Polynomial whose values divide $n!$

Question

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 $74!$).

Answer 1

Here is a completely elementary proof, inspired by Pasten's comments.

Let $P(n)=an^2+bn+c$.

Take $n=a^5x^4+2a^3(ab+2a+1)x^3+a(2a^3c+a^2b^2+6a^2b+3ab+6a^2+5a+1)x^2+(ab+2a+1)(2a^2c+2ab+b+2a+1)x+a^3c^2+2a^2bc+abc+2a^2c+ac+c+ab^2+b^2+2ab+2b+a+1$

Then

$P(n)=P_1(x)P_2(x)P_3(x)$

where

$$\begin{align*} P_1(x)=&a^2x^2+abx+2ax+ac+b+1\\ \\ P_2(x)=&a^4x^2+a^3bx+2a^3x+2a^2x+a^3c+a^2b+ab+a^2+2a+1\\ \\ P_3(x)=&a^5x^4\\ &+2a^4bx^3+4a^4x^3+2a^3x^3\\ &+2a^4cx^2+a^3b^2x^2+6a^3bx^2+3a^2bx^2+6a^3x^2+4a^2x^2+ax^2\\ &+2a^3bcx+4a^3cx+2a^2cx+2a^2b^2x+ab^2x+6a^2bx+4abx+bx+4a^2x+2ax\\ &+a^3c^2+2a^2bc+abc+2a^2c+c+ab^2+b^2+2ab+b+a\\ =&n-(a^2x^2+abx+2ax+x+ac+b+1) \end{align*}$$

Clearly for $x$ large enough $P_1(x), P_2(x), P_3(x)$ are distinct and less than $n$ in absolute value. Since the product of 3 distinct numbers $\le n$ divides $n!$, the result follows.$\qquad\qquad\blacksquare$

The idea behind this proof is that always $P(x)\;|\;P(P(x)+x)$ and applying this idea twice allows to factor $P(\text{some polynomial})$ into 3 factors all small enough.

Here is the relevant Maxima code:

(%i1)   a*x*x+b*x+c;
(%o1)   ...
(%i2)   a*x^2+(b+1)*x+c;
(%o2)   ...
(%i3)   subst(a^2*x^2+a*b*x+2*a*x+x+a*c+b+1, x, (%o2));
(%o3)   ...
(%i4)   expand((%o3));
(%o4)   ...
(%i5)   subst(%o4, x, (%o1));
(%o5)   ...
(%i6)   expand((%o5));
(%o6)   ...
(%i7)   factor((%o6));
(%o7)   (a^2*x^2+a*b*x+2*a*x+a*c+b+1)*(a^4*x^2+a^3*b*x+2*a^3*x+2*a^2*x+a^3*c+a^2*b+a*b+a^2+2*a+1)*(a^5*x^4+2*a^4*b*x^3+4*a^4*x^3+2*a^3*x^3+2*a^4*c*x^2+a^3*b^2*x^2+6*a^3*b*x^2+3*a^2*b*x^2+6*a^3*x^2+4*a^2*x^2+a*x^2+2*a^3*b*c*x+4*a^3*c*x+2*a^2*c*x+2*a^2*b^2*x+a*b^2*x+6*a^2*b*x+4*a*b*x+b*x+4*a^2*x
+2*a*x+a^3*c^2+2*a^2*b*c+a*b*c+2*a^2*c+c+a*b^2+b^2+2*a*b+b+a)

Answer 2

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.