cp's OEIS Frontend

A002145 are precisely the rational primes in the ring of Gaussian integers.

From the representation of complex numbers as 2 X 2 matrices, a(n) is also the multiplicative order of the matrix [2,-1;1,2] or [2,1;-1,2] modulo p.

a(n) is divisible by ord(5,p): If (2+-i)^n == 1 (mod p), then 5^n == 1 (mod p).

a(n) divides (p+1) * ord(5,p), since we have (2+-i)^(p+1) == 5 (mod p).

If 5 is a quadratic residue modulo p, then ord(5,p) divides (p-1)/2, and so a(n) divides (p^2-1)/2. Conversely, if a(n) divides (p^2-1)/2, then (x+-y*i)^2 == 2+-i (mod p) for some integers x, y, and so (x^2+y^2)^2 == 5 (mod p), which means that 5 is a quadratic residue modulo p.

Note that the primes congruent to 3 modulo 4 are precisely the rational primes in the ring of Gaussian integers.

Primes p == 3 (mod 4), p > 3 such that (1+-i)^((p^2-1)/24) == 1 (mod p). Note that p^2-1 is always divisible by 24 for primes p > 3.

Primes p = A002145(k) such that the multiplicative order of 1+-i modulo p (A385163(k)) divides (p^2-1)/24. Since A385165(k) = 4*ord(-4,p), this is also primes p == 3 (mod 4) such that 96*ord(-4,p) divides p^2-1, where ord(a,p) is the multiplicative order of a modulo p.

Sequence is infinite since it contains all primes congruent to 95 modulo 96.

Primes p == 3 (mod 4), p > 3 such that [1,-1;1,1]^((p^2-1)/24) or [1,1;-1,1]^((p^2-1)/24) == I_2 (mod p).

Since 96 divides p^2-1 for p being a term of this sequence, we must have p == 15 (mod 16).