We need to show that for sufficiently large $n$, there exists prime number $p\leqslant n$ such that $p\nmid i^2+n$ for all $i\leqslant n$. For any even positive integer $\ell$, we have $4n-(\ell^2 +1)^2\equiv 3\pmod{4}$. So, there exists prime number $p\equiv 3\pmod{4}$ that $p\mid 4n-(\ell^2+1)^2$. If this prime number is not greater than $n$. Suppose there exists $i\leqslant n$ that $p\mid i^2+n\implies p\mid (2i)^2+4n$. We get $p\mid (2i)^2+(\ell^2+1)^2 \implies p\mid 2i$ and $p\mid \ell^2+1$. Again, since $p\equiv 3\pmod{4}$, we get $p\mid \ell$ and $p\mid 1$. The latter gives us the contradiction. So, it's enough to show there exists even $\ell$ that $0<4n- (\ell^2+1)^2\leqslant n$. Let $k$ be the greatest even positive integer that $4n\geqslant (k^2+1)^2$. (We need and will consider $n\geqslant (2^2+1)^2/4$ so that there exist at least one.) By definition of $k$, $((k+2)^2+1)^2>4n$. If $4n- (k^2+1)^2>n$, we get $3n>(k^2+1)^2$. Hence, $$((k+2)^2+1)^2>4n>\frac{4(k^2+1)^2}{3}.$$Easy to see that the above inequality is false for sufficiently large $k$. In other words, there exists $C$ that it is not true for $k>C$. Hence, $4n-(k^2+1)^2 >n$ is false for all $n>(C^2+1)^2/3$.

could you please explain this part ThE-dArK-lOrD wrote: Quote: Hence $ 4n-(l^2+1)^2\equiv_4 3$

@above Sorry, that was likely wrong. I've edited my sol in #2; should be correct now.

The given condition is that \begin{align*} n! \mid (n + 1)(n + 4)(n + 9) \cdots (n + n^2). \end{align*}For $n \equiv 1 \pmod{4}$ and $n \equiv 2 \pmod{4}$, note that $v_2(n + i^2) \le 1$ for all $i$. Furthermore, $v_2(n + i^2) = 1$ happens for only at most $\frac{n}{2} + 1$ values of $i$. Also, $v_2(n!) \ge \frac{n}{2} + 2$ for large $n$ (in fact, $v_2(n!)$ is always at least $n - \log_2(n)$ or so). This yields a contradiction for large $n$. For $n \equiv 0 \pmod{4}$, we have $n - 1 \equiv 3 \pmod{4}$, so we may take a prime divisor $p \equiv 3 \pmod{4}$ of $n-1$. For this $p$ the number $n$ is a quadratic residue modulo $p$, so $-n$ is a quadratic nonresidue. This proves that $p \nmid n + i^2$ for all $i$. However, $p \mid n!$, as $p \le n-1 < n$. For $n \equiv 3 \pmod{4}$, proceed as in the case $n \equiv 0\pmod{4}$, but take a prime divisor $p \equiv 3 \pmod{4}$ of $n-4$ instead of $n-1$.