OFFSET
0,7
COMMENTS
From Jianing Song, Dec 06 2025: (Start)
Note that a is irreducible if and only if (a) is either a prime ideal of the product of two (not necessarily distinct) non-principal prime ideals; the same applies in every quadratic field whose class group has exponent 2. We have:
- {N(P) : P principal prime ideal} = {5} U {p : p prime == 1, 9 (mod 20)} U {p^2 : p prime == 11, 13, 17, 19 (mod 20)}. There are two prime ideals with norm p for each prime p == 1, 9 (mod 20), and only one with norm 5 or p^2;
- {N(P) : P non-principal prime ideal} = {2} U {p : p prime == 3, 7 (mod 20)}. There are two prime ideals with norm p for each prime p == 3, 7 (mod 20), and only one with norm 2.
As a result:
- a(4) = a(5) = a(p^2) = 1 for primes p == 11, 13, 17, 19 (mod 20);
- a(p) = 2 for primes p == 1, 9 (mod 20);
- a(2*p) = 2 for primes p == 3, 7 (mod 20);
- a(p^2) = 3 for primes p == 3, 7 (mod 20);
- a(p*q) = 4 for distinct primes p == 3, 7 (mod 20);
- a(n) = 0 for every other n.
(End)
REFERENCES
David A. Cox, Primes of the form x^2+ny^2, Wiley, 1989.
A. Frohlich and M. J. Taylor, Algebraic number theory, Cambridge university press, 1991.
LINKS
Jianing Song, Table of n, a(n) for n = 0..10000
EXAMPLE
From Jianing Song, Dec 08 2025: (Start)
Write Pp = (p, x+y*sqrt(-5)) and Pp' = (p, x-y*sqrt(-5)) for p == 3, 7 (mod 20), where (x,y) is the unique positive solution to x^2 + 5*y^2 = p^2. Write P2 = (2, 1+sqrt(-5)).
There are 2 irreducible elements with norm 6, corresponding to P2*P3 = (1-sqrt(-5)) and P2*P3' = (1+sqrt(-5)).
There are 3 irreducible elements with norm 9, corresponding to P3^2 = (2+sqrt(-5)), P3*P3' = (3), and P3^2' = (2-sqrt(-5)).
There are 2 irreducible elements with norm 14, corresponding to P2*P7 = (3+sqrt(-5)) and P2*P7' = (3-sqrt(-5)).
There are 4 irreducible elements with norm 21, corresponding to P3*P7 = (4-sqrt(-5)), P3*P7' = (1+2*sqrt(-5)), P3'*P7 = (1-2*sqrt(-5)), and P3'*P7' = (4+sqrt(-5)).
There are 2 irreducible elements with norm 46, corresponding to P2*P23 = (1-3*sqrt(-5)) and P2*P23' = (1+3*sqrt(-5)).
There are 3 irreducible elements with norm 49, corresponding to P7^2 = (2+3*sqrt(-5)), P7*P7' = (7), and P7^2' = (2-3*sqrt(-5)).
There are 4 irreducible elements with norm 69, corresponding to P3*P23 = (7+2*sqrt(-5)), P3*P23' = (8+sqrt(-5)), P3'*P23 = (8-sqrt(-5)), and P3'*P23' = (7-2*sqrt(-5)).
There are 2 irreducible elements with norm 86, corresponding to P2*P43 = (9+sqrt(-5)) and P2*P43' = (9-sqrt(-5)).
There are 2 irreducible elements with norm 94, corresponding to P2*P47 = (7+3*sqrt(-5)) and P2*P47' = (7-3*sqrt(-5)). (End)
PROG
(PARI) A091731(n) = {
if(bigomega(n) >= 3, return(0));
my(f = factor(n));
if(n==4 || n==5, return(1));
if(#f~ == 1, if((f[1, 1] % 20 == 1 || f[1, 1] % 20 == 9) && f[1, 2] == 1, return(2)); if(kronecker(-20, f[1, 1]) == -1 && f[1, 2] == 2, return(1)); if((f[1, 1] % 20 == 3 || f[1, 1] % 20 == 7) && f[1, 2] == 2, return(3)));
if(#f~ == 2 && (f[2, 1] % 20 == 3 || f[2, 1] % 20 == 7), if(f[1, 1] == 2, return(2)); if(f[1, 1] % 20 == 3 || f[1, 1] % 20 == 7, return(4)));
return(0);
} \\ Jianing Song, Dec 07 2025
CROSSREFS
Cf. A139513 (primes decomposing in Q(sqrt(-5))), A003626 (primes remaining inert in Q(sqrt(-5))), A033205 ((p) is the product of two principal ideals), A106865 ((p) is the product of two non-principal ideals).
Other sequences related to factorizations in Z[sqrt(-5)]: A091727, A091728, A091729, A091730, A262828.
KEYWORD
nonn,easy
AUTHOR
Paul Boddington, Feb 02 2004
STATUS
approved