A209877 - OEIS

A209877

a(n) = A209874(n)/2: Least m > 0 such that 4*m^2 = -1 modulo the Pythagorean prime A002144(n).

1, 4, 2, 6, 3, 16, 15, 25, 23, 17, 11, 5, 38, 49, 50, 22, 14, 40, 81, 56, 7, 61, 72, 32, 8, 41, 30, 114, 69, 144, 57, 74, 68, 21, 52, 137, 167, 10, 133, 196, 127, 191, 174, 24, 104, 143, 26, 59, 43, 12, 258, 238, 289, 97, 77, 252, 53, 29, 13, 283, 48, 190, 335, 361, 31, 228, 291, 159, 263, 123, 260, 325, 363, 247, 162

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OFFSET

1,2

COMMENTS

Also: Square root of -1/4 in Z/pZ, for Pythagorean primes p=A002144(n).

Also: Least m>0 such that the Pythagorean prime p=A002144(n) divides 4(kp +/- m)^2+1 for all k>=0.

In practice these can also be determined by searching the least N^2+1 whose least prime factor is p=A002144(n): For given p, all of these N will have a(n) or p-a(n) as remainder mod 2p.

LINKS

Robert Israel, Table of n, a(n) for n = 1..10000

EXAMPLE

a(1)=1 since A002144(1)=5 and 4*1^2+1 is divisible by 5; as a consequence 4*(5k+/-1)^2+1 = 100k^2 +/- 40k + 5 is divisible by 5 for all k.

a(2)=4 since A002144(2)=13 and 4*4^2+1 = 65 is divisible by 13, while 4*1^1+1=5, 4*2^2+1=17 and 4*3^2+1=37 are not. As a consequence, 4*(13k+/-4)^2+1 = 13(...)+4*4^1+1 is divisible by 13 for all k.

MAPLE

f:= proc(p) local m;

if not isprime(p) then return NULL fi;

m:= numtheory:-msqrt(-1/4, p);

min(m, p-m);

end proc:

map(f, [seq(i, i=5..1000, 4)]); # Robert Israel, Mar 13 2018

MATHEMATICA

f[p_] := Module[{r}, r /. Solve[4 r^2 == -1, r, Modulus -> p] // Min];

f /@ Select[4 Range[300] + 1, PrimeQ] (* Jean-François Alcover, Jul 27 2020 *)

PROG

(PARI) apply(p->lift(sqrt(Mod(-1, p)/4)), A002144)