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 $(\phi-1)(\phi-2)...(\phi-p) \bmod p^2$? Surprisingly, this can be expressed as $$(\phi-1)(\phi-2)...(\phi-p) = \sqrt{5} + p\left(\frac{1}{2}+A\sqrt{5}\right) \bmod p^2$$ empirically (I verify this rule by a code here ideone. $a$ and $b$ of the output corresponds to $a=\frac{1}{2}p$ and $b=1+pA$). 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?