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a(n) = A209874(n)/2: Least m > 0 such that 4*m^2 = -1 modulo the Pythagorean prime A002144(n).
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%I #21 Jul 27 2020 15:16:24

%S 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,

%T 41,30,114,69,144,57,74,68,21,52,137,167,10,133,196,127,191,174,24,

%U 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

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

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

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

%C 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.

%H Robert Israel, <a href="/A209877/b209877.txt">Table of n, a(n) for n = 1..10000</a>

%e 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.

%e 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.

%p f:= proc(p) local m;

%p if not isprime(p) then return NULL fi;

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

%p min(m,p-m);

%p end proc:

%p map(f, [seq(i,i=5..1000,4)]); # _Robert Israel_, Mar 13 2018

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

%t f /@ Select[4 Range[300] + 1, PrimeQ] (* _Jean-François Alcover_, Jul 27 2020 *)

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

%Y Cf. A209874.

%Y Cf. A002496, A014442, A085722, A144255, A089120, A193432.

%Y Cf. A002314, A177979.

%K nonn

%O 1,2

%A _M. F. Hasler_, Mar 14 2012