brun's method and primitive roots

77 Views Asked by At

let be $N(x) = \sum\limits_{d|p - 1} {\mu \left( d \right)} {F_x}(d)$ that $p$ is a prime number and ${F_x}(d) = \frac{{{x^2}}}{{4d}} + O(x{p^{\frac{1}{2}}})$ that $x+1=g(p)$ and $g(p)$ is the least primitive root modulo $p$. Applying Brun's method to $N(x)$ in conjunction with ${F_x}(d)$ in order to make a lower estimate for $N(x)$, one obtain $$N(x) > \frac{{{x^2}}}{4}\sum\limits_{} {\frac{{\mu (d)}}{d}} + O({m^c}{p^{{\textstyle{1 \over 2}}}}x)$$ that $m$ is the number of distinct primes that divide $p-1$ and $c$ is a constant.can you explain these steps?also i dont know what is Brun's method. for more details you can see this (4.1) and (4.2)