SEARCH FOR GOOD EXAMPLES OF HALL S CONJECTURE. 1. Introduction. x 3 y 2 = k

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1 MATHEMATICS OF COMPUTATION Volume 00, Number 0, Pages S (XX) SEARCH FOR GOOD EXAMPLES OF HALL S CONJECTURE STÅL AANDERAA, LARS KRISTIANSEN, AND HANS KRISTIAN RUUD Abstract. A good example of Hall s Conjecture is a pair of natural numbers x, y such that 0 < x 3 y 2 < x /2. We have implemented a new algorithm and found nine not previously known good examples. Moreover, we have verified that all good examples with x < 0 29 are now found. (.) Consider the equation. Introduction x 3 y 2 = k where x, y N and k Z. It is easy to see that the equation has infinitely many solutions when k = 0 (let x = t 2 and y = t 3 where t is a natural number). By Siegel s theorem, there will be only finitely many solutions of the equation when k 0. Moreover, it is hard to find solutions where k is small compared to x and y. Hall [4] conjectured that there is a constant C such that k > Cx /2 for any solution of (.) where k 0. This conjecture is discussed in e.g. Danilov [2] and Elkies [3]. The general opinion is that Hall s original conjecture is too strong, and his conjecture has been reformulated to a weaker modern variant: For any exponent e < 2, there exists a constant C e > 0 such that k > C e x e. For more on Hall s conjecture and further references, see Calvo et al. []. Hall s Conjecture is neither proved nor disproved. To shed some light on the conjecture, researchers have searched for solutions of (.) where 0 < k < x /2. We follow the terminology of Calvo et al. [] and refer to such solutions as good examples of Hall s conjecture. We will say that a pair of natural numbers (x, y) is a good example when 0 < x 3 y 2 < x /2. We have implemented a new algorithm for finding good examples. We have found 9 not previously known good examples, and we have checked that all good examples (x, y) where x is less than 0 29 are now found. Other algorithms for finding good examples can be found in Elkies [3] and Calvo et al. []. The algorithm of Elkies has an asymptotic running time similar to ours. The algorithm of Calvo et al. seems to have slightly better asymptotic running time, but the authors offer no proof that their algorithm will detect all good examples within its search space. However, it is easily verified that the algorithm will find all the new good examples presented in this paper if it is given enough running time. Received by the editor November 5, 205 and, in revised form, November 6, Mathematics Subject Classification. Primary Y50, 65A05; Secondary D25. c 207 American Mathematical Society

2 2 STÅL AANDERAA, LARS KRISTIANSEN, AND HANS KRISTIAN RUUD 2. The Algorithm Definition 2.. We define the polynomials B, C, F and H by where x, y, p, q are positive integers. B(q, p, x) = p 2 q 2 x C(q, p, x, y) = p 3 3pq 2 x + 2q 3 y F (q, p, x, y) = 4pC 3B 2 H(q, p, x, y) = 9F B 8C 2 Our algorithm is based on the four polynomials given by the definition above. The values of these polynomials will be small when (x, y) is a good example and p q is the approximation to x /2 given by the next theorem. Theorem 2.2. Let (x, y) be a good example. Then, there exist p, q, Q N and δ R such that (i) p = qx /2 ( + δ), (ii) 0 < q < x /6 < Q and (iii) qx /2 (Q + q) Moreover, p and q are co-prime. < δ < qx /2 Q. Proof. In this proof we use continued fractions. For more on continued fractions, see e.g. Niven et al. [7] or Khintchine [6]. First we note that x /2 is an irrational number when (x, y) is a good example. (If x /2 is a natural number, then (x, y) will not be a good example as x 3 y 2 = 0. But x /2 is either a natural number or an irrational number. Thus, x /2 is irrational.) Let a 0, a, a 2,... be the coefficients for the continued fraction for x /2, that is x /2 = lim n [a 0; a,..., a n ]. The list of coefficients will be infinite as x /2 is irrational. Let h i and k i be, respectively, the nominator and the denominator of the convergent [a 0 ; a,..., a i ], that is hi k i = [a 0 ; a,..., a i ]. Then, for any i N, we have < h i k i (k i + k i+ ) x /2 k i < k i k i+ and k i < k i+. Now, pick the least j such that k j+ > x /6. Let q = k j, let p = h j and let Q = k j+. Then, we have < p q(q + Q) q x/2 < qq where q < x /6 < Q (we cannot have q = x /6 as x /6 N). Next, let δ be the real number such that p = qx /2 ( + δ). Then, we have < qx /2 ( + δ) q(q + Q) x /2 < q qq. Thus < δ < qx /2 (q + Q) qx /2 Q. Note that p and q are co-prime since h j and k j are co-prime for any convergent h j k j.

