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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.

%I A232682 #30 Dec 13 2013 14:54:32
%S A232682 3,5,6,7,10,11,12,13,14,15,17,19,20,22,23,24,26,27,28,30,31,33,34,35,
%T A232682 38,39,40,41,43,44,45,46,47,48,51,52,54,55,56,59,60,61,62,63,65,66,67,
%U A232682 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
%N 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.
%C A232682 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).
%C A232682 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).
%C A232682 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.
%C A232682 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.
%C A232682 If n is in this sequence, so is nk^2 for any positive integer k. - _Charles R Greathouse IV_, Dec 13 2013
%C A232682 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
%H A232682 V. Raman, <a href="/A232682/a232682.txt">Proof for individual terms</a>
%e A232682 n = 2 is not a member of this sequence because 15 = 1^2 + 14*1^2 = 7*1^2 + 2*2^2.
%e A232682 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.
%Y A232682 Cf. A232531, A232532, A232681.
%K A232682 nonn
%O A232682 1,1
%A A232682 _V. Raman_, Nov 27 2013