Let $p_1, \dots, p_l \in \mathbb{Z}$ be pairwise disjoint odd primes.
By the Chinese remainder follows there exists $x \in \mathbb{Z}$ satisfying
\begin{align*} x & = 1 \text{ mod } 4 \cdot p_2 \cdot \dots \cdot p_l\\ x & = a \text{ mod } p_1 \end{align*} for some $0 \leq a < p_1$.
But is there always a solution if we want $a$ not to be a square in $F_{p_1}^*$?
For this we can rewrite the system of equations using the Legendre symbol:
\begin{align*} x & = 1 \text{ mod } 4 \cdot p_2 \cdot \dots \cdot p_l\\ \left( \frac{x}{p_1} \right)& = -1. \end{align*}