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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

Primes p such that x^4 == 2 has more than two (in fact four) solutions mod p. This is the sequence of terms common to A040098 (primes p such that x^4 == 2 has a solution mod p) and A007519 (primes of form 8n+1). Solutions mod p are represented by integers from 0 to p - 1. For p > 2, i is a solution mod p of x^4 == 2 iff p - i is a solution mod p of x^4 == 2, thus the sum of first and fourth solution is p and so is the sum of second and third solution. The solutions are given in A065909, A065910, A065911 and A065912. - Klaus Brockhaus, Nov 28 2001

Primes of the form x^2+64y^2. - T. D. Noe, May 13 2005

n^3 divides a(n) for n in A121707.

It appears that p^(3k-1) divides a(p^k) for all integer k > 1 and prime p > 2:

for prime p > 2, p^2 divides a(p), p^5 divides a(p^2) and p^8 divides a(p^3).

Additionally, p^3 divides a(3p) for prime p > 2.

For prime p > 3, p divides a(p+1) and p^3 divides a(2p+1);

for prime p > 5, p divides a(3p+1) and p^3 divides a(4p+1);

for prime p > 7, p divides a(5p+1) and p^3 divides a(6p+1):

It appears that p divides a((2k+1)p+1) for integer k >= 0 and prime p > 2k+3, and p^3 divides a(2kp+1) for integer k > 0 and prime p > 2k+2.

p divides a((p+1)/2) for primes in A002145: primes of the form 4n+3, n >= 1.

p^2 divides a((p+1)/2) for primes in A007522: primes of the form 8n+7, n >= 0.

n*(2*n+1) divides a(2*n+1) for n >= 1. - Franz Vrabec, Dec 20 2020

Nonnegative numbers of the form 8x^2 - y^2. - Jon E. Schoenfield, Jun 03 2022

Every a(n) from a((p-1)/2) to a(p-1) is divisible by prime p for p = {7, 47, 191, 383, 439, 1151, 1399, 2351, 2879, 3119, 3511, 3559, ...} = A167860, apparently a subset of primes of the form 8n+7 (A007522).

7^3 divides a(13) and 7^2 divides a(10)-a(13).

Every a(n) from a(kp-1 - (p-1)/2) to a(kp-1) is divisible by prime p from A167860.

Every a(n) from a((p^2-1)/2) to a(p^2-1) is divisible by prime p from A167860. For p=7 every a(n) from a((p^3-1)/2) to a(p^3-1) and from a((p^4-1)/2) to a(p^4-1)is divisible by p^2.