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

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.

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}

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.]

• Can you justify the main claims about S more formally? Nov 21, 2022 at 21:29
• Yes @user33096, what about $P(x)=x(x-1) \ldots (x-10)$.
– Mike
Nov 21, 2022 at 21:44
• Anyways @user33096 I revised my post....[I also corrected my mistake, infact $-1$ is not in $S$ if $S$ is finite.]
– Mike
Nov 21, 2022 at 22:07