

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 is the complement of A031363. Proof: From the equation in the name follows a^2  5c^2 = n(d^2  5b^2). This equation has positive integer solutions if n is of the form x^2  5y^2, because A031363 is closed under multiplication. If there is no positive integer solution for the equation, it is because n is not a member of A031363. Thus n belongs to the present sequence, which was to be proved. This sequence contains no squares, but all odd powers of a term belong to the sequence.  Klaus Purath, Jul 31 2023


LINKS



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



KEYWORD

nonn


AUTHOR



STATUS

approved



