Justify that a certain set depending on a parameter is large

Question

If there are some square free numbers, a finite amount of them $\{t_1,t_2,...t_k\}$ and we define the set $\mathcal{N}_L=\{l\in\mathbb{N}/L-l^2\notin\bigcup t_j\mathbb{Z}^2\}$, where $n\notin\bigcup t_j\mathbb{Z}^2$ means that if we write $n$ as the product of its square part and square free part as $n=tm^2$ then $t\notin\{t_1,t_2,...t_k\}$.

It seems intuitive to me the fact that as $L\rightarrow\infty$ then $\#\mathcal{N}_L\rightarrow\infty$, being $\#$ the amount of elements in the set.

However, I don't know how to prove it rigorously as I have almost no control on the explicit form of the elements of the set. Thank you for any help.

Answer 1

Let $p_1,\dots,p_s$ be the distinct primes in $t_1,\dots,t_k$.

We claim that, for any prime $p_i$, there is $a_i$ such that whenever $\ell \equiv a_i \pmod{p_i}$, we have $L - \ell^2 \not\equiv 0 \pmod{p_i}$. This is clear, since there are at least two different squares modulo $p_i$ (in other words, $a_i$ can always be chosen to be either 0 or 1).

By the Chinese Reminder Theorem, the system $\ell \equiv a_i \pmod{p_i}$, for $i=1,\dots,s$, has a solution.

If $\ell$ is a solution to the system of congruences above, then $L - \ell^2$ is not divisible by any of the primes $p_i$ so it cannot have its square-free part equal to any of the numbers $t_1,\dots,t_k$.

The solution of the system has the form $\ell \equiv a \pmod{p_1p_2 \dots p_s}$, for some $a<p_1p_2\dots p_s$. Therefore the number of choices for $\ell$ in the interval $[0,L]$ is at least $\lfloor L/(p_1\dots p_s) \rfloor$, so it grows without a bound.