Computational Complexity of a special case of Integer Programming
It is NP-hard, by reduction from 3SAT.

Your problem is equivalent to the following: given a system of inequalities of the form $x \not\equiv A_i \pmod{B_i}$, where the $A_i,B_i$ are given, test whether any solution exists.  (Why?  If $0 &lt; c_i^2 &lt; B_i^2$ then $c_i \not\equiv 0 \pmod{B_i}$, and vice versa.)

Now 3SAT can be reduced to this problem (satisfiability of a system of inequalities), so it follows that this problem is NP-hard.  Here's how the reduction works.

Let $\varphi$ be a 3SAT instance, with variables $v_1,\dots, v_n$.  Let $p_1,\dots,p_n$ be the first $n$ primes.  Add inequalities of the form $x \not\equiv 2 \pmod{p_i}$, $x \not\equiv 3 \pmod{p_i}$, ..., $x \not\equiv p_i-1 \pmod{p_i}$ for each $i$, forcing $x$ to be either 0 or 1 modulo each prime.  Now, suppose we have a clause $v_i \lor v_j \lor v_k$.  This translates to an inequality $x \not\equiv 0 \pmod{p_i p_j p_k}$.  The clause $v_i \lor \neg v_j \lor v_k$ translates to $x \not\equiv \alpha \pmod{p_i p_j p_k}$ where $\alpha$ is the unique value such that $\alpha \equiv 0 \pmod{p_i}$, $\alpha \equiv 1 \pmod{p_i}$, and $\alpha \equiv 0 \pmod{p_i}$ (you can find $\alpha$ using the Chinese remainder theorem).  And so on -- in this way, each clause can be translated into an inequality.

Finally, there exists a value $x$ that satisfies all of these inequalities if and only if there exists an assignment that satisfies $\varphi$.  (Why?  $x\equiv 1 \pmod{p_i}$ corresponds to setting $v_i$ to true, and $x\equiv 0 \pmod{p_i}$ corresponds to setting $v_i$ to false.  You can then verify equivalence between clauses and inequalities.)  Since the resulting system has size polynomial in $n$, this gives a polynomial-time reduction from 3SAT to your problem.