$$p$$ is a odd prime,  then

1. there are infinite primes $$q$$ such that $$q$$ is a quadratic residue modulo $$p$$;
2. there are infinite primes $$q$$ such that $$q$$ is a quadratic nonresidue modulo $$p$$.

$a_0=q, a_1=q+p,a_2=q+pa_1,a_3=q+pa_1a_2,\dotsc, a_n=q+pa_1a_2\dotsb a_{n-1},\dotsc$

$\bigg(\dfrac{kq_1q_2q_3\dotsm q_n+1}p\bigg)=-1.$

$\bigg(\frac qp\bigg) = \bigg(\frac pq\bigg).$

$a_1 = 4^k – p,$
$a_2 = 4^ka_1^2 -p,$
$a_3 = 4^ka_1^2a_2^2 -p,$
$\dotsc$
$a_n = 4^ka_1^2a_2^2\dotsm a_{n-1}^2-p,$
$\dotsc$

$\bigg(\frac qp\bigg)=\bigg(\frac {-p}q\bigg).$

$a_1 = 4+p,$
$a_2 = 4a_1^2 + p,$
$a_3 = 4a_1^2a_2^2 + p,$
$a_4 = 4 a_1^2a_2^2 a_3^2 + p,$
$\dotsc$
$a_n = 4 a_1^2a_2^2\dotsm a_{n-1}^2+ p,$
$\dotsc$