A Lucas-Lehmer approach to generalised Lebesgue-Ramanujan-Nagell equations
aa r X i v : . [ m a t h . N T ] D ec A LUCAS-LEHMER APPROACH TO GENERALISEDLEBESGUE-RAMANUJAN-NAGELL EQUATIONS
VANDITA PATEL
Abstract.
We describe a computationally efficient approach to resolving equa-tions of the form C x + C = y n in coprime integers, for fixed values of C , C subject to further conditions. We make use of a factorisation argumentand the Primitive Divisor Theorem due to Bilu, Hanrot and Voutier. Introduction
Ramanujan [12], in 1913, conjectured that the only positive integral solutions tothe equation x + 7 = 2 n are (1 , , (3 , , (5 , , (11 , , (181 , . This was proven by Nagell [11] in 1948, and the equation is now called the Ramanujan–Nagell equation. More generally, equations of the form(1) C x + C = C n where C , C , C are fixed non-zero integers are referred to as generalised Ramanujan–Nagell equations. Various special cases of (1) have been considered by many authorsusing a variety of methods [3]. For any such C , C , C , it is straightforward toreduce (1) to solving S -unit equations. This allows us to conclude that the set ofsolutions to (1) is finite by a famous theorem of Siegel. It also gives an effectivealgorithm for solving the equation.In this paper we consider the generalisation(2) C x + C = y n where C , C are fixed, but x , | y | > n ≥ x, y, n ), but does not alonegive a practical method for determining them. In fact, the earliest special case of(2) appears to be due to Victor Lebesgue [9] who in 1850 solved (2) for C = C = 1.In 1948, Nagell [11] solved the cases C = 1, C = 3, 5, and it is now usual to referto the equation(3) x + C = y n as the Lebesgue–Nagell equation. In a series of papers (culminating in [7]), J.H.E.Cohn solved (3) for many values of C >
0. After the appearance of the celebrated
Date : December 23, 2019.2010
Mathematics Subject Classification.
Primary 11D61, Secondary 11D41, 11D59.
Key words and phrases.
Exponential equation, Lehmer sequences, primitive divisor theorem,Thue equation. theorem of Bilu, Hanrot and Voutier (BHV) on primitive divisors of Lucas andLehmer sequences [4], Cohn revisited (3) in [8], showing that BHV allows for aneasy resolution for 77 values in the range 1 ≤ C ≤ C = 74 and C = 86 were solved by Mignotte and de Weger [10]. Using the modular approachbased on Galois representations of elliptic curves and modular forms, the cases C = 55 and C = 95 were solved by Bennett and Skinner [2]. The remaining 19values were dealt with in a pioneering paper due Bugeaud, Mignotte and Siksek [5],which combines Baker’s theory with the modular approach. A related work whichrelies heavily on BHV is due to Abu Muriefah, Luca, Siksek and Tengely [1], andadapts Cohn’s method to the equation x + C = 2 y n (see also [14], [15] for relatedequations).In view of Cohn’s work, it is natural to consider (2), which we refer to as thegeneralised Lebesgue–Ramanujan–Nagell equation. We extend Cohn’s method sothat it applies in far greater generality.More precisely, we study equations of the form:(4) C x + C = y n , x, y ∈ Z + , gcd( C x , C , y n ) = 1 , n ≥ . We may assume without loss of generality that n is an odd prime, or that n = 4.We prove the following. Theorem 1.
Let C be a positive squarefree integer and C a positive integer.Write C C = cd where c is squarefree. We assume that C C . Let p be an odd prime for which the equation (5) C x + C = y p , x, y ∈ Z + , gcd( C x , C , y p ) = 1 , has a solution ( x, y ) . Then either,(i) p ≤ , or(ii) p = 7 and y = 3 , or , or(iii) p divides the class number of Q ( √− c ) , or(iv) p | (cid:16) q − (cid:16) − cq (cid:17)(cid:17) , where q is some prime q | d and q ∤ c . In Section 6, we give an effective method that solves (4) for a given value of n ≥
3. Our algorithm relies on standard algorithms for solving Thue equationsand determining integral points on elliptic curves. We implemented our methodin
Magma [6] which has inbuilt implementation of these algorithms and we usedthis, together with Theorem 1, to determine the solutions to (4) for 2 ≤ C ≤ ≤ C ≤
80 subject to the restrictions: C is squarefree, gcd( C , C ) = 1, and C C C = 1 and 1 ≤ C ≤
100 is completely solved in [5], which incorporates the earlierwork of Cohn, Bennett and Skinner, and Mignotte and de Weger.The author thanks Yann Bugeaud and Szabolcs Tengely for useful conversations.2.
