Let $R=\mathbb{Z}[\sqrt{-13}]$. Let $p$ be a prime integer, $p\neq 2,13$ and suppose that $p$ divides an integer of the form $a^2+13b^2$, where $a$ and $b$ are in $\mathbb{Z}$ and are coprime. Let $\mathfrak{P}$ be the ideal generated in $R$ by $p$ and $a+b\sqrt{-13}$ and $\overline{\mathfrak{P}}$ the ideal generated by $p$ and $a-b\sqrt{-13}$: $$\mathfrak{P}=(p,a+b\sqrt{-13})\qquad;\qquad\overline{\mathfrak{P}}=(p,a-b\sqrt{-13}).$$
I want to show that $\mathfrak{P}\cdot\overline{\mathfrak{P}}=pR$.
I have already proved one inclusion: certainly $\mathfrak{P}\cdot\overline{\mathfrak{P}}$ can be generated by $p^2,p(a\pm\sqrt{-13}),a^2+13b^2$ which are all $p$-multiples, hence $\mathfrak{P}\cdot\overline{\mathfrak{P}}\subseteq pR$.
For the reverse inclusion, I am tring to express $p$ in terms of product of elements of the two ideals, but with no results up to now. Any help would be appreciated, thanks.
Hint#1: Why are you done, if $a^2+13b^2$ is not divisible by $p^2$?
Hint#2: Consider the case, when $a^2+13b^2$ is divisible by $p^2$. The integers $a$ and $b$ were assumed to be coprime, so at least one of $a,b$ is not divisible by $p$. Subtract $p$ from that coefficient, and replace $a+b\sqrt{-13}$ with either $(a-p)+b\sqrt{-13}$ or $a+(b-p)\sqrt{-13}$. Go back to Hint #1.