Find all polynomials \(P(x)\) satisfying $$ P(x-1)\,P(x+1)=P(x^2-1) \qquad\text{for all }x\in\mathbb{R}. $$

Solution

Suppose that \(\alpha\) is a root of \(P(x)\). Then

$$ P(\alpha)=0. $$

Substituting \(x=\alpha+1\) into the given equation,

$$ P((\alpha+1)-1)\,P((\alpha+1)+1) = P((\alpha+1)^2-1). $$

Since \(P(\alpha)=0\), it follows that

$$ 0\cdot P(\alpha+2)=P(\alpha^2+2\alpha), $$

and hence

$$ P(\alpha^2+2\alpha)=0. $$

Following,

Solution

Let \(\alpha\) be a root of \(P(x)\). Then

$$ P(\alpha)=0. $$

Given

$$ P(x-1)P(x+1)=P(x^2-1). \qquad (*) $$

Substitute \(x=\alpha+1\) into \((*)\). Then

$$ P(\alpha)\,P(\alpha+2)=P\!\left((\alpha+1)^2-1\right). $$

Since \(P(\alpha)=0\), we obtain

$$ P\!\left((\alpha+1)^2-1\right)=0. $$

Hence,

$$ (\alpha+1)^2-1 $$

is also a root of \(P(x)\).

Applying the same argument to this new root, we substitute

$$ x=(\alpha+1)^2 $$

into \((*)\). Since

$$ P\!\left((\alpha+1)^2-1\right)=0, $$

it follows that

$$ P\!\left((\alpha+1)^4-1\right)=0. $$

Thus,

$$ (\alpha+1)^4-1 $$

is another root of \(P(x)\).

Continuing inductively, we conclude that

$$ (\alpha+1)^{2^k}-1 $$

is a root of \(P(x)\) for every integer

$$ k\ge0. $$

Since a nonzero polynomial has only finitely many roots, the sequence

$$ (\alpha+1)^{2^k}-1 $$

must eventually repeat. This is possible only when

$$ \alpha+1\in\{-1,0,1\}. $$

Hence every root of \(P(x)\) belongs to the set

$$ \{-2,-1,0\}. $$

Therefore,

$$ P(x)=A\,x^l(x+1)^m(x+2)^n, $$

where \(A\neq0\) and \(l,m,n\) are nonnegative integers.

Substituting this form into \((*)\), we obtain

$$ \begin{aligned} P(x-1)&=A(x-1)^l x^m(x+1)^n,\\[2mm] P(x+1)&=A(x+1)^l(x+2)^m(x+3)^n,\\[2mm] P(x^2-1)&=A(x^2-1)^l(x^2)^m(x^2+1)^n. \end{aligned} $$

Comparing the irreducible factors on both sides, we must have

$$ l=n=0. $$

Hence

$$ P(x)=Ax^m. $$

Substituting this back into \((*)\),

$$ A^2x^m(x^2-1)^m = Ax^{2m}. $$

This identity holds only if either \(A=0\) (the zero polynomial) or \(m=0\) and \(A=1\).

Therefore, the only polynomial solutions are

$$ \boxed{P(x)\equiv0 \quad\text{or}\quad P(x)\equiv1.} $$