

A232681


Numbers n such that the equation a^2 + 5*n*b^2 = 5*c^2 + n*d^2 has no solutions in positive integers for a, b, c, d.


4



2, 3, 6, 7, 8, 10, 12, 13, 14, 15, 17, 18, 21, 22, 23, 24, 26, 27, 28, 30, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 46, 47, 48, 50, 51, 52, 53, 54, 56, 57, 58, 60, 62, 63, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 77, 78, 82, 83, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 96, 97, 98
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OFFSET

1,1


COMMENTS

With n = 2, the equation a^2 + 10*b^2 = 2*d^2 + 5*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 5 arithmetic, we have a^2  2*d^2 == 0 (mod 5). Since a square is always congruent to 0 (mod 5), 1 (mod 5) or 4 (mod 5), this is possible if and only if a == 0 (mod 5) and d == 0 (mod 5). Now let a = 5*p, d = 5*q, so a^2 = 25*p^2, d^2 = 25*q^2. Substituting this into the equation a^2 + 10*b^2 = 2*d^2 + 5*c^2 gives 25*p^2 + 10*b^2 = 50*q^2 + 5*c^2, i.e. 5*p^2 + 2*b^2 = 10*q^2 + c^2. Taking modulo 5 arithmetic with this equation again gives 2*b^2  c^2 == 0 (mod 5). By using the same argument as above, this is possible if and only if b == 0 (mod 5) and c == 0 (mod 5). We already showed that a == 0 (mod 5) and d == 0 (mod 5), so gcd(a, b, d, c) should be a multiple of 5. This contradicts our assumption that gcd(a, b, d, c) = 1 and a/5, b/5, d/5, c/5 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 5 arithmetic if the only solution to a^2  n*d^2 == 0 (mod 5) is a == 0 (mod 5) and d == 0 (mod 5). This happens if n == 2, 3 (mod 5).
On the other hand, if we take modulo n arithmetic and if a^2  5*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  5*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 = 5*k is a multiple of 5 and not a multiple of 25, taking modulo 5 arithmetic yields 'a' to be a multiple of 5. Putting a = 5*p, and dividing the equation by 5 gives 5*(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 5 times another number that can be written by the same form x^2+k*y^2.
If n is a multiple of 25, then n = 25*m is a member of this sequence if and only if m is a member of this sequence.
This appears to be the complement of A031363. If so, the definition could be simplified.  Franklin T. AdamsWatters, Apr 02 2016


LINKS

Table of n, a(n) for n=1..71.
V. Raman, Proof for individual terms


EXAMPLE

n = 2 is a member of this sequence because there is no positive integer m which can be simultaneously written as both x^2+10*y^2 and 5*x^2+2*y^2. The former requires the sum of {2, 5, 7, 13, 23, 37} mod 40 prime factors of m to be even, while the latter requires the sum of {2, 5, 7, 13, 23, 37} mod 40 prime factors of m to be odd.
n = 3 is a member of this sequence because there is no positive integer m which can be simultaneously written as both x^2+15*y^2 and 5*x^2+3*y^2. The former requires the sum of {2, 3, 5, 8} mod 15 prime factors of m to be even, while the latter requires the sum of {2, 3, 5, 8} mod 15 prime factors of m to be odd.


CROSSREFS

Cf. A031363, A232531, A232532, A232682.
Sequence in context: A047560 A032899 A193528 * A296350 A076436 A028768
Adjacent sequences: A232678 A232679 A232680 * A232682 A232683 A232684


KEYWORD

nonn


AUTHOR

V. Raman, Nov 27 2013


STATUS

approved



