Polynomial whose values divide $n!$ Let $P(n)$ be an irreducible polynomial of degree $2$ over the positive integers. Do  there exist infinitely many positive integers $n$ such that $P(n)$ divides $n!$?
Edit: motivation  by examples:
A) $p(n)=n^2+1$ (true, $21^2+1$ divides $21!$).
B) $p(n)=n^2+n+1$ (true, $74^2+74+1$ divides $74!$).
 A: If the question is asking whether for a given irreducible quadratic polynomial $f$ whether there exist infinitely many positive integers $n$ for which $f(n) | n!$, then one argues as follows: we may assume that $f(n) > 0$, since there are finitely many $n$ for which $f(n) < 0$. If $n$ is such that whenever $p | f(n)$ we have $p \leq n$ then the desired outcome is true. Much more is true in fact, as shown by Bober, Fretwell, Martin, and Wooley: for any $\varepsilon > 0$ there exists infinitely many $n$ for which $f(n)$ is free of prime factors exceeding $n^\varepsilon$.
Edit: As Mark Sapir points out (but not explicitly), the above argument is too cavalier. It assumes implicitly that $f(n)$ is also square-free.
To fix the argument, we need to dig deeper into the proof given in the B-F-M-W paper. We describe their construction. First choose a large parameter $X$, and take $k$ to be the product of all primes $p < X$ and co-prime to $2a \phi(a)$, where $a$ is the leading coefficient of $f$ and $\phi$ is the Euler-totient function. In particular, $k$ is odd and square-free. For each positive integer $d$ let $\Omega_d$ be the set of primitive $d$-th roots of unity. Let $\alpha, \alpha^\prime$ be the two roots of $f$. Suppose $m,n, A,B$ are integers such that
$$\displaystyle (ma \alpha + n)^k = A\alpha + B.$$
Then put
$$\displaystyle h_d(t) = \prod_{\zeta \in \Omega_d} (t - (m a \alpha + n)\zeta)(t - (m a \alpha^\prime + n) \zeta),$$
and for $G(t) = (t^k - B)/A$ one finds
$$\displaystyle f(G(t)) = C \prod_{d | k} h_d(t).$$
In particular, the $G$ is square-free. They go on to show that an affine transformation of $G$, which is denoted $g$, satisfies the requirement of their theorem.
It follows that for the $g$ given in their theorem, one has $f(g(t)) = f_1(t) \cdots f_m (t)$ with the $f_j$'s pairwise co-prime. Thus, if $p$ is a prime and there exists an integer $k$ for which $p | \gcd(f_i(k), f_j(k))$ for distinct $i,j$ then $p$ must divide the resultant of $f_i, f_j$. It follows that there are only finitely many such primes and this possibility does not affect our argument. We may then assume that if $p^\ell | f(g(n))$ then $p^\ell | f_j(n)$ for exactly one $1 \leq j \leq m$. Then
$$p^\ell \leq |f_j(t)| \ll n^{ck/\sqrt{\log \log k}} < g(n)^\varepsilon,$$
say. By Polignac's formula, the largest power of $p$ dividing $(g(n))!$ is at least
$$\displaystyle \left \lfloor \frac{g(n)}{p} \right \rfloor \geq \frac{g(n)}{p} - 1 \gg g(n)^{1 - \varepsilon}.$$
If $p^\ell \nmid (g(n))!$ then we obtain an inequality of the form
$$\ell \gg_{\varepsilon} g(n)^{1 - \varepsilon},$$
and since $\ell \ll_{\varepsilon} \log g(n)$, this implies
$$\log g(n) \gg_{\varepsilon} g(n)^{1 - \varepsilon}$$
which can only hold for finitely many $n$. Hence for sufficiently large $n$ we see that $p^{\ell} | f(g(n))$ implies that $p^{\ell}$ divides $(g(n))!$, as desired.