Primitive prime divisors of Lehmer sequences A Lehmer pair is a pair of algebraic integers α, β , such that ( α + β ) and αβ are non–zero coprime rational integers and α/β is not a root of unity. The Lehmersequence associated to the Lehmer pair ( α, β ) is˜ u n = ˜ u n ( α, β ) = ( α n − β n α − β , if n is odd , α n − β n α − β , if n is even . A prime p is called a primitive divisor of ˜ u n if it divides ˜ u n but does not divide( α − β ) · ˜ u · · · ˜ u n − . We shall make use of the following celebrated theorem [4]. Theorem 2 (Bilu, Hanrot and Voutier) . Let α , β be a Lehmer pair. Then ˜ u n ( α, β ) has a primitive divisor for all n > , and for all prime n > . A Lehmer pair ( α, β ) is called n - defective if ˜ u n ( α, β ) does not have a primitivedivisor. Two Lehmer pairs ( α, β ) and ( α ′ , β ′ ) are said to be equivalent if α/α ′ = β/β ′ ∈ {± , ±√− } . Table 2 of [4] gives all equivalence classes of n - defective Lehmer pairs for all 6 < n ≤
30 except n = 8, 10, 12. In particular, • there are no 11-defective Lehmer pairs; • every 13-defective Lehmer pair is equivalent to (( √ a + √ b ) / , ( √ a − √ b ) / a, b ) = (1 , − • every 7-defective Lehmer pair is equivalent to (( √ a + √ b ) / , ( √ a − √ b ) / a, b ) = (1 , − , − , − , − , − , − Preliminary Descent
Throughout Sections 3 and 4 we maintain the following assumptions and nota-tion:(a) C is a squarefree positive integer, C is a positive integer and gcd( C , C ) = 1.We moreover suppose that C C C C = cd where c , d are positive integers and c is squarefree.(b) ( x, y ) satisfies (5).(c) p is an odd prime. Moreover, if p = 3 then we suppose additionally that C C / p does not divide the class number of Q ( √− c ). Lemma 3.1.
Let ( x, y ) be a solution to (5) . Let O K be the ring of integers for thenumber field K = Q ( √− c ) . Then there is some δ ∈ O K such that (6) C x + d √− c = δ p C ( p − / . Moreover, we have (7) δ p C p/ − ¯ δ p C p/ = 2 d · √− c √ C . Proof.
Let K = Q ( √− c ) and O K its ring of integers. Let h K be the class numberof K and we assume that p ∤ h K . As C C y is odd.As C , c are both squarefree, gcd( C , C ) = 1 and C C = cd it follows that C | c . Let C = p · · · p r where we note that the primes p , . . . p r ramify in K .We factorise equation (5) in O K as follows: (cid:0) C x + d √− c (cid:1) (cid:0) C x − d √− c (cid:1) = C · y p = p · · · p r · y p . Let us write p i for the prime ideal above p i where 1 ≤ i ≤ r . Let a = p · · · p r and we obtain: ( C x + d √− c ) O K = p · · · p r · y p = a − p · ( ay ) p = ( C (1 − p ) / ) · ( ay ) p VANDITA PATEL where ay is a principal ideal of O K . Indeed, [ ay ] p = 1 in the class group. Thereforethe class [ ay ] has order dividing p . By assumption p ∤ h K . Thus ay is principal.Therefore, we write ay = δ O K . The unit group of O K has order 2, 4 or 6, andis therefore p -divisible, unless p = 3. However, for p = 3 we have assumed that C C / K = Q ( √− p -divisible. Thus adjusting δ byan appropriate unit we obtain (6). Subtracting the conjugate from (6), we get δ p C ( p − / − ¯ δ p C ( p − / = 2 d √− c, which is equivalent to (7). This completes the proof of the lemma. (cid:3) Remark. If C C ≡ y to be even. In that caseit is no longer true that we can express ( C x + d √− c ) O K in the form ay p where a = C O K . 4. Satisfying the Lehmer condition
Let K = Q ( √− c ) as before, and consider the extension, L/K , where L = Q ( √− c, √ C ). Observe that L/K is trivial if C = 1, and is quadratic other-wise. We write O L for its ring of integers and set α = δ/ √ C , β = ¯ δ/ √ C . Thusequation (7) becomes(8) α p − β p = 2 d · √− c √ C . For the remainder of this section, in the case − c δ = r + s √− c, ¯ δ = r − s √− c, where r , s are integers. In the case − c ≡ δ = r + s √− c , ¯ δ = r − s √− c , where r and s are either both odd or both even. Lemma 4.1. α and β are algebraic integers. Moreover, αβ = y, p C x + p − C = α p , p C x − p − C = β p . Proof.
