A261246 Positive integers D such that the generalized Pell equation X^2 - D Y^2 = 2 is soluble.

2, 7, 14, 23, 31, 34, 46, 47, 62, 71, 79, 94, 98, 103, 119, 127, 142, 151, 158, 167, 191, 194, 199, 206, 223, 238, 239, 254, 263, 271, 287, 302, 311, 322, 334, 343, 359, 367, 382, 383, 386, 391, 398, 431, 439, 446, 463, 478, 479, 482, 487, 503, 511

Offset

1,1

Comments

For the fundamental positive solution x(n)^2 - a(n)*y(n)^2 = 2 see (x(n) = A261247(n), y(n) = A261248(n)), for n >= 1.

Conjecture: The sequence consists of all numbers D not a square and even D = 2*d has odd d with prime factors of the form 1 or 7 (mod 8). Odd D has prime factors of the form 1 or 7 (mod 8) but there is an odd number of primes of the form 7 (mod 8). The following will prove that these conditions for D are necessary in order to have solutions.

This conjecture is false. For the odd D case see the counterexamples in A263010, and for the even D in A264352. - Wolfdieter Lang, Nov 12 2015

If there is a solution for D, D not a square, then only one class of solution exists due to Nagell's Theorem 110, p. 208, because then 2 divides 2*D. All solutions will be proper because 2 is a prime.

For the even prime D = p = 2 the positive fundamental solution is [x(1) = 2, y(1) = 1].

For odd primes D = p there can be solutions only for p == +7 (mod 8), that is p from A007522. Then x and y are both odd. Proof: Consider a solution of x^2 - p*y^2 = 2. The parities of x and y have to be either even and even or odd and odd. For odd x one has x^2 == +1 (mod 8) (because x^2 = 8*T(X) + 1 with x = 2*X+1 and the triangular numbers T = A000217); similarly for y^2 if y is odd. In the even-even case x^2 and y^2 are both congruent to 4 (mod 8). The even-even case leads to 4 - 4*p = 2 (mod 8), excluding all odd p, namely p == 1, 3, 5, 7 (mod 8). The odd-odd case is 1 - p*1 = 2 (mod 8), and p == 1, 3, 5 (mod 8) are excluded. Therefore, only p == 7 (mod 8) qualifies for a solution, and then x and y will be both odd.

For D = p == 7 (mod 8) from A007522 one can test if there exists a fundamental positive solution (at most one class can exist, therefore there is either no solution or just one) [2*U(p)+1, 2*V(p)+1] by checking the two inequalities (see Nagell, eq. (4) and (5), p. 206) 0 <= V(p) < floor((Y(p)/sqrt(X(p) + 1) - 1)/2) and 0 <= U(p) <= floor((sqrt(X(p) + 1) - 1)/2), with the positive fundamental solution [X(p), Y(p)] of X^2 - p*Y^2 = +1. These solutions can be found in (A033313(k), A033317(k)) if A000037(k) is the prime p == 7 (mod 8) one is testing.

For composite even D there are solutions only if D/2 is odd. Proof: If D is even then x has to be even, hence x^2 == 0 (mod 4) and then D*y^2 == -2 (mod 4), hence D cannot be 0 (mod 4). Thus an even D can only be of the form D = 2*d with d odd. The modulo 3 and modulo 5 argument used in the next case will show that d can have only prime factors of the form +1 or -1 (mod 8).

For composite odd D one finds like above that the even-even x and y case is excluded, and the odd-odd case needs D == -1 (mod 8) == 7 (mod 8). Hence a candidate for D is from A004771 - A007522. D cannot have any prime factor p of the form 3 or 5 (mod 8) because otherwise x^2 == 2 (mod p), but the Legendre symbol (2/p) = -1 for such p's (see, e.g., Nagell, Theorem 81, p. 136). For example, D = 15 = 3*5 cannot have a solution. Thus the only candidates for D have prime factors p of the form +1 or +7 (mod 8), with the number of the latter ones being odd. E.g., D = 7*17 = 119 qualifies as a candidate and it has indeed solutions, namely the ones obtainable from the fundamental one [11, 1].

The general proper positive solutions for D(n) = a(n) are obtained from the fundamental ones [x(n), y(n)] given in A261247 and A261248 with the help of powers of the matrix M(n) = [[u(n), D(n)*v(n)], [v(n), u(n)]], where u(n) and v(n) are the positive fundamental solutions of U(n) - D(n)*V(n) = 1, by (x(n; k), y(n; k))^T = M(n)^k (x(n), y(n))^T (T for transposed), for k >= 0. [u(n), v(n)] = [A033313(j(n)), A033317(j(n))] if A000037(j(n)) = D(n) = a(n).

Observation: All degrees (7, 47, 79, 103, 119, 127) of the modular equations derived for solving Ramanujan's question 699 by Galkin & Kozirev (see reference and A318732) are terms of this sequence. - Hugo Pfoertner, Sep 24 2023

References

J. W. S. Cassels, Rational Quadratic Forms, Cambridge, 1978; see Chap. 3.

V. M. Galkin, O. R. Kozyrev, On an algebraic problem of Ramanujan, pp. 89-94 in Number Theoretic And Algebraic Methods In Computer Science - Proceedings Of The International Conference, Moscow 1993, Ed. Horst G. Zimmer, World Scientific, 31 Aug 1995

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

Links

Giovanni Resta, Table of n, a(n) for n = 1..1000

Wolfdieter Lang, Binary Quadratic Forms (indefinite case).

Example

The first fundamental solutions [x(n), y(n)] are (the first entry gives D(n)=a(n)):
[2, [2, 1]], [7, [3, 1]], [14, [4, 1]],
[23, [5, 1]], [31, [39, 7]], [34, [6, 1]],
[46, [156, 23]], [47, [7, 1]], [62, [8, 1]],
[71, [59, 7]], [79, [9, 1]], [94, [1464, 151]],
[98, [10, 1]], [103, [477, 47]], [119, [11, 1]],
[127, [2175, 193]], [142, [12, 1]],
[151, [41571, 3383]], [158, [88, 7]],
[167, [13, 1]], [191, [2999, 217]],
[194, [14, 1]], [199, [127539, 9041]],
[206, [244, 17]], [223, [15, 1]], [238, [108, 7]],
[239, [2489, 161]], ...

Mathematica

Select[Range[600], False =!= Reduce[x^2 - # y^2 == 2, {x, y}, Integers] &] (* Giovanni Resta, Aug 12 2017 *)

Crossrefs

Cf. A261247, A261248, A007522, A004771, A000217, A000037, A033313, A033317, A263010.

See also A038873 (2 and primes == +-1 mod 8), A001132.

Sequence in context: A325159 A087324 A340664 * A008865 A227582 A249852

Adjacent sequences: A261243 A261244 A261245 * A261247 A261248 A261249

Keyword

nonn

Author

Wolfdieter Lang, Sep 06 2015

Status

approved