We consider the solution of $x^2=x+1$ and denote them as $\phi=\frac{1}{2}(1-\sqrt{5}),\bar\phi=\frac{1}{2}(1+\sqrt{5})$.
Suppose $\phi \not\in \mathbb{F}_p$. In other words, $\sqrt{5} \not \in \mathbb{F}_p\Leftrightarrow p = \pm 2 \bmod 5$. Arbitrary element of $\mathbb{F}_p(\phi)$, $a$ and $b$ satisfies $ab=0\Leftrightarrow a=0 \lor b=0$. From Fermat's little theorem and factor theorem, 
$$(x-1)(x-2)...(x-p) = x^p -x \bmod p .$$
Then, put $x=\phi$. Since Frobenius map,$x \mapsto x^p$, transfer $\phi$ to conjugate of itself, $\bar\phi$,
\begin{align}
&(\phi-1)(\phi-2)...(\phi-p) = \bar\phi -\phi \bmod p \\
\Leftrightarrow &(\phi-1)(\phi-2)...(\phi-p) = \sqrt{5}\bmod p
\end{align}
Then, what about $(\phi-1)(\phi-2)...(\phi-p) \bmod p^2\mathbb{Z}[\phi]$?
Surprisingly, this can be expressed as 
$$(\phi-1)(\phi-2)...(\phi-p) = \sqrt{5} + p\left(\frac{1}{2}+A\sqrt{5}\right) \bmod p^2\mathbb{Z}[\phi]$$
empirically (I verify this rule by a code here [ideone][1]. $a$ and $b$ of the output corresponds to $a=\frac{1}{2}p$ and $b=1+pA$). Here $\sqrt{5}$ is defined as $1-2\phi$. $\frac{1}{2}$ is the inverse of $2$ at $\bmod p^2\mathbb{Z}$. Such element can be found by Extended Euclidean algorithm because $2$ and $p^2$ are coprime to each other. I cannot find the rule for $A$, but everytime the coeffieicnt of $p$ of $\mathbb{F}_p$ is $\frac{1}{2}$ at $\bmod p^2$.

How to prove this rule?

My idea:
Consider the coefficient, $c_i$ of $(x-1)(x-2)...(x-(p-1))(x-p)=\sum_{i=0}^{p-1}c_ix^i$. It seems $c_i \bmod p=0$ for all $i$ and especially, $c_{2i}=0\bmod p^2$ for $2 \leq 2i \leq p-3$ if $p$ is prime. If $\phi$ is in the form, $a\sqrt{5}$, we only need to consider the term $c_0, c_{p-1}$ because other terms are divisible by $p^2$ or in the form $a'\sqrt{5}$.  
  [1]: https://ideone.com/pw0QsN