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A343238
All positive integer moduli a(n) for which the congruence x^2 == -5 (mod a(n)) is solvable for integer x (representatives from {0, 1, ..., a(n)-1}); ordered increasingly.
6
1, 2, 3, 5, 6, 7, 9, 10, 14, 15, 18, 21, 23, 27, 29, 30, 35, 41, 42, 43, 45, 46, 47, 49, 54, 58, 61, 63, 67, 69, 70, 81, 82, 83, 86, 87, 89, 90, 94, 98, 101, 103, 105, 107, 109, 115, 122, 123, 126, 127, 129, 134, 135, 138, 141, 145, 147, 149, 161, 162, 163, 166, 167, 174, 178, 181, 183, 189, 201, 202
OFFSET
1,2
COMMENTS
This sequence includes A139513, that is, Legendre(-5, p) = +1 for odd primes not 5, that is, primes congruent to {1, 3, 7, 9} mod 20. Here 5 is a member of the sequence with solution x = 0.
The primes of this sequence are given in A240920.
The present sequence gives the numbers of the form 2^a*5^b*Product_{j=1..m} (p_j)^e(j), with a and b from {0, 1}, p_j a prime from {1, 3, 7, 9} (mod 20), i.e., from A139513, m >= 0 and e(j) >= 0 (this includes the number 1). These numbers are ordered increasingly.
This follows from the Legendre-symbol(-5, p)= +1 and the lifting theorem (see, e.g., Apostol, Theorem 5.30, p. 121-2) for p = 2 and 5 (no lifting for the solutions for p = 2 and p = 5), and the unique lifting for the primes satisfying Legendre-symbol(-5, p) = +1.
Therefore the number of representative solutions x from {0, 1, ..., a(n)-1}, denoted by M(a(n)), is 1 for precisely four cases: a(1) = 1 (x = 0), a(2) = 2 (x = 1), a(4) = 5 (x = 0) and a(8) = 10 = 2*5 (x = 5). For each of the mentioned prime powers there are just 2 solutions. This implies that for the number of solutions in the general a(n) case, n not 1, 2, 4, 8, only the primes p_j are of interest: M(a(n)) = 2^m(n).
For these solutions x see A343239, and for the multiplicity M(a(n)) see A343240.
This congruence is needed to find all proper solutions of the positive definite binary quadratic form of discriminant Disc = -20 = -4*5 representing k = a(n). The solutions x lead to the so-called representative parallel primitive forms (rpapfs). See A344231 for more details.
For a bisection see A344231 and A344232, related to integer solutions of X^2 + 5*Y^2 = A344231(k) and 2*X^2 + 2*X*Y + 3*Y^2 = A344232(k).
REFERENCES
Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, pp 121, 122.
FORMULA
There exists at least one x from {0, 1, ..., m-1} satisfying x^2 + 5 == 0 (mod m), for positive integer m. These m values are then ordered increasingly as (a(n))_{n>=1}.
EXAMPLE
a(3) = 3: two solutions 1 and 2.
a(7) = 3^2 = 9: two solutions 2 and 7.
a(8) = 10 = 2*5 only one solution 5.
a(53) = 135 = 5*3^3: two solutions 20 and 115.
PROG
(PARI) isok(k) = issquare(Mod(-5, k)); \\ Michel Marcus, Sep 17 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, May 16 2021
STATUS
approved