3 SEARCH FOR GOOD EXAMPLES OF HALL S CONJECTURE 3 The proof of the next theorem will be given in Section 5. Theorem 2.3. Let (x, y) be a good example where x 2. Then, there exist coprime natural numbers p and q such that q < x /6 and 0 < C < 3qx /6 +, F < 8q + and H < 72q 4 +. In order to explain our algorithm, we introduce the notion of a good septuple. Definition 2.4. Let x, y, p, q, c N and f, h Z. The sequence (x, y, p, q, f, c, h) is a good septuple if p and q are co-prime and q < x /6, f = F (q, p, x, y), c = C(q, p, x, y), h = H(q, p, x, y) and x 3 y 2 < x /2. We define the function E by (x, y) if there exists p such that (x, y, p, q, f, c, h) is a good septuple E(q, f, c, h) = (0, 0) otherwise. An algorithm for computing E(q, f, c, h). It follows from Definition 2. that (2.) (2.2) (2.3) (2.4) B = H + 8C2 9F p = F + 3B2 4C x = p2 B q 2 y = 3pq2 x p 3 + C 2q 3. If (x, y, p, q, f, c, h) is a good septuple, we have f = F (q, p, x, y) and c = C(q, p, x, y) and h = H(q, p, x, y). Thus, we can compute E(q, f, c, h) by the following procedure: Compute b such that b = B and equality (2.) holds. Use f, c, h for respectively F, C, H. Compute p such that equality (2.2) holds. Use f, c, b for respectively F, C, B. Compute x such that equality (2.3) holds. Use p, q, and use b for B. Compute y such that equality (2.4) holds. Use p, q, and use c for C. Check if (x, y, p, q, f, c, h) is a good septuple, that is, check if p, x, y are natural numbers and check if x 3 y 2 < x /2. If it is a good septuple, the output is (x, y); otherwise, the output is (0, 0). Overview of the algorithm. We are now ready to give an overview of our algorithm for finding good examples (where x 2). A pair (x, y) is a good example if and only if there exist p, q, f, c, h such that (x, y, p, q, f, c, h) is good septuple (this follows straightforwardly from Theorem 2.3). Since q < x /6 when (x, y, p, q, f, c, h) is a good septuple, we can decide if (x, y) is a good example checking all good septuples where q < x /6. Our algorithm works by generating good septuples by a sieve method. The input to the algorithm is a natural number x max. The algorithm outputs all good examples (x, y) where x < x max. Figure gives a high-level description of the algorithm.

