OFFSET
1,1
COMMENTS
Primes p == 3 (mod 4) are precisely the rational primes in the ring of Gaussian integers.
Let ord(a,m) be the multiplicative order of a modulo m. (Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i]). For a prime p == 3 (mod 4), we have that ord(2+-i,p) is divisible by ord(5,p), and that ord(2+-i,p) divides (p+1) * ord(5,p). What's more, ord(2+-i,p) divides (p^2-1)/2 if and only if 5 is a quadratic residue of integers modulo p. (See A385165).
As a result, if ord(2+-i,p) is not divisible by 8, then ord(5,p) is odd:
- Of course this is true if ord(2+-i,p) is odd.
- If ord(2+-i,p) == 2 (mod 4) and ord(5,p) is even, then ord(2+-i,p)/ord(5,p) is odd, and so ord(2+-i,p) divides ((p+1)/4) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) is odd, a contradiction.
- If ord(2+-i,p) == 4 (mod 8) and ord(5,p) is even (we have ord(5,p) == 2 (mod 4) since p == 3 (mod 4)), then ord(2+-i,p)/ord(5,p) == 2 (mod 4), and so ord(2+-i,p) divides ((p+1)/2) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) == 2 (mod 4), a contradiction.
From the above paragraph, this sequence is also primes p == 3 (mod 4) such that ord(2+-i,p)/ord(5,p) is odd.
LINKS
Jianing Song, Table of n, a(n) for n = 1..10000
EXAMPLE
8731 is a term since (2+-i)^635253 == 1 (mod 8731), and 635253 is odd.
8839 is a term since (2+-i)^57447 == 1 (mod 8839), and 57447 is odd.
9011 is a term since (2+-i)^2029953 == 1 (mod 9011), and 2029953 is odd.
PROG
(PARI) ord(p) = my(d = divisors((p+1)*znorder(Mod(5, p)))); for(i=1, #d, if(Mod([2, -1; 1, 2], p)^d[i] == 1, return(d[i]))) \\ for a prime p == 3 (mod 4), returns ord(2+-i, p)
isA385169(p) = isprime(p) && p%4==3 && ord(p)%2
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jianing Song, Jun 20 2025
STATUS
approved
