

A297620


Positive numbers n such that n^2 == p (mod q) and n^2 == q (mod p) for some consecutive primes p,q.


1



6, 10, 12, 24, 42, 48, 62, 72, 76, 84, 90, 93, 108, 110, 120, 122, 128, 145, 146, 174, 187, 188, 194, 204, 208, 215, 220, 228, 232, 240, 241, 264, 297, 306, 310, 314, 317, 326, 329, 336, 349, 357, 366, 372, 386, 408, 410, 423, 426, 431, 444, 454, 456, 468, 470, 474, 518, 522, 535, 538, 546, 548
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OFFSET

1,1


COMMENTS

Positive numbers n such that n^2 == p+q mod (p*q) for some consecutive primes p, q.
Each pair of consecutive primes p,q such that p is a quadratic residue mod q and p and q are not both == 3 (mod 4) contributes infinitely many members to the sequence.
Odd terms of this sequence are 93, 145, 187, 215, 241, 297, 317, 329, 349, 357, 423, 431, 535, ...  Altug Alkan, Jan 01 2018


LINKS



EXAMPLE

a(3) = 12 is in the sequence because 71 and 73 are consecutive primes with 12^2 == 73 (mod 71) and 12^2 == 71 (mod 73).


MAPLE

N:= 1000: # to get all terms <= N
R:= {}:
q:= 3:
while q < N^2 do
p:= q;
q:= nextprime(q);
if ((p mod 4 <> 3) or (q mod 4 <> 3)) and numtheory:quadres(q, p) = 1 then
xp:= numtheory:msqrt(q, p); xq:= numtheory:msqrt(p, q);
for sp in [1, 1] do for sq in [1, 1] do
v:= chrem([sp*xp, sq*xq], [p, q]);
R:= R union {seq(v+k*p*q, k = 0..(Nv)/(p*q))}
od od;
fi;
od:
sort(convert(R, list));


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