4 4 STÅL AANDERAA, LARS KRISTIANSEN, AND HANS KRISTIAN RUUD q max := xmax. /6 for q :=, 2,..., q max do begin Generate a set F q of possible values for f. for each f F q do begin Generate a set C qf of possible values of c. for each c C qf do begin Generate a set H qfc of possible values for h. for each h H qfc do begin (x, y) := E(q, f, c, h) (see Definition 2.4). Output (x, y) if (x, y) (0, 0). end end end end Figure. The high-level description of our algorithm. Our algorithm might also output good examples (x, y) where x x max. If x x max, there might exist a good septuple (x, y, p, q, f, c, h) where q < x /6 max and c < 3qx /6 max. This is why our algorithm has found good examples with x > More on the Algorithm In this section we explain how the algorithm in Figure computes the sets F q, C qf and H qfc. We will also discuss the running time. The correctness of our algorithm is based on the next couple of lemmas. Lemma 3.. We have (i) C p 3, F p 4 and H p 6 (mod q 2 ) (ii) H 8C 2 (mod 9 F ) (iii) p 4 2pC + F 0 and H 5p 3 C 4p 6 (mod q 3 ) (iv) either F 0 or F (mod 4) (v) F and q are co-prime if p and q are co-prime. Proof. Clause (i), (ii) and (iii) follow straightforwardly from Definition 2.. In the case when B is even there exists i Z such that F = 3B 2 = 3(2i) 2 = 2i 2. Hence, we have F 0 (mod 4) when B is even. A similar argument shows that F (mod 4) when B is odd. This proves (iv). In order to see that (v) holds, assume that F and q are not co-prime. Then there is a prime m that divides both F and q. By (i), m will also divide p. Hence, p and q are not co-prime. The proof of the next lemma is straightforward.

5 SEARCH FOR GOOD EXAMPLES OF HALL S CONJECTURE 5 Lemma 3.2. Let m and n be natural numbers such that m n, and let P (x) be a polynomial. Furthermore, let x 0 x (mod q m ), and let y P (x) (mod q n ). Then we have y P (x 0 ) (mod q m ). 3.. The Construction of the Set F q. Assume that (x, y, p, q, f, c, h) is a good septuple. By Lemma 3. (i), we have f p 4 (mod q 2 ). Now, let p 0 be any natural number such that p 0 p (mod q). By Lemma 3.2, we have f p 4 0 (mod q). Thus, f can be written of the form (3.) f = (p 4 0 mod q) + iq for some i Z and some p 0 where p 0 < q. By Theorem 2.3 we have (3.2) (8q + ) < f < 8q + and by Lemma 3. (iv), we have (3.3) f 0 or f ( mod 4). If q >, our algorithm computes the set F q such that f F q iff (3.), (3.2) and (3.3) hold. In the case when q =, the algorithm let F q = { 8, 7, 4, 3, 0,, 4, 5, 8 }. Note that we only need q to determine the set F q (in particular we do not need the value of p) The Construction of the Set C qf. Assume that (x, y, p, q, f, c, h) is a good septuple where x < x max. By Lemma 3. (i), we have f p 4 (mod q 2 ). Let p be any natural number such that p p (mod q 2 ). By Lemma 3.2, we have f p 4 ( mod q 2 ). Hence, (3.4) p 4 = f + iq 2 for some i Z. Let p P iff 0 < p < q 2 and (3.4) holds. Our algorithm computes the set P from q and f (the set will contain at most four elements). By Lemma 3. (i), we have c p 3 (mod q 2 ). By Lemma 3.2, we have c p 3 ( mod q 2 ). Hence, c can be written of the form (3.5) c = (p 3 mod q 2 ) + iq 2 for some p P and some i Z. By Theorem 2.3, we have 0 < c < 3qx /6 +. Recall that q max = x /6 max. Hence (3.6) 0 < c 3qq max +. If q >, our algorithm computes the set C qf such that c C qf iff (3.5) and (3.6) holds. If q =, the algorithm computes the set C qf such that c C qf iff (3.6) holds. We need q, q max and the set P to determine the set C qf (and we need q and f F q to determine P ).

