Polynomial solutions to a difference equation

Question

This question may look unmotivated, but is connected with continued fractions for $\pi^2$. Let $n$ be a nonnegative integer, and consider the difference equation $$(x+2n+4)(x+n+1)P(x+1)-(x-1)(x+n-1)P(x-1)=((2n^2+8n+7)x+(2n+3)(n^2+3n+1))P(x)$$ Prove (experimentally it is true) that there exists a unique polynomial $P$ (of course up to a multiplicative constant) satisfying this, and that it has degree $n(n+3)$.

Same question for the similar equation $$(x+2n+4)(x+n+4)P(x+1)-(x-1)(x+n+2)P(x-1)=((2n^2+8n+7)x+(2n+3)(n+2)(n+3))P(x)$$

Probably much more difficult question: where do these difference equations come from (I only obtained them via pattern recognition).

Answer 1

A few minor comments which don't fit conveniently into the comment fields (and would resist to future edits):

1. Assuming that a monic solution $P$ has degree $m$, then comparing coefficients at $x^{m+1}$ yields $m = n(n+3)$, so the degree of a putative solution is as expected.

2. Write $P(x)=a_0+a_1x+\ldots a_{m-1}x^{m-1}+x^m$. For $i=0,1,\ldots,m-1$ the coefficient of $x^{m-i}$ of the defining relation expresses $a_{m-i-1}$ as a linear combination of the $a_j$'s for $j>m-i-1$. Therefore there is at most one solution.

3. The problem is: There is one more condition for $i=m$. So in order to prove existence, one has to show that it is automatically fulfilled. I guess that's the hard part ...

4. Your first and second example is essentially the same one. For if you replace $x$ in the first case by $x-n-1/2$ and in the second case $x$ by $x-n-5/2$, you get essentially the same system, where the putative solutions $P$ differ by an affine transformation of the argument.

5. In fact if one replaces $x$ with $x - n - 1/2$ and afterwards $n$ with $n-2$, then one arrives at the somewhat simpler looking difference equation \begin{equation} (2x + 1)(2x + 2n + 3)Q(x+1)-(2x - 3)(2x - 2n + 1)Q(x-1)=2((4n^2 - 2)x - 1)Q(x). \end{equation}