/3 3y 2 i D 1/3 + y 3 i y i + a i 2 i 2D 1/3 = y i + 3a i y i D 2/3 3a i y 2 i D 1/3 + a i y 3 i 2D 1/3. If we have y 1 < 1 then, since y 2 y 1 = x 2 x 1 = a 1 2 1 1, we thus have y 2 > 1 and hence, in all cases, may write y 2 = D θ for θ > 0. Since, again by the arithmetic-geometric mean inequality, the function fθ = 3D 2/3+θ 3D 1/3+2θ + D 3θ is monotone increasing as a function of θ, we thus have 2 2 > 3D 2/3 3D 1/3 + 1 > 2D 2/3, at least assuming that D 27. From 49, it follows that 2 3 > 2D 2/3 2 3 3 2D + 32D 2/3 2D 1/3 > 4 2 3 3 D 5/3 > 1.8 D 5/3. We therefore have x 3 > D 5/9 and so, again alying 49, we find that 2 4 > 1.5 D 11/3 and so x 3 4 > 1.5 D 11/3 D > D 11/3, since we may assume that D 2. We thus have, from Corollary 4 where we tae x = 3x 4, that D = x 3 4 2 4 4/3 x 4 > D 11/3 4/9 = D 44/27, a contradiction. Suose next that D < 0 so that x 2 > x 1 > D 1/3. From 48, we have that 2 2 > 2 D 2/3 and hence 2 3 > 2 2 3x 2 2 + 3 2 2 x 2 + 2 22 > 8 D 2. A final aeal to 48 imlies that 2 4 > 2 3 3x 2 3 + 3 2 3 x 3 + 2 23 > 512 D 6.

RAMANUJAN-NAGELL CUBICS 25 We thus have, again aealing to Corollary 4, that D = x 3 4 2 4 x 4/3 4 = 2 4 + D 4/9 > 512 D 6 4/9 > D 8/3, a contradiction. This comletes the roof of Theorem 6. It is erhas worthwhile noting that we now of only three odd values of D for which the equation x 3 +D = 2 n has even as many as two solutions in integers x and n, namely D = 215 with x, n = 6, 0 and 7, 7, D = 1 with x, n = 0, 0 and 1, 1, and D = 3, with x, n = 1, 1, 1, 2 and 5, 7. 10. Thue-Mahler equations. As noted earlier, the equation 50 x 3 xy 2 + 8y 3 = 2, which generalizes 1, has itself at most finitely many solutions in integers x, y, n, which may be determined effectively following arguments of Tzanais and de Weger [13], based uon lower bounds for linear forms in comlex and -adic logarithms, together with comutational techniques from Diohantine aroximation. Hambroo [6] has an imlementation of such an aroach which wors well in the case of few rimes and low degree forms i.e. recisely the situation in which we find ourselves; aealing to his Thue-Mahler solver, the only corime solutions to 50 are with x, y one of 113, 53, 19, 9, 2, 1, 1, 1, 1, 5, 0, 1, 1, 0, 1, 1, 3, 1, 3, 1, 5, 1, 7, 3, 8, 1, 13, 6. Similarly, the equation x 3 13xy 2 + 20y 3 = 2. has corresonding solutions with x, y among 21, 5, 4, 1, 3, 1, 1, 1, 1, 0, 1, 1, 2, 1, 3, 1, 4, 1, 7, 3, 11, 5, 13, 1, 19, 3. There exist comletely general bounds for the number of solutions to Thue-Mahler equations that deend only uon the degree of the given form F and the number of rimes on the right hand side of the equation F x, y = α1 1 α. Along these lines, let us note that Evertse [5] has shown that if F is an irreducible cubic form and is a fixed rime,

26 MARK BAUER AND MICHAEL A. BENNETT then the equation F x, y = n has at most 7 60 + 6 7 4 solutions in integers. This bound, while admirably uniform, exceeds 10 50. REFERENCES 1. M. Bauer and M. Bennett, Alication of the hyergeometric method to the generalized Ramanujan-Nagell equation, Ramanujan J. 6 2002, 209 270. 2. M. Bennett, G. Martin, K. O Bryant and A. Rechnitzer, Exlicit bounds for rimes in arithmetic rogressions, to aear. 3. F. Beuers, On the generalized Ramanujan-Nagell equation. I., Acta Arith. 38 1980/81, no. 4, 389 410. 4. G. V. Chudnovsy, On the method of Thue-Siegel, Ann. Math. 117 1983, 325 382. 5. J.-H. Evertse, Uer Bounds for the Numbers of Solutions of Diohantine Equations, Thesis, Amsterdam, 1983. 6. K. Hambroo, Imlementation of a Thue-Mahler solver, htt: //www.math.rochester.edu/eole/faculty/hambroo/research/ubc_2011_ fall_hambroo_yle_udated.df 7. S. V. Kotov, The greatest rime factor of a olynomial Russian, Mat. Zameti 13 1973, 515 522. 8. K. Mahler, Zur Aroximation algebraischer Zahlen, II., Math. Ann. 108 1933, 37 55. 9. T. Nagell, The diohantine equation x 2 +7 = 2 n, Ar. Mat. 4 1961, 185 187. 10. Physics Forum, htt://www.hysicsforums.com/showthread.h?t=443958 11. T. N. Shorey, A. J. van der Poorten, R. Tijdeman and A. Schinzel, Alications of the Gel fond-baer method to diohantine equations, Transcendence Theory : Advances and Alications, Academic Press, London, 1977, 59 77. 12. C. L. Siegel under the seudonym X, The integer solutions of the equation y 2 = ax n + bx n 1 +... +, J. London Math. Soc. 1 1926, 66-68. 13. N. Tzanais and B. M. M. de Weger, How to exlicitly solve a Thue-Mahler equation, Comositio Math. 84 1992, no. 3, 223 288. Deartment of Mathematics, University of Calgary, Calgary, AB, Canada Email address: mbauer@math.ucalgary.ca Deartment of Mathematics, University of British Columbia, Vancouver, B.C., V6T 1Z2 Canada Email address: bennett@math.ubc.ca