In the sequel, $\mathbb{Z}_p$ denotes the cyclic group of integers modulo $p$.
Working to a problem in Algebraic Geometry, I was led to state the following
Conjecture. Given every positive integer $b \geq 2$ and every odd prime $p$, it is possible to find non-zero elements $$\lambda_1, \ldots, \lambda_b, \, \; \mu_1, \ldots, \mu_b \in \mathbb{Z}_p$$ such that \begin{cases} \sum_{i=1}^b \lambda_i =1 \\ \sum_{i=1}^b \mu_i =1 \\ \lambda_i\mu_i \neq 1 \; \mathrm{for \, all \, } i. \end{cases}
I was able to verify it in many cases, but so far I could not find a general and easy proof. I suspect that an easy counting argument (that at the moment I do not see) might do the job, so I apologize in advance if the answer to the following question turns out to be trivial.
Q. Is the conjecture above true? If so, what is a clean and simple proof for it?
Choose arbitrarily $\lambda_i,\,i=1,\cdots,b-1$ and $\mu_i,\,i=1,\cdots,b-2$ such that $\lambda_i\cdot\mu_i\ne1,\,\forall i=1,\cdots,b-2$. (Then $\lambda_b$ is uniquely determined by $\lambda_b=1-\sum_{i=1}^{b-1}\lambda_i$.) Then $\mu_{b-1}$ and $\mu_b$ are subject to conditions: $\mu_{b-1}+\mu_b$ is a constant ($c:=1-\sum_{i=1}^{b-2}\mu_i$), and $\mu_{b-1}\ne\lambda_{b-1}^{-1},\,\mu_{b}\ne\lambda_{b}^{-1}$. The last two conditions are equivalent with $\mu_{b-1}\ne\lambda_{b-1}^{-1},\,c-\lambda_b^{-1}$. If $p>3$ then clearly this is possible as there are more than two non-zero elements in $\mathbb Z_p$. The only case where this is impossible is when $p=3$ and $\left\{\lambda_{b-1}^{-1},\,c-\lambda_b^{-1}\right\}=\left\{1,2\right\}$.
In fact when $p=3$, the condition $\lambda_i\cdot\mu_i\ne1$ implies that $\mu_i=-\lambda_i,\,\forall i$, and hence $\sum_{i=1}^{b}\mu_i=-\sum_{i=1}^{b}\lambda_i=-1$, so it is impossible to find elements satisfying the requirements when $p=3$.
Hope this helps.