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Find all polynomials $P(x)$ with real coefficients satisfying $P(x-1)P(x+1)=P(P(x))$

I tried $P(x) = c$ (const) and got $c=0$ or $c=1$. But when I tried $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}$, normally I would be able to find $a_{n}$ but in this case, both highest coefficients on the two sides is $a_{n}^{2}$ so I can't find it.

I don't know how to progress in this problem. Any help would be highly appreciated!

P/s: With the help of @Theo Bendit, I come up with this solution.

Let $\deg(P(x))=n$, then $\deg(LHS)=2n$ while $\deg(RHS)=n^{2}$ => $n=2$ (as $n=0$ then I have done it above). Therefore, $P(x)=ax^{2}+bx+c$.

We see that for $x^{4}$, we have $a^2=a^3 => a=1$ and we can do the same for $b$ and $c$.

In the end, we get $P(x)=x^2-2x+1$ is a solution.

pluton
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Kii
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    Instead of trying $P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0$, try figuring out the possible degrees of $P(x)$. If $P$ has degree $n$, then what is the degree of $P(x - 1)P(x + 1)$? What is the degree of $P(P(x))$? You should find that there are only finitely many degrees $n$ that you could possibly have, and they are all relatively small. – Theo Bendit Feb 24 '24 at 04:13
  • @TheoBendit I thought that if deg(P)=n then deg(RHS)=$n^{2}$ and so is the LHS. Then this is nothing to do with it. Did I mistake something? – Kii Feb 24 '24 at 04:16
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    Try it with a monomial. If $P(x) = x^n$, then $P(P(x)) = (x^n)^n = x^{n^2}$, as you say, but $P(x - 1)(P(x + 1) = (x - 1)^n(x + 1)^n = (x^2 - 1)^n$, which has degree $x^{2n}$. – Theo Bendit Feb 24 '24 at 04:18
  • I thought $x^{n}^{2}=x^{2n}$ or am I wrong? – Kii Feb 24 '24 at 04:22
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    @Kil Exponentiation is not associative. In particular, $(x^n)^2 \neq x^{(n^2)}$. When we write $x^{n^2}$, it means the latter, by convention, because the former is $x^n \cdot x^n = x^{n + n} = x^{2n}$. – Theo Bendit Feb 24 '24 at 04:23
  • @TheoBendit for a monomial, I think n=2 satisfied the degree. Thank you for that and do you have any ideas on the polynomial? – Kii Feb 24 '24 at 04:32
  • Well, my point with the degree thing is that a degree $n$ solution must satisfy $n^2 = 2n$, i.e. $n = 0$ or $n = 2$. That is, $P$ is constant, or $P$ is quadratic. If you sub in $P(x) = ax^2 + bx + c$ into your equation, you should be able to find the remaining solutions through equating coefficients. – Theo Bendit Feb 24 '24 at 04:33
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    Thank you a lot @TheoBendit! – Kii Feb 24 '24 at 04:35

1 Answers1

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

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

Let's say that $P(x)$ has degree $n$. That means that $P(x+1)$ and $P(x-1)$ will each have degree $n$ too. Their product will consequently be of degree $2n$ (because $x^n$ times $x^n$ is $x^{2n}$).

The right-hand side will have a degree of $n^2$ because $P(x)$ is of degree $n$ and $P(\textit{something of degree n})$ will return something of degree $n^2$.

We can therefore conclude since both sides must have the same degree, that $n^2=2n$ and $n$ either equals $2$ or $0$. A zero-degree polynomial is a cheeky answer, so let's just say that $P(x)$ is of degree $2$, or a quadratic.

$P(x)=ax^2+bx+c$

$P(x+1)=a(x+1)^2+b(x+1)+c=a(x^2+2x+1)+b(x+1)+c=ax^2+2ax+a+bx+b+c$

$P(x-1)=a(x-1)^2+b(x-1)+c=a(x^2-2x+1)+b(x-1)+c=ax^2-2ax+a+bx-b+c$

$P(x+1)P(x-1)=(ax^2+2ax+a+bx+b+c)(ax^2-2ax+a+bx-b+c)$

Getting a calculator to evaluate this...

$\space=(a^2 - b^2 + 2 a c + c^2)+ (- 2 a b + 2 b c )x+( - 2 a^2 + b^2 + 2 a c)x^2 + (2 a b) x^3 + (a^2 )x^4$

Next up is evaluating that $P(P(x))$

Since $P(x)=ax^2+bx+c$,

$P(P(x))=a(ax^2+bx+c)^2+b(ax^2+bx+c)+c$

$=(a^3) x^4 + (2 a^2 b )x^3 + (2 a^2 c + a b^2 + a b)x^2 + (2abc+ + b^2 )x + (b c + c+ac^2)$

$ $

$(a^2 - b^2 + 2 a c + c^2)+ (- 2 a b + 2 b c )x+( - 2 a^2 + b^2 + 2 a c)x^2 + (2 a b) x^3 + (a^2 )x^4=(a^3) x^4 + (2 a^2 b )x^3 + (2 a^2 c + a b^2 + a b)x^2 + (2abc + b^2 )x + (b c + c+ac^2)$

Now I just have to match the $x^4$ coefficients with each other, then the $x^3$ coefficients, and so on.

$a^3=a^2$ so $a=1$

Matching the $x^2$ coefficients gives:

$2c+b^2+b=-2+b^2+2c$ so $b=-2$

Finally, evaluating the constant gives $-2c+c+c^2=1-4+2c+c^2$. Simplifying gives $c=1$

And so finally, we have our coefficients giving the polynomial $P(x)=x^2-2x+1$

Honestly, $P(x)=0$ was a better answer tho.

Lucien Jaccon
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