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A209877
a(n) = A209874(n)/2: Least m > 0 such that 4*m^2 = -1 modulo the Pythagorean prime A002144(n).
2
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
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
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)
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Mar 14 2012
STATUS
approved