6 6 STÅL AANDERAA, LARS KRISTIANSEN, AND HANS KRISTIAN RUUD 3.3. The Construction of the Set H qcf. Assume that (x, y, p, q, f, c, h) is a good septuple. Let p 2 be such that p 2 p (mod q 3 ). By Lemma 3. (iii), we have p 4 2pc + f 0 (mod q 3 ). By Lemma 3.2, we have (3.7) p 4 2 2p 2 c + f 0 ( mod q 3 ). Our algorithm computes the set P 2 such that p 2 P 2 iff 0 p 2 < q 3 and (3.7) holds. We need q, c C qf and f F q to determine the set P 2. By clause (ii) and (iii) of Lemma 3. and Lemma 3.2, there is p 2 P 2 such that (3.8) h 8c 2 ( mod 9 f ) and h 5p 3 2c 4p 6 2 ( mod q 3 ). By Theorem 2.3, we have (3.9) (72q 4 + ) < h < 72q 4 +. The case when 3 does not divide q. We will now explain how our algorithm computes the set H qfc when 3 does not divide q. We know that p and q are co-prime (since (x, y, p, q, f, c, h) is a good septuple). By Lemma 3. (v), f and q are co-prime. Thus, 9 f and q 3 will also be co-prime since 3 does not divide q. The Chinese Reminder Theorem and (3.8) yield h such that (3.0) 0 h < 9 f q 3 and h = h + i9 f q 3 for some i Z. Let S(q, p 2, f, c) denote the unique h such that (3.0) holds. The number S(q, p 2, f, c) can be computed from q, p 2, f, c by the Euclidean method. If q >, our algorithm computes the set H qfc such that h H qfc iff (3.9) holds and h = S(q, p 2, f, c) + i9 f q 3 for some p 2 P 2 and some i Z. The set H qfc will be computed by a tailored algorithm when q =. The case when 3 divides q. By Lemma 3. (v), f and q are co-prime. Thus, if 3 divides q, the greatest common divisor of 9 f and q 3 will be 9. The equations in (3.8) have a solution iff h 8c 2 (mod 9). If there is a solution, the Chinese Reminder Theorem yields h such that (3.) 0 h < f q 3 and h = h + i f q 3 for some i Z, and our algorithm can proceed as described in the case when 3 does not divide q. We need q, f F q, c C qf and the set P 2 to determine H qfc (and we need q, f F q, c C qf to determine P 2 ) The Running Time. Let S denote the number of elements in the set S. It is easy to see that F q 7q. It follows from (3.5) and (3.6) that C qf P (3qq max + ) q 2. We have computed the set P from q and f, but the number of elements in the set will not depend on q or f (tedious considerations will show that P 4). Hence there is a constant k such that C qf (kq max )/q. We can now determine an upper

7 SEARCH FOR GOOD EXAMPLES OF HALL S CONJECTURE 7 p # x r q Comments. ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) ) The first 40 entries of the table can be found in []. These entries are found by Hall [4], Gebel et al. [5], Elkies [3], Calvo et al. [] and Johan Bosman using the software of Calvo et al. [] 2) Found by Jiménez Calvo [8]. 3) Found by Calvo et al. []. 4) Found by the authors of this paper. 5) From the Danilov-Elkies infinite Fermat-Pell family, see [3] or []. Table. Good examples of Hall s conjecture. bound for how many times the third loop in Figure will be executed. We have q max q= f F q C qf q max q= kq max q f F q q max 7q max q= q max = kq max 7q max q= kq max q q = O(q 2 max log q max ). This shows that the number of times the loop for each c C qf... will be executed is of order O(qmax 2 log q max ). It follows from (3.9), (3.0) and (3.) that there exists a constant k such that the number of elements in H qfc is bounded by kq. Thus the number of elements in H qfc is of order O(q max ). This entails that the innermost loop of the algorithm in Figure will be executed O(qmax 2 log q max ) O(q max ) times. In the innermost loop numbers will be added, multiplied and divided. The running time of these arithmetical computations will be of order O(log 2 x). Thus, as q max = x /6 max and x < x max, our algorithm runs in time O(q 2 max log q max ) O(q max ) O(log 2 x max ) = O(q 3 max log q max ) O(log 2 x max ) = O((x /6 max) 3 log x /6 max) O(log 2 x max ) The algorithm of Elkies [3] has a similar running time. = O(x /2 max log O() x max ).

