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Find all polynomials $P(x) \in \mathbb R\left[x\right]$, satisfying: $$xP(x-1) = (x+1)P(x)$$

I tried, but I am getting something like this: $$P(x)=\ldots(x-3)(x-2)(x-1)(x)(x+1)(x+2)(x+3)\ldots$$ (My teacher gave me this math question and I need help because my answer is not looking like a polynomial)

TShiong
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ady
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    What have you tried, and where did you get stuck? – dxiv Jun 05 '23 at 04:33
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  • I am getting P(x)=...........(x-3)(x-2)(x-1)(x)(x+1)(x+2)(x+3)........... – ady Jun 05 '23 at 04:39
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    @ady An infinite product of linear factors is not a polynomial. But you should edit that into the question, as well as how you derived that form, rather than posting it as a comment. – dxiv Jun 05 '23 at 04:44
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    Hint. You didn't write how you arrive to that, but you're probably using without noticing that $p(x)$ is not the $0$ polynomial. – jjagmath Jun 05 '23 at 04:55
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    Let $Q(x)=xP(x-1)$. Then, the question is equivalent to asking whether $Q(x)=Q(x+1)$. – Geoffrey Trang Jun 05 '23 at 05:27

2 Answers2

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One can easily see that zero polynomial satisfies the given condition $xp(x-1)=(x+1)p(x)$.

Now suppose $p(x)$ is a non-trivial polynomial then substituting $x+1$ in place of $x$ in the above constraint gives $(x+1)p(x)=(x+2)p(x+1)$. Thus, moving inductively, one can see that $xp(x-1)=(x+n)p(x+n-1)$ holds for each $n$. Since, $p(x)$ is a polynomial, the right hand side of the above equation tends to $\pm\infty$ as $n$ becomes larger and larger, which is absurd because $xp(x-1)$ is supposed to be a fixed quantity. Thus, a non-trivial polynomial cannot satisfy the given constraint.

NewB
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    That works (+1), or you could argue that setting $x=0$ in the sequence of equalities implies $p(0)=0=p(1)=p(2)=\ldots$ but a non-zero polynomial can only have finitely many roots. – dxiv Jun 05 '23 at 05:14
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    @dxiv You argument is even more lucid, I should have noticed that. – NewB Jun 05 '23 at 05:16
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    @dxiv ... Or by setting $Q(x):=(x-1)P(x)$ giving $Q(x)=Q(x+1)$ which means that $Q$ is periodical with period $1$, which is impossible for a non-zero polynomial. – Jean Marie Jun 05 '23 at 07:04
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Suppose $P(x) \in \mathbb{R}[x]$ is a polynomial. Then $Q(x) = (x+1)P(x) \in \mathbb{R}[x]$ is a polynomial such that $$Q(x-1) = ((x-1)+1)P(x-1) = xP(x-1) = (x+1)P(x) = Q(x)$$ for every $x$ and therefore $Q(x) - Q(0)$ has a zero at every integer-value of $x$. The only way a polynomial in $\mathbb{R}[x]$ can have infinitely many zeros is if it's the zero polynomial, so $Q(x) - Q(0) = 0,$ or $$Q(0) = Q(x) = (x+1)P(x).$$

Setting $x=-1$ implies $Q(0) = 0$ so $(x+1)P(x) = 0$, or $P(x) = 0$ is the zero polynomial.

Brian Moehring
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