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%I #8 Mar 25 2020 08:02:40
%S 25601,50177,59393,65537,96001,115201,195457,262657,266369,267649,
%T 279553,286721,295937,299393,306689,331777,366593,425857,460289,
%U 495617,509569,525313,528001,556289,595073,710273,730753,790529,819457,868993,1021697,1022977,1049089
%N Primes p such that the equation x^64 == -2 (mod p) has a solution, and ord_p(-2) is even.
%C Such primes are the exceptional p for which x^64 == -2 (mod p) has a solution, as x^64 == -2 (mod p) is soluble for *every* p with ord_p(-2) odd.
%C But if ord_p(-2) is even and p - 1 = 2^r.j with j odd, then x^64 == -2 (mod p) is soluble if and only if ord_p(-2) is not divisible by 2^(r-5). See comment at A163185 for explanation.
%C Most primes p for which x^64 == -2 (mod p) has a solution (A051101) have ord_p(-2) odd (so belong to A163183). Thus 25601 (first element of current sequence, and 827th element of A051101) is the first element where A051101 and A163183 differ.
%H Jinyuan Wang, <a href="/A163186/b163186.txt">Table of n, a(n) for n = 1..1000</a>
%e For p = 25601, 562^64 == -2 (mod p), the 2-power part of p-1 is 2^10 and ord_p(-2) = 400, which is even but has 2-power part 2^4, which is not divisible by 2^(10-5).
%p with(numtheory):k:=6: A:=NULL:p:=2: for c to 30000 do p:=nextprime(p); o:=order(-2, p); R:=gcd(2^100, p-1); if o mod 2=0 and p mod 2^(k+1) = 1 and o mod R/2^(k-1)<>0 then A:=A, p; fi; od:A;
%o (PARI) lista(nn) = forprime(p=3, nn, if(znorder(Mod(-2, p))%2==0 && []~!=polrootsmod(x^64+2, p), print1(p, ", "))); \\ _Jinyuan Wang_, Mar 24 2020
%Y A051101 (all primes p for which x^62 == -2 (mod p) has a solution) is a union of A163183 (primes p with ord_p(-2) odd) and the current sequence.
%K nonn,easy
%O 1,1
%A _Christopher J. Smyth_, Jul 24 2009
%E More terms from _Jinyuan Wang_, Mar 24 2020
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