By the proof of Lemma 3.1, a = C O K and so √ C O L = a O L which divides ay O L = δ O L . Hence α = δ/ √ C is an algebraic integer.Dividing (6) by √ C gives √ C x + √− C = α p and applying complex conjugationgives √ C x − √− C = β p . Multiplying the two equations gives y p = ( αβ ) p . Butas α , β are complex conjugates, y , αβ are both positive, so y = αβ as required. (cid:3) Lemma 4.2. ( α + β ) is a non–zero rational integer.Proof. By Lemma 4.1, ( α + β ) is an algebraic integer. However,( α + β ) = (cid:18) δ + ¯ δ √ C (cid:19) = ( r /C if − c r /C if − c ≡ , thus ( α + β ) is a rational number as well as being an algebraic integer. Thus it isa rational integer.Next, we suppose that ( α + β ) = 0. Then δ is purely imaginary, and (6) impliesthat x = 0. This contradicts our assumption that x is positive. (cid:3) The following is immediate from Lemma 4.1.
Lemma 4.3. αβ is a non-zero rational integer. Lemma 4.4. ( α + β ) and αβ are coprime. Moreover α/β is not a unit.Proof. Suppose that ( α + β ) and αβ are not coprime. Then there exists a prime q of O L which divides both. Thus, q | α, β . By Lemma 4.1, q | y and q | (2 √ C x ).As we saw previously, y must be odd. Hence q | y and q | C x , contradicting ourcoprimality assumption.Finally suppose α/β is a unit. In particular α | β and β | α . We claim that α is a unit. Suppose otherwise, and let q | α be a prime of O L . Then q | β andwe obtain a contradiction as above. Hence α must be a unit and so β is a unit.Therefore y = αβ is a unit in Z . Thus y = ±
1. This contradicts C x + C = y p and the positivity assumption for the solution. (cid:3) Lemmata 4.1, 4.2, 4.3, 4.4 provide a proof to the following:
Proposition 4.5.
Let α, β be as above. Then α and β are algebraic integers.Moreover, ( α + β ) and αβ are non–zero, coprime, rational integers and α/β is nota unit. Proof of Theorem 1
In this section we prove Theorem 1. We suppose p > p ∤ h K . We wouldlike to show that ( p, y ) = (7 , , ,
9) or there is some prime q | d , q ∤ c suchthat p | B q where B q = q − (cid:16) − cq (cid:17) = 1 q + 1 if (cid:16) − cq (cid:17) = − . Let ( α, β ) be as above. Proposition 4.5 tells us that ( α, β ) is indeed a Lehmer pair.We denote by ˜ u k the associated Lehmer sequence. From (9), (10) we have(11) α − β = ( s √− c √ C if − c s √− c √ C if − c ≡ . Combining with (8) gives(12) ˜ u p = α p − β p α − β = ( ds if − c ds if − c ≡ . We suppose first that ( α, β ) is not p -defective. Thus there is a prime q | ˜ u p suchthat q ∤ ( α − β ) and q ∤ ˜ u ˜ u · · · ˜ u p − . We claim that q = 2. Suppose q = 2. Let q be a prime of O L dividing q . Then α p ≡ β p (mod q ) , α β (mod q ) . Hence α/β has order p in ( O L / q ) ∗ . This group has order Norm( q ) −
1. As L hasdegree 4, Norm( q ) = 2 or 4 or 16. Thus p = 3 or 5 which contradicts p > q = 2.Next we claim that q ∤ C . Suppose q | C . Let q be a prime of O L dividing q .Then α p ≡ β p (mod q ) and √ C ≡ q ). By Lemma 4.1, q | √− C . Hence q | C and q | (2 C ). But C , C are coprime and q = 2 giving a contradiction.Thus q ∤ C . VANDITA PATEL
From (11), the fact that q ∤ C and q ∤ ( α − β ) we deduce that q ∤ c as required.Let q be a prime of K above q . Then δ/δ q ) and ( δ/δ ) p ≡ q ).If ( − c/q ) = 1 then F q = F q and so p | ( q − − c/q ) = − F q = F q .However, δ/δ (mod q ) belongs to the kernel of the norm map F ∗ q → F ∗ q which hasorder q + 1. Thus in this case, p | ( q + 1). Hence p | B q .To complete the proof we need to consider the case where ( α, β ) is p -defective.By Theorem 2 and the discussion following it, we know that p = 7 or 13. Moreover( α, β ) is equivalent to ( α ′ , β ′ ) = (( √ a + √ b ) / , ( √ a − √ b ) /
2) where the possibilitiesfor ( a, b ) are listed in that discussion. Recall α/α ′ = β/β ′ ∈ {± , ±√− } . More-over, y = αβ . Thus if α/α ′ = β/β ′ = ±√− y = − α ′ β ′ . However, y is positive and α ′ β ′ is also positive in all cases. Thus α/α ′ = β/β ′ = ±
