Prime and primary ideals in $\mathbb{Z}[\sqrt{5}]$ 
Let $R=\mathbb{Z}[\sqrt{5}]$. The ideal $(2, 1-\sqrt{5})$ is prime in $R$, right (as $R/(2, 1-\sqrt{5})=\mathbb{F}_2$)?
  1. Is then $(2^n, 1-\sqrt{5})$ primary for some $n\geq2$?
  2. Are $(2)$ and $(3)$ prime ideals and $(2^n)$ and $(3^k)$ primary ideals in $R$?

I was thinking that at least $(2^n)$ should be primary as the zero divisors of $R/(2^n)$ are $2, 1+\sqrt{5}, 1-\sqrt{5}$, i.e. nilpotent...
Or are there some problems as $-4=(1+\sqrt{5})(1-\sqrt{5})$...
 A: $P=(2,1-\sqrt{5})$ is a prime:
First show it is not $R$. Suppose $1=2(a+b\sqrt{5})+(1-\sqrt{5})(c+d\sqrt{5})$ for some $a,b,c,d\in \mathbb{Z}$. Simplify we have that
$$
2a+c-5d=1, \quad 2b-c+d=0,
$$
add these two we have $2a+2b-4d=1$, contradiction. (By this argument, we can show that $2\notin (4,1-\sqrt{5})$, so $(2,1-\sqrt{5})\neq (4,1-\sqrt{5})$.)
Then as you said, we prove $R/P=\{\bar{0},\bar{1}\}$. Let $a+b\sqrt{5}+P\in R/P$. Then $a,b$ can only be $0$ or $1$. Since $1+\sqrt{5}\in P$, $1+P=\sqrt{5}+P$, we have $R/P=\{P,1+P\}$. These two elements are different since $1\notin P$.
$(2)$ is $P$-primary:
$(2)$ is not prime as mentioned in the comment above. Let $(a+b\sqrt{5})(c+d\sqrt{5})=2(e+f\sqrt{5})$, and $a+b\sqrt{5}\notin (2)$, which means $a,b$ not both even. Then we have 
$$
ac+5bd+\sqrt{5}(ad+bc)=2e+2f\sqrt{5}.
$$
Case 1: $a$ even, $b$ odd.
Then $d$ even and $c$ even. So $c+d\sqrt{5}\in (2)$, done.
Case 2: $a$ odd, $b$ even. (similar as case 1)
Case 3: $a,b$ both odd.
If $c$ is even, then $d$ is even, done. If $c$ is odd, then $d$ is odd. Now, $(c+d\sqrt{5})^{2}=c^{2}+5d^{2}+2cd\sqrt{5}\in (2)$, done. So, $(2)$ is primary. 
$\sqrt{(2)}=(2,1-\sqrt{5})$: $2\in\sqrt{(2)}$. Since $(1-\sqrt{5})^{2}\in (2)$, $1-\sqrt{5}\in \sqrt{(2)}$. So, $P\subseteq\sqrt{(2)}$. Since $P$ is maximal and $\sqrt{(2)}$ is prime (so not $R$), $\sqrt{(2)}=P$.
Edit:
$(2^{n})$ is primary:
I just saw a theorem: if $\sqrt Q$ is maximal, then $Q$ is primary. (Dummit Foote page 682 prop 19).
$\sqrt{(2^{n})}=\sqrt{(2)}=(2,1-\sqrt{5})$ is maximal, so $(2^{n})$ is primary. Its associated prime is also $(2,1-\sqrt{5})$.
Edit 2:
$(2^{n}, 1-\sqrt{5})$ is primary:
first show $\sqrt{(2^{n},1-\sqrt{5})}=(2,1-\sqrt{5})$. We have $(2,1-\sqrt{5})\subseteq \sqrt{(2^{n},1-\sqrt{5})}$. Conversely, since $(2^{n},1-\sqrt{5})\subseteq (2,1-\sqrt{5})$, $\sqrt{(2^{n},1-\sqrt{5})}\subseteq \sqrt{(2,1-\sqrt{5})}=(2,1-\sqrt{5})$. Now, $\sqrt{(2^{n},1-\sqrt{5})}=(2,1-\sqrt{5})$ is maximal, so by the theorem above, $(2^{n}, 1-\sqrt{5})$ is primary with associated prime $(2,1-\sqrt{5})$.
$(3)$ is prime:
First, $(3)\neq R$ since otherwise $1=3a+3b\sqrt{5}$, contradiction. Next, suppose $(a+b\sqrt{5})(c+d\sqrt{5})=3(e+f\sqrt{5})$, and $a,b$ not both multiple of $3$. Then 
$$
ac+5bd=3e, \quad ad+bc=3f.
$$
If $3\mid a$ and $3\nmid b$, then $3\mid c,d$, done. (Similarly, we are done if $3\mid b,3\nmid a$.)
If $3\nmid a$ and $3\nmid b$, then if $3\mid c$ then $3\mid d$, and if $3\mid d$ then $3\mid c$, so we can assume $3\nmid c$ and $3\nmid d$.
case 1: $a\equiv 1 \bmod 3$, $b\equiv 1 \bmod 3$. 
If $c\equiv 1$, then $d\equiv 1$ by the first equation above. This contradicts the second equation. If $c\equiv 2$, then $d\equiv 2$, also contradiction. The other $3$ cases are similar. So, $(3)$ is prime.
$(3^{k})$ is primary:
$R/(3)=\{a+b\sqrt{5}+(3):a,b\in\{0,1,2\}\}$ finite integral domain, so a field. Hence $(3)$ is maximal. Also, $\sqrt{(3^{k})}=(3)$ maximal, so $(3^{k})$ is primary with associated prime $(3)$. 
