A non-trivial solution for $\lambda_1 + 2 \lambda_2 + \cdots + (p-2)\lambda_{p-2} = 0$ in $\mathbb{Z}_p$ Let $K = \mathbb{Z}_p$, where $p$ is a prime number and $p \neq 2,3$ . Find a non-trivial solution for the system:
$$\left\{ \begin{align}
  & \lambda_1 + \lambda_2 + \cdots + \lambda_{p-2} = 0 \ (\mod p) \\ 
 & \lambda_1 + 2 \lambda_2 + \cdots + (p-2)\lambda_{p-2} = 0(\mod p) \\ 
\end{align} \right.,$$
where $\lambda_i \in K$, for all $i=1, \cdots, p-2$.
Comments: In the first part of the exercise I solved the system
$$\left\{ \begin{align}
  & \lambda_1 + \lambda_2 + \cdots + \lambda_{p-2}+ \lambda_{p-1} = 0 (\mod p)\\ &\lambda_1 + 2 \lambda_2 + \cdots + (p-2)\lambda_{p-2} + (p-1)\lambda_{p-1}= 0
(\mod p) \\ 
\end{align} \right.,$$
using that
$$1+2+ \cdot + (p-1) = \frac{p(p-1)}{2}$$ and 
$$1+2^2+3^2+ \cdots + (p-1)^2 = \frac{(p-1)p(2p-1)}{6}$$.
I can not see a logic look like the one in this part to solve the next one, for exemple, if $p = 5$, a solution is $\lambda_1 = 2$, $\lambda_2 = 1$ and $\lambda_3 = 2$, but I can not see a logic to the general case.

If we have $$\lambda_2 + 2\lambda_3 + \cdots + (p-3) \lambda_{p-2} \equiv 0 \pmod{p} \tag{$\ast$}$$ then setting $\lambda_1 = -(\lambda_2 +  \cdots+\lambda_{p-2})$ gives a solution $(\lambda_1, \cdots, \lambda_{p-2})$ to the described system of two equations. So, we just need to find a nontrivial solution to the single equation $(\ast)$. We can do this by choosing $\lambda_2, \cdots, \lambda_{p-3}$ arbitrarily and then taking $\lambda_{p-2}$ to be the unique residue forcing the equation to hold.
For example, choose $\lambda_2 = \cdots = \lambda_{p-3} = 1$. Then we require $$1+2+\cdots + (p-4) + (p-3) \lambda_{p-2} \equiv 0 \pmod{p}$$ $$\implies \frac{(p-4)(p-3)}{2} + (p-3) \lambda_{p-2} \equiv 0 \pmod{p}$$ $$ \implies \lambda_{p-2} = \frac{4-p}{2} \pmod{p}$$ where we have assumed $p \neq 3$ so we can divide by $p-3$. Then we can solve for $\lambda_1$ as well, yielding a solution to the original system.
