I am asking for guidance on how to think about the following problem. If there is a clear-cut answer, or literature on the subject, that would be useful.
$p_n$ is the $n$th prime number. Let $R_{i}=\{0\le r_{ij} \le p_i-1\}$ be the set of $p_i$ residue classes $\bmod p_i$. The Chinese Remainder Theorem (CRT) allows us to identify numbers that are characterized as belonging simultaneously to certain residue classes with respect to various prime numbers.
I wish to apply this thinking to squares and quadratic residues. For $i\ge 2$, let $S_i=\{s_{ij} \equiv (r_{ij})^2 \bmod p_i\}$ be the set of $\frac{p_i+1}{2}$ quadratic residues with respect to the prime $p_i$, and $M=m^2$ be the square of any positive integer. Plainly, being a square, $M$ is equivalent $\bmod p_i$ to some $s_{ij}$ in each corresponding $S_i$.
I choose one particular quadratic residue for each of multiple prime numbers, for example by a formula or algorithm. Call a chosen quadratic residue $s_{iJ}$. I wish to identify $M\not \equiv s_{iJ}$. CRT says this should always be possible, even if the number of primes used becomes very large.
On the other hand, since there are $\frac{p_i+1}{2}$ quadratic residues for each $p_i$, a randomly chosen $M$ should have a $\frac{\frac{p_i+1}{2}-1}{\frac{p_i+1}{2}}=\frac{p_i-1}{p_i+1}$ chance of being an acceptable choice for each $p_i$. As I include more and more $p_i$, the likelihood that $M$ is not equivalent to at least one $s_{iJ}$ becomes $\prod \frac{p_i-1}{p_i+1}$ which looks like it should tend to $0$ (note that it is term by term smaller than $\prod \frac{p_i-1}{p_i}$).
If that is so, then no matter how I choose $s_{iJ}$, for some $M_0$, $M>M_0 \Rightarrow \exists (i,J) | M\equiv s_{iJ} \bmod p_i$. This seems to conflict with the prediction of CRT. How do I reconcile this apparent paradox?