Polynomials that satisfy $(x-1)(p(x+1))=(x+2)(p(x))$ where $p(2)=12$? I am taking a graduate class on Equation Theory and one of my homework questions asks me to "Determine all polynomials $p(x)$ such that $(x-1)(p(x+1))=(x+2)(p(x))$ and $p(2)=12$. A provided hint is to try carefully selected values of $x$.
I've been able to use Excel to "plug and chug" to find values for $x\in\mathbb{Z}\geq1$. ($p(x)=0$ for $x\in\mathbb{Z}<1$.) I used Excel to graph the values I found and I was given a trend line that approximated $p(x)=2x^3-2x$.
What might some approaches be to be able to find all such polynomials?
Thanks!
 A: You found a solution $p(x) = 2x^3 - 2x$.
For all $n \in \mathbb{N}, n \geq 2$ holds
$p(n+1) = \frac{n+2}{n-1}p(n)$ with $p(2) = 12$
hence by induction $p$ is determined for all $n\geq 2$.
This means that the solution is unique, and is $2x^3 - 2x$.
A: We are given that
$$ \frac{ p(x+1) } { p(x) } = \frac{ x+2}{x-1}$$
Let $ p(x) = A \prod ( x - \alpha_i)$.
Hint:  If $ \alpha_i \neq 1$, show that for some $j$, $\alpha_j = \alpha_i - 1$.
Hint: If $ \alpha_i \neq -1$, show that for some $k$, $ \alpha_k = \alpha_i + 1$.
Hint: Use the fact that any polynomial has finite degree.
Conclude that $ p(x) = A [(x-1) x (x+1)] q(x)$, where $ \frac{q(x+1)}{q(x)} = 1 $ for some polynomial $q(x)$.
Show that $q(x)$ is a constant.
Hence, $p(x) = A(x-1)x(x+1)$, and since $p(2) = 12$, we can find $A=2$.

In particular, this helps you characterize all possible equations of the form 
$$ f(x) p(x) = g(x) p(x+1)$$
which have a non-trivial solution.
A: Think about roots of this polynomial. Notice that, $x=-1,0$ are roots of this polynomial. 
What can you tell about the polynomial, if you know one of its roots? That it must be divisible by $(x-a)$ where $a$ is root of this polynomial. Try to use this in this problem, i am sure you will  be able to complete it.  :)
A: Given the condition
$$
(x-1) p(x+1) = (x+2) p(x) \wedge p(2) = 12.
$$
We find
$$
p(x-1) = \frac{x-2}{x+1} p(x) \wedge p(2) = 12 \Rightarrow p(1)=0, p(0)=0, p(-1)=0,
$$
i.e. we have the zeros $-1, 0, +1$. So we can write
$$
p(x) = (x-1)x(x+1) f(x),
$$
where
$$
f(2) = 2.
$$
Whence
$$
(x-1)x(x+1)(x+2) f(x+1) = (x-1)x(x+1)(x+2) f(x),
$$
so
$$
f(x) = f(x+1).
$$
So the function $f(x)$ is periodic.
When we consider "infinite" polynomials, we can think of $f(x) = 2\cos(2 k \pi x)$.
So we obtain for example
$$
p(x) = 2(x-1)x(x+1) \cos(2 k \pi x),
$$
as a solution.
The case $k=0$ yields
$$
p(x) = 2(x-1)x(x+1) = 2x^3-2x,
$$
but this is not the only solution!