8 8 STÅL AANDERAA, LARS KRISTIANSEN, AND HANS KRISTIAN RUUD 4. Computations and Results A first implementation of the algorithm was written in Python. The results of test runs looked promising, and we reimplemented the algorithm in C using the Gnu Multi-Precision library to carry out operations with arbitrary-length integers. The C implementation was run with q max = 0000 (corresponding to a x max of 0 24 ). The run lasted for 368 processor-hours, and after 840 processor-hours the algorithm found example #43 in Table. A subsequent run, with q max = (corresponding to a x max of ), lasted for processor-hours and found an example that was earlier found by Calvo et al. []. After these experiences we modified our C code for being run on the Norwegian national computing facilities (Notur). The computations at national facilities took about.3 millions processor-hours. Our computations have detected 9 previously unknown good examples and verified that all good examples where x is less than 0 29 are included in Table. The second column of Table shows the x of a good example (x, y). The r appearing in the third column of the table, is given by r = x /2 /(x 3 y 2 ). High values of r indicate that (the original) Hall s Conjecture is false. The fourth column shows the rational approximation p q to x/2. 5. The Proof of Theorem 2.3 Throughout this section we will use w to denote x /2 and k to denote x 3 y 2. Lemma 5.. Let (x, y) be a good example. Then there exists γ R such that y = x 3/2 ( + γ) and k 2x 3 < γ < k + 2x 3. Proof. Let γ R be such that y = x 3/2 ( + γ). Then we have (5.) y = x 3/2 ( + γ) = w 3 ( + γ). Furthermore, we have γw 3 = w 3 ( + γ) w 3 = y w 3 (5.) = (y 2 w 6 )/(y + w 3 ) = (y 2 x 3 )/(y + w 3 ) since w = x /2 = k/(y + w 3 ) since x 3 y 2 = k This establishes that γw 3 = k/(y + w 3 ), and thus γ = k/w 3 (y + w 3 ). Furthermore, since k < w = x /2 and x 2, we have (5.2) We also have (5.3) γ = k w 3 (y + w 3 ) < k w 6 < w 5 <. γ = k 2w 6 ( + 2 γ)

9 SEARCH FOR GOOD EXAMPLES OF HALL S CONJECTURE 9 since γ (5.4) (5.2) = k w 3 (y + w 3 ) Next we prove that (5.) = γ < k w 3 (w 3 ( + γ) + w 3 ) k = w 6 (2 + γ) ( + γ 2 ) < + γ. Let r be a real number such that 0 < r <. Then we have = k 2w 6 ( + 2 γ). < + r 2 r2 = ( + r)( r 2 2 ) and thus /( r 2 ) < + r. By (5.2), we have γ <. Hence, if γ > 0, we have and if γ < 0, we have + γ 2 < γ 2 < + γ = + γ + γ = < + γ. 2 γ 2 This shows that /( + γ 2 ) < + γ. A symmetric argument shows that γ < /( + γ 2 ). Use that > r 2 r2 = ( r)( + r 2 2 ) when 0 < r <. This concludes the proof of (5.4). By (5.3) and (5.4), we have k k γ 2w 6 = k ( γ ) 2w 6 < γ < Finally, as k γ < w <, we have 4 This proves the lemma as w = x /2. k ( + γ ) 2w 6 = k 2w 6 < γ < k + 2w 6. k + k γ 2w 6. We are now ready to prove Theorem 2.3. Let (x, y) be a good example where x 2. We need to prove that there exist co-prime natural numbers p and q such that q x /6 and (Claim ) 0 < C(q, p, x, y) < 3qx /6 + (Claim 2) F (q, p, x, y) < 8q + (Claim 3) H(q, p, x, y) < 72q 4 + By Theorem 2.2 and Lemma 5., we have δ, γ R and p, q, Q N such that p and q are co-prime and (5.5) p = qw( + δ) and y = w 3 ( + γ).

