Find all polynomials $P(x)$ so that $P(x)(x+1)=(x-10)P(x+1)$

Question

Find all polynomials $P(x)$ so that $P(x)(x+1)=(x-10)P(x+1)$.

I'm looking for a general solution to the above problem. For instance, say I was trying to find all polynomials $P$ satisfying $(x+1) P(x) = (x-9)P(x+1)$ or all polynomials $P$ satisfying $(x+2)P(x)=(x-9)P(x+2).$

Would it be feasible to solve the following very general problem: Find all integers a and b for which there exists a nonconstant polynomial $P$ so that $(x+a)P(x) = (x+b)P(x+1)$. Obviously $a = b$ would not work because $P(x)=P(x+1)$ implies $P$ is constant. I'm not sure if my approach below can be useful for solving this.

Let $(*)$ be the equation $(x+1)P(x)=(x-10)P(x+1)$. Assume $P$ is nonconstant, for if it is constant, one can easily show that it must be zero and if it is zero, it is clearly a solution.

Throughout our solution, we will use the following key facts: if $p(x),q(x),r(x)$ are complex polynomials and $$\gcd(p(x), q(x)) = 1, p(x) | q(x)r(x)\Rightarrow p(x) | r(x)\quad (a)$$

$$\text{if } p(x),q(x) \text{ are complex polynomials so that there exists a nonzero complex polynomial } r(x) \text{ with } p(x)r(x)=q(x)r(x)\text{, then} p(x)=q(x) \quad (b)$$

From fact (a), we have that $(x+1)P(x) = (x-10)P(x+1)$ implies $x+1$ divides $P(x+1)$ and $x-10$ divides $P(x)$. Write $$P(x+1) = (x+1)^r Q(x)\quad (1)$$ and $$P(x) = (x-10)^s R(x) \quad (2)$$ for some $r,s> 0$ and some polynomials $Q$ and $R$ with complex coefficients so that $Q(-1)\neq 0, R(10)\neq 0$. From (1) we have $P(x)=x^r Q(x-1),$ which equals $(x-10)^s R(x)$ by (2). Again since $x-10$ and $x$ are coprime and $r,s>0,$ we have that $(x-10)^s$ divides $Q(x-1)$ and $x^r$ divides $R(x).$ So we may write $R(x) = x^{r} R_2(x)$ for some $R_2(x)$ and $Q(x-1) = (x-10)^{s} Q_2(x)$ for some $Q_2(x)$. Plugging these into the equation for P gives $x^r (x-10)^{s} Q_2(x) = (x-10)^s x^{r} R_2(x)\Rightarrow Q_2(x) = R_2(x)$ by fact (b) and we get that $R(x) = x^r R_2(x)$ so $P(x) = (x-10)^s x^r R_2(x)$ for some polynomial $R_2(x)$. Substituting into the original equation, we get that $(x+1)(x-10)^s x^r R_2(x) = (x-10) (x-9)^s (x+1)^r R_2(x+1)$. So by Euclid's lemma again, $R_2(x+1)$ is divisible by $x^r$ and $(x-10)^{s-1}$ and $R_2(x)$ is divisible by $(x+1)^{r-1}$ and $(x-9)^s.$ Hence the above equation can be rewritten as $(x+1)^r (x-10)^s x^r (x-9)^s Q_3(x)= (x-10)^s (x-9)^s (x+1)^r x^r R_3(x)$ for some polynomials $Q_3, R_3$ where $R_2(x) = (x+1)^{r-1}(x-9)^s Q_3(x)$ and $R_2(x+1)=x^r (x-10)^{s-1} R_3(x)$.

