An application of Koike's Trace Formula

Question

Koike's Trace Formula states that \begin{equation} \mbox{Tr}((U_p^{\kappa})^n) = - \sum_{0 \leq u < \sqrt{p^n}\\ (u,p)=1}H(u^2-4p^n)\frac{\gamma(u)^\kappa}{\gamma(u)^2 - p^n}-1, \end{equation} where $H(D)$ is the Hurwitz class number of $D$ and $\gamma(u)$ is the unique $p$-adic unit root of the equation $$x^2-ux+p^n=0.$$ Let $\kappa\in \mathbb{C}_5$ with $v_5(\kappa) \geq 0$ and $v_5(\kappa - 8) < 2$. How can I prove that $$v_5(\mbox{Tr}(U_5^{\kappa}) - 60 + 20\kappa - 25\kappa^2) \geq 3 ?$$ Thank you in advance. Edit: For $p=5,n=1$, we get $$\mbox{Tr}(U_5^{\kappa}) = -\frac{\gamma(1)^\kappa}{\gamma(1)^2 - 5}- \frac{3\gamma(2)^\kappa}{2(\gamma(2)^2 - 5)}-1$$ Since $\gamma(1)\equiv 1\mod 5$, we have that $$\gamma(1)^\kappa = (1+(\gamma(1)-1))^\kappa \equiv 1 +(\gamma(1)-1)\kappa + \frac{\kappa^2-\kappa}{2}(\gamma(1)-1)^2 \mod 5^3 .$$ Observe that $$\frac{1}{1-5} \equiv 1 + 5 + 5^2 \mod 5^3,$$ Then $$\frac{1}{\gamma(1)^2 - 5} = \frac{1}{\gamma(1)^2}\frac{1}{1 - \frac{5}{\gamma(1)^2}} \equiv \frac{1}{\gamma(1)^2}+ \frac{5}{\gamma(1)^4} + \frac{5^2}{\gamma(1)^6}\mod 5^3$$ I don't know how to remove the term $\gamma(1)$ in the resulting expression and to work with $\gamma(2)$ since it is not congruence with $1 \mod 5$.