10 0 STÅL AANDERAA, LARS KRISTIANSEN, AND HANS KRISTIAN RUUD Moreover, we have the following bounds: (5.6) 0 < q < w /3 < Q (5.7) δ < qwq (5.8) γ < k 2w 6. Note that we also have k < w as (x, y) is a good example. Furthermore, note that (x, y) is a good example where x 2. There are no other good examples where x is less than 5234 except the one where x = 2. It follows that w > 72 and Q > 4. The Proof of (Claim ). The definition of C says that (5.9) C = p 3 3pq 2 x + 2q 3 y By (5.5) and (5.9), we have C = q 3 w 3 (δ 3 + 3δ 2 + 2γ) (substitute qw( + δ) and w 3 ( + γ) for respectively p and y and simplify the expression). First we prove 0 < C. It follows from (5.7) that δ < 2, and thus we have (5.0) δ 3 + 3δ 2 = 2δ 2 + δ 2 ( + δ) > 0. If γ 0, it follows trivially from (5.0) that 0 < C. If γ < 0, then y < w 3, i.e. x 3 y 2 = k > 0. Then we have C = q 3 w 3 (δ 3 + 3δ 2 + 2γ) (5.0) > q 3 w 3 2γ (5.8) > q 3 w 3 2 k 2w 6 = q3 k w 3 (5.6) > k w 2 >. This proves that C >. We will now prove that C 0 (and thus we have C > 0). Assume that C = 0 (we prove that (x, y) is not a good example). By (5.9), we have p 3 = q(2q 2 y 3pqx). This shows that any prime divisor of q divides p, but p and q are co-prime, and thus q =. Equation (5.9) with C = 0 and q = gives (5.) 0 = p 3 3px + 2y We will prove that a solution of (5.) cannot yield a good example. The proof splits into the three cases: the case when x = p 2, the case when x > p 2, and the case when x < p 2. Assume x = p 2. Then y = p 3 gives a solution of (5.), but x 3 y 2 = 0. Thus, (x, y) is not a good example. Assume x > p 2. Then we have x = p 2 + m for some positive natural number m. Thus x 3 = (p 2 + m) 3 = p 6 + 3p 4 m + 3p 2 m 2 + m 3 By (5.) we have and y = 3px p3 2 = 3p(p2 + m) p 3 2 y 2 = p 6 + 3p 4 m + 9p2 m 2. 4 = p 3 + 3pm 2

11 SEARCH FOR GOOD EXAMPLES OF HALL S CONJECTURE Hence x 3 y 2 = 3p2 m 2 + m 3 > p 4 and we conclude that (x, y) is not a good example as p x /2. A similar argument shows that (x, y) is not a good example if x < p 2. This proves that C > 0. Next we prove that C < 3qw /3 +. This will complete the proof of (Claim ) since w /3 = x /6. We have C = q 3 w 3 (δ 3 + 3δ 2 + 2γ) q 3 w 3 ( δ δ γ ) ( [ ] (5.7,5.8) 3 [ ] 2 [ ] ) k < q 3 w qwq qwq 2w 6 = Q 3 + 3qw Q 2 + q3 k w 3 (5.6) < w + 3qw/3 + k w 2 < w + 3qw/3 + w. Thus, we have C < 3qw /3 + as w = x /2 > 72 in any good example where x 2. The Proof of (Claim 2). It follows straightforwardly from Definition 2. that F = p 4 6p 2 q 2 w 2 + 8pq 3 y 3q 4 w 4. By (5.5), we have F = q 4 w 4 (δ 4 + 4δ 3 + 8γ + 8γδ). We have F q 4 w 4 ( δ δ γ + 8 γ δ ) ( [ ] (5.7,5.8) 4 [ ] 3 [ ] k < q 4 w qwq qwq 2w 6 [ ] [ ] ) k + 8 2w 6 qwq = Q 4 + 4qw Q 3 + 4q4 k w 2 + 4q3 k w 3 Q < Q 4 + 4qw Q 3 + 4q4 w + 4q3 w 2 Q Thus, F < 8q + as w > 72 and Q > 4. (5.6) < 4 + 4q + 4q + Q4 wq. The Proof of (Claim 3). It follows straightforwardly from Definition 2. that H = p 6 5p 4 q 2 w p 3 q 3 y 45p 2 q 4 w q 5 pw 2 y + 27q 6 w 6 32q 6 y 2. By (5.5), we have H = q 6 w 6 (44δγ 32γ δ 3 γ + 20δ 2 γ + 6 δ 5 + δ 6 ). In order to prove that that H < 72q 4 +, we need (5.2) and (5.3) q 6 w 6 (32 γ δ 3 γ + 20 δ 2 γ + 6 δ 5 + δ 6 ) < 44q 6 w 6 δ γ [ ] [ ] (5.7,5.8) k < 44q 6 w 6 qwq 2w 6 = 72q5 k wq (5.6) < 72q 4. The proof of (5.2) is tedious, but straightforward. Use that k < w, that q < w /3 < Q, that w > 72 and that Q > 4.

