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A186814 a(n) = smallest number m such that A002144(n) divides gcd(A002314(n)^2+1,(A002314(n)+m)^2+1).

%I A186814 #9 Mar 30 2012 18:35:54
%S A186814 1,3,9,5,25,23,7,39,19,21,53,81,43,83,63,61,101,13,143,31,169,15,55,
%T A186814 113,225,105,157,175,17,263,89,41,77,269,165,159,271,361,123,363,75,
%U A186814 315,239,365,93,51,437,321,397,529,439,351,543,229,333,355,449,557,625,431,517,27,583
%N A186814 a(n) = smallest number m such that A002144(n) divides gcd(A002314(n)^2+1,(A002314(n)+m)^2+1).
%C A186814 Sequence A002314 gives the minimal integer square root of -1 modulo p(n),where p(n) = n-th prime of form 4k+1.
%e A186814 for n=1, k = A002314(1) = 2 => a(1) = 1, because 2^2+1 = 5 and (2+1)^2+1 = 2*5 ;
%e A186814 for n=2, k = A002314(2) = 5 => a(2) = 3, because 5^2+1 = 2*13 and (5+3)^2+1 = 5*13 ;
%e A186814 for n=3, k = A002314(3) = 4 => a(3) = 9, because 4^2+1 = 17 and (4+9)^2+1 = 2*5*17;
%e A186814 for n=4, k = A002314(4)= 12 => a(4)= 5, because 12^2+1
%e A186814 = 5*29 and (12+5)^2+1 = 2*5*29, and 29 divides
%e A186814 GCD(5*29, 2*5*29)=145.
%p A186814 with(numtheory):T:=array(1..90):j:=1:for i from 1 to 250 do:x:=4*i+1:if type(x,prime)=true
%p A186814   then T[j]:=x:j:=j+1:else fi:od:for p from 1 to j do:u:=T[p]:id:=0: for m from
%p A186814   1 to 1000 while(id=0) do: z:=m^2+1:for d from 1 to u while(id=0) do: z1:=(m+d)^2+1:zz:=
%p A186814   gcd(z,z1):if irem(zz,u)=0 then id:=1:printf(`%d, `,d):else fi:od:od:od:
%Y A186814 Cf. A002314, A002144, A186710, A186713.
%K A186814 nonn
%O A186814 1,2
%A A186814 _Michel Lagneau_, Feb 27 2011