2

I need to calculate the Dirichlet density of the set of primes $p$ of the form $p = n^2 +1$ (in fact show it is zero), but I have no idea how to go about it. My definition of Dirichlet density of a set of primes X is:

$dens(X)= \lim_{s\rightarrow 1^+}\frac{\Sigma_{p\in X}\frac{1}{p^s}}{log(\frac{1}{s-1})}$

I cant figure out how on earth to calculate that sum over primes. I suspect I'm not supposed to calculate it explicitly, so I thought of putting a bound on the sum which converges to a finite limit, but got nowhere.

My other thought is that the set of primes in question is a proper subset of the set of primes p such that (-1) is a square modulo p, and this has density of 1/2. But I don't know where to go from here. Please could anyone point me in the right direction?

Thanks!

MattBurrows
  • 689

1 Answers1

2

It is clear that the set of such primes is contained in the set of all integers of the form $m^2 + 1$, hence for any $\sigma > 1$, \begin{equation*} \sum_{p \in X} \frac{1}{p^{\sigma}} \leq \sum_{n \geq 1} \frac{1}{m^{2\sigma}} < \infty, \end{equation*} by the integral test. Hence, since $-\log(\sigma - 1) \rightarrow \infty$ as $\sigma \rightarrow 1^+$, the ratio $\frac{\sum_{p \in X} \frac{1}{p^{\sigma}}}{-\log(\sigma - 1)} \rightarrow 0$ as $\sigma \rightarrow 1^+$. In other words, the Dirichlet density is 0.

Dead-End
  • 610