cp's OEIS Frontend

A318527 Numbers k such that -3 is a quadratic residue (not necessarily coprime) modulo k, k + 1, k + 2 and k + 3.

%I A318527 #49 Sep 26 2022 06:18:25
%S A318527 1,721,4681,6121,6481,7201,7561,9001,11161,14401,16921,18361,19441,
%T A318527 20521,24481,24841,27361,29881,32761,36721,39241,39601,42121,42841,
%U A318527 43201,47161,47521,48241,49681,51121,52201,53641,60481,61561,62641,63361,64441,65521,65881,68041,73441,73801,74521,74881,75961,76321,78481,82441,82801,84241,88201,91081,94681
%N A318527 Numbers k such that -3 is a quadratic residue (not necessarily coprime) modulo k, k + 1, k + 2 and k + 3.
%C A318527 Start of 4 consecutive terms in A057128. Note that there are no 5 consecutive terms there.
%C A318527 Two interesting properties:
%C A318527 (a) All terms are congruent to 1 mod 360.
%C A318527 Proof: since -3 is not a quadratic residue modulo any prime == 5 (mod 6) we have k == 1 (mod 5) and k == 0 or 1 (mod 6). If k is even, then k == 6 (mod 30). -3 is not a quadratic residue modulo 9 so k == 3 (mod 9), then k == 66 (mod 90), k + 3 = 90*t + 69 = 3*(30*t + 23) but 30*t + 23 == 5 (mod 6), so -3 is not a quadratic residue modulo 30*t + 23, a contradiction. Thus k must be odd, then k == 1 (mod 30). For the same reason k == 1 or 4 (mod 9). If k == 4 (mod 9), then k == 31 (mod 90), k + 2 = 90*t + 33 = 3*(30t + 11) but 30*t + 11 == 5 (mod 6), a contradiction. So k == 1 (mod 9), then k == 1 (mod 90). If k == 91 (mod 180), then k + 3 = 180*t + 94 = 2*(90*t + 47) but 90*t + 47 == 5 (mod 6), a contradiction. So k == 1 (mod 180). -3 is not a quadratic residue modulo 8 so k == 1, 2, 3 or 4 (mod 8), thus k == 1 (mod 360) which is what we wanted.
%C A318527 (b) k is a term iff -3 is a quadratic residue modulo k*(k + 1)*(k + 2)*(k + 3)/2.
%C A318527 Proof: "<=" is obvious, since k*(k + 1)*(k + 2)*(k + 3)/2 is multiple of k, k + 1, k + 2 and k + 3. "=>": Note that -3 is a quadratic residue modulo lcm(k, k + 1, k + 2, k + 3). Now we show that lcm(k, k + 1, k + 2, k + 3) = k*(k + 1)*(k + 2)*(k + 3)/2. If not, then k is a multiple of 3, but by (a) we have k == 1 (mod 3), a contradiction.
%H A318527 Jianing Song, <a href="/A318527/b318527.txt">Table of n, a(n) for n = 1..10000</a> (using data from A318911)
%F A318527 a(n) = 360*A318911(n) + 1.
%e A318527 721 is a term since 93^2 == -3 (mod 721), 137^2 == -3 (mod 722), 210^2 == -3 (mod 723) and 97^2 == -3 (mod 724).
%o A318527 (PARI) isA057128(n) = issquare(Mod(-3, n));
%o A318527 isA318527(n) = isA057128(n)&&isA057128(n+1)&&isA057128(n+2)&&isA057128(n+3);
%o A318527 for(n=1, 100000, if(isA318527(n), print1(n, ", ")))
%Y A318527 Cf. A057128.
%Y A318527 Cf. A305864 (start of 3 consecutive terms in A057128), A318911.
%K A318527 nonn
%O A318527 1,2
%A A318527 _Jianing Song_, Aug 30 2018