A232682 Numbers n such that the equation a^2 + 7*n*b^2 = 7*c^2 + n*d^2 has no solutions in positive integers for a, b, c, d.

3, 5, 6, 7, 10, 11, 12, 13, 14, 15, 17, 19, 20, 22, 23, 24, 26, 27, 28, 30, 31, 33, 34, 35, 38, 39, 40, 41, 43, 44, 45, 46, 47, 48, 51, 52, 54, 55, 56, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 73, 75, 76, 77, 78, 79, 80, 82, 83, 85, 86, 87, 88, 89, 90, 91, 92, 94, 95, 96, 97, 99

Offset: 1

Author

V. Raman, Nov 27 2013

Keywords

• nonn

Comments

With n = 3, the equation a^2 + 21*b^2 = 3*d^2 + 7*c^2 has no solutions in positive integers for a, b, d, c as the following proof shows: Let's assume that gcd(a, b, d, c) = 1, otherwise if gcd(a, b, d, c) = g, then a/g, b/g, d/g, c/g would be a smaller set of solutions to the equation. Considering modulo 7 arithmetic, we have a^2 - 3*d^2 == 0 (mod 7). Since a square is always congruent to 0 (mod 7), 1 (mod 7), 2 (mod 7) or 4 (mod 7), this is possible if and only if a == 0 (mod 7) and d == 0 (mod 7). Now let a = 7*p, d = 7*q, so a^2 = 49*p^2, d^2 = 49*q^2. Substituting this into the equation a^2 + 21*b^2 = 3*d^2 + 7*c^2 gives 49*p^2 + 21*b^2 = 147*q^2 + 7*c^2, i.e. 7*p^2 + 3*b^2 = 21*q^2 + c^2. Taking modulo 7 arithmetic with this equation again gives 3*b^2 - c^2 == 0 (mod 7). By using the same argument as above, this is possible if and only if b == 0 (mod 7) and c == 0 (mod 7). We already showed that a == 0 (mod 7) and d == 0 (mod 7), so gcd(a, b, d, c) should be a multiple of 7. This contradicts our assumption that gcd(a, b, d, c) = 1 and a/7, b/7, d/7, c/7 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, d, c) is (0, 0, 0, 0).

We can similarly prove for the other values of n by taking modulo 7 arithmetic if the only solution to a^2 - n*d^2 == 0 (mod 7) is a == 0 (mod 7) and d == 0 (mod 7). This happens if n == 3, 5, 6 (mod 7).

On the other hand, if we take modulo n arithmetic and if a^2 - 7*d^2 == 0 (mod n) has the only solution a == 0 (mod n) and d == 0 (mod n), then n is a member of this sequence. If r is a prime factor of n and if r^2 does not divide n and the equation a^2 - 7*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 = 7*k is a multiple of 7 and not a multiple of 49, taking modulo 7 arithmetic yields 'a' to be a multiple of 7. Putting a = 7*p, and dividing the equation by 7 gives 7*(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 7 times another number that can be written by the same form x^2+k*y^2.

If n is in this sequence, so is nk^2 for any positive integer k. - Charles R Greathouse IV, Dec 13 2013

If a prime p divides n and 7 is a quadratic non-residue mod p then n is not in the sequence. - Charles R Greathouse IV, Dec 13 2013

Examples

			n = 2 is not a member of this sequence because 15 = 1^2 + 14*1^2 = 7*1^2 + 2*2^2.
n = 3 is a member of this sequence because there is no positive integer m which can be simultaneously written as both x^2+21*y^2 and 7*x^2+3*y^2.
		

Links

• V. Raman, Proof for individual terms

Crossrefs

Cf. A232531, A232532, A232681.