

A306310


Odd numbers k > 1 such that 2^((k1)/2) == (2/k) = A091337(k) (mod k), where (2/k) is the Jacobi (or Kronecker) symbol.


1



341, 5461, 10261, 15709, 31621, 49981, 65077, 83333, 137149, 176149, 194221, 215749, 219781, 276013, 282133, 534061, 587861, 611701, 653333, 657901, 665333, 688213, 710533, 722261, 738541, 742813, 769757, 950797, 1064053, 1073021, 1109461, 1141141, 1357621, 1398101
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OFFSET

1,1


COMMENTS

All terms are composite because for odd primes p we always have 2^((p1)/2) == (2/p) = A091337(p) (mod p).
All terms are congruent to 5 modulo 8. So this sequence is equivalent to "Numbers k == 5 (mod 8) such that 2^((k1)/2) == 1 (mod k)".
Proof: in the following proof let p be any prime factor of k. Note that 2 is a quadratic residue modulo p if and only if p == 1, 7 (mod 8).
(a) if k == 3 (mod 8), then 2^((k1)/2) == 1 (mod p), so 2^((k+1)/2) == 2 (mod p). Let x = 2^((k+1)/4), then x is an integer such that x^2 == 2 (mod p), so 2 is a quadratic residue modulo p => p == 1, 7 (mod 8). Since any prime factor of k is congruent to 1, 7 modulo 8 we have k == 1, 7 (mod 8), a contradiction.
(b) if k == 7 (mod 8), then 2^((k1)/2) == 1 (mod p). If p == 7 (mod 8), then 2 is a quadratic residue modulo p while 1 is not, a contradiction. If p == 5 (mod 8), let d be the multiplicative order of 2 modulo p. 2 is not a quadratic residue modulo p, so d divides p  1 but d does not divide (p  1)/2, so v2(d) = v2(p1) = 2, where v2 = A007814 is the 2adic valuation. On the other hand, 2^(k1) == 1 (mod p), so d divides k  1, but k  1 == 2 (mod 4), a contradiction. So p == 1, 3 (mod 8). Since any prime factor of k is congruent to 1, 3 modulo 8 we have k == 1, 3 (mod 8), a contradiction.
(c) if k == 1 (mod 8), then 2^((k1)/2) == 1 (mod p). Let x = 2^((k1)/8), then x^4 == 1 (mod p), so 1 is a quartic residue modulo p => p == 1 (mod 8). Let d be the multiplicative order of 2 modulo p, then d divides k  1 but d does not divide (k1)/2, so v2(d) = v2(k1). On the other hand, 2^((p1)/2) == 1 (mod p), so d divides (p1)/2. So v2(p1) >= v2(d) + 1 = v2(k1) + 1. Let t = v2(k1), then k  1 is not divisible by 2^(t+1), but any prime factor p of k should have p  1 is divisible by 2^(t+1), a contradiction.
Also numbers k in A001567 and congruent to 5 modulo 8 such that k  1 divided by the multiplicative order of 2 modulo k is an even number.


LINKS



EXAMPLE

341 is a term because (2/341) = 1, and 2^((3411)/2) == 1 (mod 341).


PROG

(PARI) isA306310(k)=(k%8==5) && Mod(2, k)^((k1)/2)==1
(PARI) isok(k) = (k>1) && (k%2) && (Mod(2, k)^((k1)/2) == Mod(kronecker(2, k), k)); \\ Michel Marcus, Feb 07 2019


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



