Compact representation of quadratic integers and integer points on some elliptic curves Filip Najman, University of Zagreb, Croatia 1 Introduction p Let Q( d) be a real quadratic field. Followingp [Mau] we define a compact representation of an algebraic number β 2 Q( d) to be k 2k−j Y αj β = ; (1) d j=1 j p p (aj +bj d) where dj 2 Z; αj = 2 2 Q( d); aj; bj 2 Z; j = 1; : : : ; k. Bounds on k, α and dj are given in [Mau], and all depend polynomially on log d. Com- pact representations are used to store the fundamental unit of the quadratic order OK . The reason for doing this is that, as is shown in [Lag], there is an infinite set of quadratic orders, such that the binary length of the fun- damental unit is exponential in log d. This makes it impossible to create an algorithm for solving the Pell equation with complexity less than expo- nential. Compact representations are polynomial in log d, and allow faster algorithms for solving the Pell equation. This representation is an extension of a compact representationp as de- ∗ fined in [BTW] from algebraic integers to all elements of Q ( d): It is often useful to do modular arithmetic on compact representations, for example for determening the solvability of certain Diophantine equations, as seen in [JW]. We present an algorithm for computing the value of a quadratic inte- ger represented by a compact representation as defined in [Mau]. In [BTW] and [JW] there are algorithms for doing modular arithmetic on compact representations as defined in [BTW], but to our knowledge there are no algorithms for doing modular arithmetic on compact representations as de- fined in [Mau]. The main problem is that the [BTW] representation requires that the partial products j−1 2j−i Y αi γ = α (2) j j d i=1 i be quadratic integers. Using Maurer's methods, we obtain representations that do not satisfy this requirement. On such a representation, using algo- rithms from [BTW] and [JW] is not possible. 1 It is expected that the number of integer points on an elliptic curve E in Weierstrass form depends on the rank E(Q). More precisely, Lang conjectured that it grows exponentially with the rank (see [JSil]). Since not much is known on the distribution of ranks in parametric families of elliptic curves, this makes it hard to expect to find (or even predict) all integer points on a family of elliptic curves in Weierstrass form. However, for some families of elliptic curves not in Weierstarss form, there are results which give evidence that the number of integer points might not depend on rank, and that actually the number of points can be the same for all curves in a family. Several such results involve so called D(n)-m-tuples. A set of positive integers fa1; a2; : : : ; amg is called a Diophantine D(n) m- tuple if aiaj + n is a perfect square for all 1 ≤ i < j ≤ m. Using our algorithm we shall improve the following results. In [Fuj1], the following theorem is proved: Theorem 1. Let f1; 2; cg be a D(−1)-triple such that c = c = 1 ((1 + p p k 8 2)4k + (1 − 2)4k + 6), and E the elliptic curve given by 2 Ek : y = (x + 1)(2x + 1)(cx + 1): (3) Assume that c − 2 is square-free and that the rank of E over Q equals two. Then, the integer points on E are given by c − 3 (x; y) 2 f(−1; 0); (0; ±1); ( ; ±s(c−2)); (s(3s−2t); ±(t−s)(2s−t)(st−c)); 2 (s(3s + 2t); ±(t + s)(2s + t)(st + c))g; (4) p p where s = c − 1 and t = 2c − 1. It is shown in [Fuj1] that if f1; 2; cg is a D(−1)-triple, then c = ck, where k is a positive integer. It should be mentioned that the assump- tion that rk(Ek(Q)) = 2 does not always hold. One example of this is that rk(E4(Q)) = 4. Also c − 2 does not always have to be square-free. Examples of this are c26 − 2 and c40 − 2. Fujita showed that Theorem 1 holds without the assumptions on the rank and c − 2 for k ≤ 40, except for k 2 f4; 7; 8; 11; 12; 15; 20; 24; 25; 27; 30; 36; 39g. We will exclude the cases k = 4; 7; 8; 11; 12; 15; 20; 25; 27; 30 