By fact (b), we have that $Q_3(x) = R_3(x)$. We know from above that $R(10)\neq 0, Q(-1)\neq 0\Rightarrow Q_2(0) = R_2(0), Q_2(10) = R_2(10)\neq 0\Rightarrow Q_3(0)=R_3(0), R_3(10)\neq 0,$ $R_3(9) \neq 0$ because $R_2(10)\neq 0,$ and finally $R_3(-1)\neq 0$ as $R_2(0)\neq 0$. Thus we must have $P(x) = (x+1)^r (x-10)^s (x-9)^s x^r R_3(x)$ for some polynomial $R_3(x)$ that does not have $-1,0,10,9$ as zeroes and where $r,s>0.$ Plugging this into the original equation again, we have $(x+1)^{r} (x-10)^s (x-9)^s x^r R_3(x) = (x-10)(x+2)^r (x-9)^s (x-8)^s (x+1)^r R_3(x+1)$. So $(x-10)^{s-1} x^r R_3(x) = (x+2)^r (x-8)^s R_3(x+1).$ $R_3(x) = (x+2)^r (x-8)^s Q_4(x), R_3(x+1) = (x-10)^{s-1} x^r R_4(x)$ for some polynomials $Q_4, R_4$. By fact (b), we get that $R_4(x)=Q_4(x)$. But then plugging in $x=-1$ to the equation for $R_3(x+1)$ gives $R_3(0) = 0,$ contradicting the fact that $0$ is not one of the roots of $R_3$. One could repeat the argument on $R_3(x),$ but this already seems fairly tedious.

Edit: I just realized that I made an annoying computational error. So the initial contradiction above is wrong.

Answer 1

Yes, the more general problem is solvable. If we write $$P(x) = P(x + 1) \frac{x+b}{x+a},$$ we can more clearly see what we want. That is, we take $P(x)$, shift every factor by 1, remove the $(x+a)$ factor (as long as we've made sure it exists!), and add in $(x+b)$. We want all this to cancel out. There must be an $(x + a)$ factor after the shifting, so $(x + a - 1) \mid P(x)$. Now, we must have an $(x + a - 1)$ factor, so we need an $(x + a - 2)$, etc. This chain terminates if and only if $b \le a$.

To verify, we define $$P_{a,b}(x) = \prod_{b \le n < a}(x+n).$$ Then, \begin{align} (x+b)P_{a,b}(x+1) &= (x-b)\prod_{b \le n < a}(x+1+n)\\ & = \prod_{b - 1 \le n < a}(x+1+n)\\ & = \prod_{b \le n < a + 1}(x+n)\\ & = (x+a)\prod_{b \le n < a}(x+n)\\ & = (x+a)P_{a,b}(x). \end{align}

Answer 2

There is a much simpler way to do this. Let $S$ be the roots of $P(x)$ and $S_+$ the roots of $P(x+1)$. Then

• $S_+ \cup S = S \cup \{-1,10\}$. [Indeed, the only way the equation $(x-10)P(x+1)|_{x=-1} =0$ can hold is iff $-1$ is in $S_+$, and likewise, the only way that the equation $P(x)(x+1)|_{x=10}=0$ can hold is if $10$ is in $S$. So $-,10$ is in $S \cup \{-1,10\}$. Now let $y \in S_+$; $y \not \in \{-1,10\}$. Then $P(y)(y+1)$ must be $0$ and as $y+1$ is nonzero, it follws that the equation $P(y)=0$ holds, which implies $y \in S$.]

• $S_+ = \{y-1; y \in S\}$

This implies that for $S$ to be finite, $S$ must be $\{0,1,2, \ldots, 10\}$. [Indeed, to show this, observe that there is no root $y > 10$ in $S$; if $P(y)=0$ for $y>10$, then the equation $P(y)(y+1)=(y-10)P(y+1)$ forces $P(y')=0$ for $y'=y+1$, which in turn forces $P(y'')=0$ for $y''=y'+1$, and so on and so forth. Likewise, there is no root $y<-1$ in $S$. Next, there can be no $y \in (9,10)$ such that $P(y)=0$, lest $P(y')=0$ for $y'=y+1>10$. This implies there can be no $y \in (8,9)$ such that $P(y)=0$, and so on and so forth.]

Observe then that the only such finite $S$ that satisfies the above is $S=\{0,1,\ldots, 10\}$, and that $S=\{-0,1,\ldots, 10\}$ indeed does satisfy the above. In particular, if $$P(x)=x(x-1) \ldots (x-10),$$ then

$$(x+1)P(x) = (x+1)x(x-1) \ldots (x-10) = (x-10)P(x+1).$$

So now that you have the precise set $S$ of roots of $P$ iff $P$ is nonconstant finite degree and so $S$ is finite, can you finish from there. [Yes $P \equiv 0$ i.e., $S =\mathbb{C}$] works as well.]