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.

  • $\begingroup$ Franz, I'm curious, do you have a source for this conjecture? $\endgroup$ – Charles Oct 15 '15 at 3:26
  • $\begingroup$ To add to the previous comment, it would be interesting to know how far Goldbach had checked that conjecture. $\endgroup$ – KConrad Feb 6 at 1:42
  • $\begingroup$ This is from the Euler-Goldbach correspondence available online at edoc.unibas.ch/58842 . The letter in question is dated Oct 1, 1742. As I have written there, the truth of this conjecture implies some kind of Bertrand's postulate for primes of the form $a^2+1$. $\endgroup$ – Franz Lemmermeyer Feb 6 at 16:44

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.


Franz, I wrote a paper related to an analytic heuristic on Dirichlet series associated to these prime counting problems. See http://www.math.uconn.edu/~kconrad/articles/hlconst.pdf.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.