Find $P(7/8)$ given ${P(5)}^2=P(6)$ and $(x-1)P(x+1)=(x+2)P(x)$

Question

There's a polynomial $P(x)$, we know that ${P(5)}^2=P(6)$ and $$(x-1)P(x+1)=(x+2)P(x)$$

Find the value of $P(\frac{7}{8})$.

Any hints?

I know that $P(1)=0,P(0)=0,P(-1)=0$ and $P(5)=0$ or $P(5)=\frac{7}{4}$. But what next?

Answer 1

Note that $P\equiv 0$ is a trivial solution of the equation. So, let us assume $P$ is not identically zero. By plugging $x=1$, it follows that $P(1)=0$. Knowing this, we can observe that $P$ also has $0,-1$ as its roots. Define a polynomial $Q$ as $$ Q(x)=\frac{P(x)}{x(x-1)(x+1)}. $$ From the given relation, we know that $Q(x)=Q(x+1)$. Note that if $Q$ has a root $x_0$, then it should have infinite number of roots $x_0+k$, $\ k\in\Bbb Z$, which is absurd since $Q$ is a non-zero polynomial. Thus $Q$ must be a $0$-degree polynomial. This gives $P(x)=cx(x-1)(x+1)$, and from $P(5)^2=P(6)$, we have that $ (120c)^2=210c, $ i.e. $c=\frac{7}{480}$. It follows that $$ P(\frac78)=c\frac 78(-\frac18)\frac{15}8=-\frac{49}{2^{14}}. $$

Answer 2

Hint.

Solving the recurrence equation

$$ (x-1)P(x+1)-(x+2)P(x) = 0 $$

we have

$$ P(x) = C_0(x^3-x) $$

but $P(5)^2 = P(6)$ or $C_0^2(5^3-5)^2 = C_0(6^3-6)$ giving $C_0$ etc.

Of course $P(x)\equiv 0$ is also a solution.