# Identity involving complex sigma function

When trying around with the DivisorSigma function of Mathematica, I found this Identity:

$\#\{a\mid\exists b\in\mathbb{Z}[i]: ab=n\}=\underbrace{\#\{a\mid\exists b\in\mathbb{N}: ab=n\}^2}_{\sigma(0,n)^2}\Leftrightarrow \forall p|n,\;p\text{ prime}: p\equiv 1\mod 4$

(In words: the number of gauss integer divisors of n is equal to the square of the number of integer divisors of n iff n is in A004613)

I could verify it for values up to 10 million. However, I have been unable to find any Identities that could allow me to prove them. I could take several results from exactly that OEIS page, however none of those helped me out. I am missing some result that connects the gaussian integer divisors with the real ones. I would be very happy with any help on this problem.

• Are you aware that every prime $p\equiv1\pmod4$ can be written in exactly one way as $p=\pi\pi'$ where $\pi$ and $\pi'$ are Gaussian primes? – Gerry Myerson May 22 '13 at 13:26
• ... and primes $p$ with $p\equiv 3 \pmod 4$ are Gaussian primes. – martini May 22 '13 at 13:27
• @GerryMyerson I was not. Awesome! I get $\tilde{\sigma}(a)=\prod_{k=1}^n(e_k+1)(e_k+1)=(\prod_{k=1}^n(e_k+1))^2=\sigma(a)^2$ (Using $a=\prod p_k^{e_k} = \prod \pi_k^{e_k}{\pi'} _k^{e_k}$ and $\sigma(a)=\prod (e_k+1)$) – CBenni May 22 '13 at 14:46

Thanks to Gerry Myersons comment, I was able to find the general outline of the proof: Let us denote $\tilde{\sigma}$ as the divisor counting function in $\mathbb{Z}[i]$ ($\sigma$ is the divisor counting function in $\mathbb{Z}$). We use the identity $$a=\prod\limits_{k=1}^n p_k^{e_k}$$ $$\sigma(a)=\prod\limits_{k=1}^n (e_k+1)$$ and that (according to Gerry Myerson) $$p\equiv 1\mod4\Leftrightarrow \exists!\pi,\pi':\;p=\pi\pi'$$ Therefore (we set $p_k=\pi_{2k}\pi_{2k+1}$), $$a=\prod\limits_{k=1}^n (\pi_{2k}\pi_{2k+1})^{e_k}$$ and $$\tilde{\sigma}(a)=\prod\limits_{k=1}^{n} (e_k+1)(e_k+1)=\sigma(a)^2$$