12 2 STÅL AANDERAA, LARS KRISTIANSEN, AND HANS KRISTIAN RUUD Now we have H q 6 w 6 (44 δ γ + 32 γ δ 3 γ + 20 δ 2 γ + 6 δ 5 + δ 6 ) = 44q 6 w 6 δ γ + q 6 w 6 (32 γ δ 3 γ + 20 δ 2 γ + 6 δ 5 + δ 6 ). Thus, we have H < 72q 4 + by (5.2) and (5.3). Acknowledgments The authors want to thank the referees and the editor for valuable advice regarding the presentation. Furthermore, the authors want to thank N. D. Elkies for comments on an earlier version of this paper and R. H. Johansen for helpful advice on parallel programming. The authors gratefully acknowledge the support from the Norwegian meta-center for computational science (Notur). References. I. Jiménez Calvo, J. Herranz and G. Sáez, A new algorithm to search for small x 3 y 2 values. Mathematics of Computation, 78 (2009), L.V. Danilov, The Diophantine equation x 2 y 3 = k and Halls conjecture. Math. Notes Acad.Sci.USSR 32 (982), N. D. Elkies, Rational points near curves and small nonzero x 2 y 3 via lattice reduction. Pages 3363 in Algorithmic Number Theory (Proceedings of ANTS-IV;W. Bosma, ed.; Berlin: Springer, 2000; Lecture Notes in Comput. Sci. 838). 4. M. Hall, The Diophantine equation x 3 y 2. Computers in Number Theory (A. Atkin, B. Birch, eds.; Academic Press, 97), pp J. Gebel, A. Pethő and H. G. Zimmer, On Mordells equation. Compositio Math. 0 (998), A. Ya. Khintchine, Continued Fractions. (Translated by Peter Wynn.) P. Noordhoff, Ltd., Groningen, The Netherlands, I. Niven, H. S. Zuckerman and H. L. Montgomery, An Introduction to the Theory of Numbers. John Wiley & Sons, Inc Department of Mathematics, University of Oslo, P.O.Box 053 Blindern, NO-036 Oslo, Norway address: staal@math.uio.no Department of Mathematics, University of Oslo, P.O.Box 053 Blindern, NO-036 Oslo, Norway address: larsk@math.uio.no Department of Mathematics, University of Oslo, P.O.Box 053 Blindern, NO-036 Oslo, Norway

On the power-free parts of consecutive integers

ACTA ARITHMETICA XC4 (1999) On the power-free parts of consecutive integers by B M M de Weger (Krimpen aan den IJssel) and C E van de Woestijne (Leiden) 1 Introduction and main results Considering the