|
| |
|
|
A318898
|
|
a(n) = ((-4)^((p-1)/4) - 1)/p, where p is the n-th prime congruent to 1 mod 4.
|
|
1
|
|
|
|
-1, -5, 15, -565, -7085, 25575, -1266205, -17602325, 941362695, 197665011735, 2901803883615, -11147523830125, -165269711096165, 637677823344495, 2154364271382137415, -126774939137440139965, -1925041114036033717685, -447232673152232758272805, -6839447730858454557453725, 410508614063545790640124095, -1608693655111966245554191885
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
a(n) is always an integer. If p == 1 (mod 8), then (-4)^((p-1)/4) == 4^((p-1)/4) == 2^((p-1)/2) (mod p). 2 is a quadratic residue modulo p so 2^((p-1)/2) == 1 (mod p). If p == 5 (mod 8), then (-4)^((p-1)/4) == -4^((p-1)/4) == -2^((p-1)/2) (mod p). 2 is a quadratic nonresidue modulo p so 2^((p-1)/2) == -1 (mod p). Furthermore, for n > 1, a(n) is always an odd multiple of 5.
(-4)^((p-1)/4) == 1 (mod p) implies -4 is always a quartic residue modulo p. Note that x^4 + 4 = (x^2 + 2)^2 - (2x)^2 = (x^2 + 2x + 2)(x^2 - 2x + 2) = ((x + 1)^2 + 1)*((x - 1)^2 + 1), so the solutions to x^4 == -4 (mod p) are x == ((p - 1)/2)! + 1, ((p - 1)/2)! - 1, -((p - 1)/2)! + 1 and -((p - 1)/2)! - 1 (mod p).
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
The second prime congruent to 1 mod 4 is 13, so a(2) = ((-4)^3 - 1)/13 = (-65)/13 = -5. Also, the four solutions to x^4 == -4 (mod 13) are x == 4, 6, 7 and 9 (mod 13).
|
|
|
PROG
|
(PARI) forstep(p=5, 100, 4, if(isprime(p), print1(((-4)^((p-1)/4)-1)/p, ", ")))
|
|
|
CROSSREFS
|
Cf. A002144 (primes of the form 4n + 1).
Cf. A270698 (composite k == 1 (mod 4) that divides (-4)^((k-1)/4) - 1).
|
|
|
KEYWORD
|
sign
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
