Polynomials that satisfy $(x-1)(p(x+1))=(x+2)(p(x))$ where $p(2)=12$?

Question

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!

Answer 1

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. :)

Answer 2

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 $2x^3 - 2x$.

Answer 3

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 $A=2$.

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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.

Answer 4

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!