cp's OEIS Frontend

Numbers n such that the equation |x^2 - 2y^2| = n has no solutions in integers x,y. In other words, this sequence is the complement of A035251. - Thomas Ordowski and Altug Alkan, Feb 10 2017

It appears that this is the set of all numbers which contain at least one prime factor p congruent to 3 (mod 8) or 5 (mod 8) raised to an odd power.

With n = 3, the equation a^2 + 6*b^2 = 2*c^2 + 3*d^2 has no solutions in positive integers for a, b, c, d as the following proof shows: Let's assume that gcd(a, b, c, d) = 1, otherwise if gcd(a, b, c, d) = g, then a/g, b/g, c/g, d/g would be a smaller set of solutions to the equation. Considering modulo 3 arithmetic, we have a^2 - 2*c^2 == 0 (mod 3). Since a square is always congruent to 0 (mod 3) or 1 (mod 3), this is possible if and only if a == 0 (mod 3) and c == 0 (mod 3). Now let a = 3*p, c = 3*q, so a^2 = 9*p^2, c^2 = 9*q^2. Substituting this into the equation a^2 + 6*b^2 = 2*c^2 + 3*d^2 gives 9*p^2 + 6*b^2 = 18*q^2 + 3*d^2, i.e. 3*p^2 + 2*b^2 = 6*q^2 + d^2. Taking modulo 3 arithmetic with this equation again gives 2*b^2 - d^2 == 0 (mod 3). By using the same argument as above, this is possible if and only if b == 0 (mod 3) and d == 0 (mod 3). We already showed that a == 0 (mod 3) and c == 0 (mod 3), so gcd(a, b, c, d) should be a multiple of 3. This contradicts our assumption that gcd(a, b, c, d) = 1 and a/3, b/3, c/3, d/3 are a smaller set of solutions to the above mentioned equation. By using the proof of infinite descent, this implies that the only possible set of solutions to (a, b, c, d) is (0, 0, 0, 0).

We can similarly prove for the other values of n by taking modulo n arithmetic if the only solution to a^2 - 2*d^2 == 0 (mod n) is a == 0 (mod n) and d == 0 (mod n). If r is a prime factor of n and if r^2 does not divide n and the equation a^2 - 2*d^2 == 0 (mod r) has the only solution a == 0 (mod r) and d == 0 (mod r), we can also take modulo r arithmetic to prove that n is a member of this sequence. If n is odd, n is a member of this sequence if and only if 2 is a quadratic non-residue (mod n). Alternately, n should have at least one prime factor congruent to 3 (mod 8) or 5 (mod 8) raised to an odd power.

If n = 2*k is even and not a multiple of 4, taking modulo 2 arithmetic yields a to be even. Putting a = 2*p, and dividing the equation by 2 gives 2*(p^2+k*b^2) = (c^2+k*d^2). This equation will have no solution in positive integers p, b, c, d if and only if there is no number that can be written by the form x^2+k*y^2 that is twice another number that can be written by the same form x^2+k*y^2. If n is even, n is a member of this sequence if and only if n has at least one prime factor congruent to 3 (mod 8) or 5 (mod 8) raised to an odd power.

If n is a multiple of 4, then n = 4*m is a member of this sequence if and only if m is a member of this sequence.

Also numbers n such that there is no number that can be written by the form x^2+n*y^2 that is twice another number that can be written by the same form x^2+n*y^2.

Positive numbers that are not the difference between two legs of a Pythagorean right triangle. - Michael Somos, Apr 02 2017

Record values are in A240470.

The number of (x, y) pairs in row n is 1 for n = 1 and 2, and 2^P, with P = P1 + P7, where P1 and P7 are the number of prime factors 1 modulo 8 and 7 modulo 8, respectively, of A057126(n), for n >= 3.

See A057126 for comments concerning its representation by x^2 - 2*y^2.

The numbers A057126 are given by 2^e_2 * Product_{i=1..P1} p_{1,i}^e_{1,i} * Product_{j=1..P7} p_{7,j}^e_{7,j}, with the odd primes p_{1,i} and p_{7,j} congruent to 1 and 7 modulo 8, respectively. See A007519 and A007522 for these odd primes. Together with 2 these primes are given in A038873, and without 2 in A001132. The exponents are e_2 = 0 or 1, and e_{1,i} and e_{7,j} are nonnegative. The a(1) = 1 is obtained if all exponents vanish. For the proof see Lemma 18 of the linked W. Lang paper, pp. 22 - 23.

The general solutions are obtained from each fundamental solution by application of integer powers of the matrix Auto' = Mat([3,4], [2,3]). See the linked paper eq (28), p. 14, and eq. (40), p. 17 for D = 2, and k = A057126(n). For the explicit form of the powers of Auto' in terms of Chebyshev polynomials S(n, 6) = A001109(n+1) see there eq. (38), and Lemma 10, eq. (43), p. 17.

The conversion to the pair of proper solutions (X, Y) of X^2 - 2*Y^2 = A057126(n) is given by (X, Y) = (2*y - x, x - y). This may result in solutions with negative Y values. They are then transformed to the fundamental positive proper solutions via the mentioned matrix Auto'. See the right part of the example below. For this conversion see also the Nov 09 2009 comment in A035251 by Franklin T. Adams-Watters.