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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. 4
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 OFFSET 1,1 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 LINKS V. Raman, Proof for individual terms EXAMPLE 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. CROSSREFS Cf. A232531, A232532, A232681. Sequence in context: A053048 A154663 A028983 * A335657 A038550 A204232 Adjacent sequences:  A232679 A232680 A232681 * A232683 A232684 A232685 KEYWORD nonn AUTHOR V. Raman, Nov 27 2013 STATUS approved

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Last modified September 23 12:30 EDT 2020. Contains 337310 sequences. (Running on oeis4.)