# Primes of the form a^2+1

The fact that the Riemann zeta function $\zeta(s)$ and its brethren have a pole at $s=1$ is responsible for the infinitude of large classes of primes (all primes, primes in arithmetic progression; primes represented by a quadratic form). We cannot hope proving the infinitude of primes $p = a^2+1$ in this way because the series $\sum 1/p$, summed over these primes, converges. This implies that the corresponding Euler product $$\zeta_G(s)= \prod_{p = a^2+1} \frac1{1 - p^{-s}}$$ converges for $s = 1$. But if we could show that $\zeta_G(s)$ has a pole at, say, $s = \frac12$, then the desired result would follow. Now I know that there are heuristics on the number of primes of the form $p = a^2+1$ below $x$ (by Hardy and Littlewood?)

Can these heuristics be explained by hypothetical properties of $\zeta_G(s)$ (or a related Dirichlet series), or can the domain of convergence of $\zeta_G(s)$ be derived from such asymptotics?

BTW, here's a little known conjecture by Goldbach on these primes: let $A$ be the set of all numbers $a$ for which $a^2+1$ is prime ($A =${1, 2, 4, 6, 10, $\ldots$}). Then every $a \in A$ ($a > 1$) can be written in the form $a = b+c$ for $b, c \in A$. I haven't seen this discussed anywhere.

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Franz, I'm curious, do you have a source for this conjecture? – Charles Oct 15 '15 at 3:26

Hi Franz, Unfortunately I doubt this Euler product has very good behavior. If you believe the Hardy-Littlewood conjectures, then $\sum_{n\leq X}\Lambda(n^2+1) \sim cX$ where $c=\prod_{p>2}(1-\chi_{4}(p)(p-1)^{-1})$ is some positive constant which is almost certainly transcendental. If $\zeta_{G}(s)$ reflected this asymptotic behavior, then $\frac{d}{ds}\log{\zeta_{G}(s)}$ would have a pole at $s=1/2$ of residue equal to $-c$. However, that would imply $\log{\zeta_G(s)}\sim -c\log{(s-1/2)}$ in a neighborhood of $s=1/2$, so $\zeta_G(s)$ would behave like $(s-1/2)^{-c}$ near this point. In particular, it would have some kind of branch cut...
People have conjectured that $\sum_{n\leq X}\Lambda(n^2+1) = cX + O(X^{\frac{1}{2}+\varepsilon})$ is true, which would give continuation of $\zeta_G(s)$ into the halfplane $\mathrm{Re}(s)>\frac{1}{4}$ after choosing a branch, but I doubt you could get any further.