and under the Extended Riemann Hy- pothesis, also the case k = 39. In [Fuj2] it is proved Theorem 2. Let k ≥ 1 be an integer, and Ek the elliptic curve given by 2 Ek : y = (F2k+1 + 1)(F2k+3x + 1)(F2k+5x + 1): (5) If the rank of Ek over Q equals one, then the integer points on Ek are given by (x; y) 2 f(0; ±1); (4F2k+2F2k+3F2k+4; 2 2 ±(2F2k+2F2k+3 + 1)(2F2k+3 − 1)(2F2k+3F2k+4 − 1))g: (6) Note that fF2k+1;F2k+3;F2k+5g is a D(−1)-triple. As in Theorem 1, in Theorem 2 the assumption rk(Ek(Q)) = 1 does not always hold. For example, rk(Ek(Q)) 6= 1 for k = 2; 3; 4; 5; 7; 9; 10. Fujita showed that Theorem 2 holds without the assumption on the rank of Ek for f ≤ k ≤ 50, except for the cases k 2 f9; 20; 24; 25; 32; 43g. We shall eliminate the cases k = 9; 20; 24; 25 and under the Extended Riemann Hy- pothesis, also the case k = 43. In [Duj3] it is proved Theorem 3. Let Ek be the elliptic curve given by 2 Ek : y = ((k − 1)x + 1)((k + 1)x + 1)(4kx + 1): (7) If the rank of Ek over Q equals one or 3 ≤ k ≤ 1000, then all integer points on Ek are given by (x; y) 2 f(0; ±1); (16k3 − 4k; ±(128k6 − 112k4 + 20k2 − 1))g: (8) We shall extend this result to 3 ≤ k ≤ 5000. Note that fk − 1; k + 1; 4kg is a D(1)-triple. In [Duj1] it was proven that this triple extends uniquely to a Diophantine quadruple fk − 1; k + 1; 4k; 16k3 − 4kg. Note also that in [Duj3] it was shown that the statement of Theorem 3 is valid for some subfamilies with ranks 2 and 3, which makes the conjecture that for all k ≥ 3 all integer points on (7) are given by (8) plausible. In [Duj4] it is proved Theorem 4. Let Ek be the elliptic curve given by 2 Ek : y = (F2k + 1)(F2k+2x + 1)(F2k+4x + 1): (9) If the rank of Ek over Q equals one or 2 ≤ k ≤ 50, then all integer points on Ek are given by (x; y) 2 f(0; ±1); (4F2k+1F2k+2F2k+3; 2 ±(2F2k+1F2k+2 − 1)(2F2k+2 + 1)(2F2k+2F2k+3 + 1))g (10) We shall extend this result to 2 ≤ k ≤ 200. Note that fF2k;F2k+2;F2k+4g is a D(1)-triple. In [Duj2] it was proven that this triple extends uniquely to a Diophantine quadruple fF2k;F2k+2; F2k+4; 4Fk+1F2k+2F2k+3g. 3 2 An algorithm for modular arithmetic on compact representations We develop an algorithm, which for a given compact representation (1) of p 2k−j p a quadratic integer x+y ∆ = Qk αj , where x+y ∆ is the standard 2 j=1 dj 2 representation of the given quadratic integer, and a positive integer n, com- putes the values x and y modulo n. Our algorithm is a modification of the algorithm described in Theorem 5.7 in [BTW]. Algorithm 1 k−1 Qk 2 mdl:=n·2 i=1 dk a[1]:=c1 (mod mdl), b[1]:=f1 (mod mdl); rem:= 1 For (i=2,i<k−1,++i) f 2 2 xtmp:=(a[i−1] +b[i−1] ∆) ai + 2∆a[i−1]b[i−1]bi (mod mdl) 2 2 ytmp:=(a[i−1] +b[i−1] ∆) bi + 2a[i−1]b[i−1]ai (mod mdl) rem:=rem2 2 divisor:=gcd(4di−1,xtmp,ytmp) 2 4di−1·rem rem:= divisor mdl mdl:= gcd(mdl, divisor) xtmp a[i]:= divisor (mod mdl) ytmp b[i]:= divisor (mod mdl) rem reduction:= gcd(rem, mdl) while (gcd(reduction,mdl)> 1) reduction reduction= gcd(reduction;mdl) If reduction6= 1 f mult:=reduction−1 (mod mdl) rem rem:= reduction a[i]:=a[i]· mult (mod mdl) b[i]:=b[i]· mult (mod mdl) g g This algorithm would be exactly the same algorithm as the one from The- orem 5.7 in [BTW] if all the γj, as defined in (2), are quadratic integers. Algorithm 1 takes polynomial time. As in the algorithm from [BTW], we make use of the recursive equation 2 2 4di−1γi = γi−1αi: (11) 2 If all the γj are not quadratic integers, then 4di−1 does not divide the right hand side in (11), so we must remember the part that does not divide (rem in the algorithm). Also, it may occur that a factor of n is canceled, so we 4 actually get x (mod n0) and y (mod n0), where n0 divides n. To correct this we have Algorithm 2: Algorithm 2 Algorithm 2 receives as input a compact representation of a quadratic in- teger P , and a positive integer n. It also uses a version of Algorithm 1, alg1(P; n; k), which receives P and n, and stores the value of n0 in k.