Maximum number of monic polynomials $f(x)$ such that for some real $a$, $x(f(x+a) - f(x)) = nf(x)$ for a fixed $n<1000$? This was a problem that I did from a while ago that I am revisiting. I remember getting the answer as the maximum divisors of an $n < 1000$, which would be $n = 840 \rightarrow 32$ divisors and polynomials, but I can't remember how I came to that result. Anyone mind giving me a hand?
 A: Given $n$, you are trying to find monic polynomials $f$ such that for some real number $a$, we have $$x(f(x+a) - f(x)) = nf(x).$$
We will prove (proved in two ways below) that for any degree $d$, there is a unique polynomial of degree $d$ satisfying this constraint, and that this polynomial is
$$f_d(x) = x\left(x+\frac{n}d\right)\left(x+\frac{2n}d\right) \cdots \left(x+\frac{(d-1)n}d\right)$$
Observe that
$$\begin{align}
f_d\left(x+\frac{n}{d}\right) 
&= \left(x+\frac{n}{d}\right)\left(x+\frac{2n}{d}\right)\dots\left(x+\frac{dn}{d}\right)\\
&= f_d(x)\frac{(x+n)}{x}
\end{align}$$
and so with $a = n/d$, we have $x(f(x+a)-f(x)) = nf(x)$.
Now, if we want integer coefficients, note that each polynomial $f_d(x)$ has integer coefficients if and only if $d$ divides $n$. So for a given $n$, the number of polynomials satisfying the constraint is the number of divisors $d$ of $n$, as we wanted to prove.

Proofs that $f_d(x)$ is the unique polynomial of degree $d$ satisfying the constraints: I see two approaches, and probably there may exist other better ones.
First proof. Suppose $f$ is of degree $d$, say $f(x) = x^d + c_{d-1}x^{d-1} + \dots + c_1x + c_0$. Then, let us compare coefficients of $x^d, x^{d-1}, \dots$ in $x(f(x+a) - f(x))$ and in $nf(x)$, and use the fact that they must be equal. We can either


*

*expand all terms of $f(x+a)$ using the binomial theorem, or 

*use $f(x+a) = f(x) + a f'(x) + \frac{a^2}{2!}f''(x) + \dots$. 


Either way, the coefficient of $x^d$ in $x(f(x+a)-f(x))$ is $ad$, and in $nf(x)$ is $n$, and so $a = n/d$.
Next, comparing coefficients of $x^{d-1}$ gives $c_{d-1}(d-1)a + \frac{d(d-1)}{2}a^2 = nc_{d-1}$, determining $c_{d-1}$. Similarly, comparing coeficients of other powers gives successively $c_{d-2}, \dots$, each got in terms of the earlier coefficients. Thus, as all coefficients $c_k$ are determined, there can be at most one polynomial of degree $d$ satisfying the constraint. As we have found and verified $f_d(x)$ above, it must be the unique one.
Second proof. As $x$ divides the left-hand side of the given equation $x(f(x+a)-f(x)) = nf(x)$, we conclude that it must divide the right-hand side, and therefore $f(x)$ is a multiple of $x$.
Plugging $f(x) = xg(x)$ into the equation and cancelling $x$, we get $(x+a)g(x+a) = (x+n)g(x)$. So either $a = n$ (in which case the periodicity $g(x+a)=g(x)$ implies that $g$ is the constant polynomial $g(x) \equiv 1$ and so $f(x) \equiv x$), or else $(x+a)$ divides $g(x)$.
Plugging $g(x) = (x+a)h(x)$ into the equation $(x+a)g(x+a) = (x+n)g(x)$ and cancelling the $(x+a)$ factor gives $(x+2a)h(x+a) = (x+n)h(x)$. So either $n = 2a$ (in which case $h(x) \equiv 1$, and $f(x) \equiv x(x+a)$) or else $(x+2a)$ divides $h(x)$.
Proceeding this way, we get that our polynomial is
$$f(x) = x(x+a)(x+2a)\cdots(x+(d-1)a)$$
for some $d$ such that $n = da$.
