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.



Justify that a certain set depending on a parameter is large
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.