1. Hence y = α ′ β ′ = ( a − b ) /
4. When ( a, b ) = (1 , − , − , − y = 2, 4,2 respectively. This contradicts our assumption that C C p = 7, and ( a, b ) = (1 , − , − , − y = 5, 3, 9. This completes the proof.We note in passing that it is not possible to eliminate the cases p = 7, y = 5, 3,9. For example, for p = 7, y = 5, there are 59893 possibilities for a triple ( C , C , x )which satisfies C x + C = y p = 5 and all our other restrictions.6. Effectively Determining Solutions
Let C , C satisfy condition (a) of Section 3. Theorem 1 gives a list of possibleodd prime exponents n = p for which (4) might have solutions. As noted in theintroduction, we may without loss of generality suppose that n = p is an oddprime, or that n = 4. In this section, we outline a practical method to computethese solutions for fixed such value of n . We consider three cases. Case I: n is an odd prime p ∤ h K , and if p = 3 then C C / r , s be as in (9), (10).Let d ′ = ( d if − c d if − c ≡ . From (12) we obtain s | d ′ . Thus we have only a few possibilities for s . To determinethe solutions we merely have to determine the possible values of r correspondingto each s | d ′ . We shall write down an explicit polynomial f s ∈ Z [ X ] whose integerroots contain all the possible values of r corresponding to s .Fix s | d ′ . If − c f s ( X ) = ( X + s √− c ) p − ( X − s √− c ) p s √− c − d · C ( p − / s . Clearly f s ∈ Z [ X ]. Moreover, f s ( r ) = δ p − δ p δ − δ − d · C ( p − / s = 0using (7) and (9).If − c ≡ f s ( X ) = ( X + s √− c ) p − ( X − s √− c ) p s √− c − p · d · C ( p − / s . Again f s ∈ Z [ X ] and f s ( r ) = (2 δ ) p − (2 δ ) p δ − δ ) − p · d · C ( p − / s = 0using (7) and (10). Case II: n is an odd prime p , with either p | h K , or p = 3 and C C / C = p · · · p r and let p i be the unique primeideal of O K above p i . Let a = p · · · p r . We have( C x + d √− c ) O K = a · y p where y is an ideal of O K . Let b , . . . , b h be ideals of O K that form a system ofrepresentatives for the class group. Then, for some 1 ≤ i ≤ h = h K , we have yb i is principal. Therefore ab − pi must be principal. We test the ideals ab − pi forprincipality. Fix i such that ab − pi = ǫ O K where ǫ ∈ K ∗ and write yb i = δ O K ,where δ ∈ O K . Then(13) C x + d √− c = µ · ǫ · δ p where µ is a unit. If p = 3 or C C / µ is a p -th power andwe can absorb this in the the δ p factor. In this case we suppose µ = 1. Otherwisewe also consider µ = 1, ω = ( − √− / ω . We write δ as in (9), (10)depending on whether − c − c ≡ √− c and clear denominators to obtain an equationof the form F ( r, s ) = t where t is a positive integer, and F ∈ Z [ X, Y ] is a homogeneous polynomial ofdegree p ≥
3. This is a Thue equation. In our implemention we used Magma’sinbuilt Thue solver which is an implementation of the algorithm in Smart’s book[13, Chapter VII], which is based on linear forms in logarithms.
Case III: n = 4. We write X = C y , Y = C xy, and note that ( X, Y ) is now an integral point on the elliptic curve Y = X − C C X. We apply Magma’s inbuilt function for determining integral points on elliptic curveswhich is based on linear forms in elliptic logarithms, as described in Smart’s book[13, Chapter XIII].
VANDITA PATEL Solutions
We are interested in solving (4) for 2 ≤ C ≤
10, 1 ≤ C ≤
80 subject to therestrictions: C is squarefree, gcd( C , C ) = 1, and C C n = 4 or that n = p isan odd prime. For each such pair ( C , C ), Theorem 1 yields a finite set S ( C , C )of odd primes p for which we need to solve (5). Thus for each such pair ( C , C )we need only solve (4) for n ∈ S ( C , C ) ∪ { } , and for each such value n we mayapply one of the methods explained in Section 6. We implemented our approach in Magma [6]. The results of our computation are given below. C C x y n C C x y n C C x y n References [1] F. S. Abu Muriefah and F. Luca and S. Siksek and Sz. Tengely,
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