Find all polynomials `P(x)` such that: `P(x-1).P(x+1)=P(x^2-1)`

 Find all polynomials `P(x)` such that: `P(x-1).P(x+1)=P(x^2-1)`

Solution

Suppose that `\alpha` is a root of `P(x)`  then, `P(\alpha)=0`

Therefore, `P((\alpha+1)-1)=0`

Following, `P(x-1).P(x+1)=P(x^2-1)`  `(*)`. We replace the value of `x=\alpha+1`

                      `P((\alpha+1)^2-1)=0`

We easily see that `(\alpha+1)^2-1` is also a root of `P(x)`

We will again replace `x=(alpha+1)^2`in to the equation `(*)`

                    `P((\alpha+1)^4-1)=0`

We easily see that `(\alpha+1)^4-1)` is another root of `P(x)`

Inductively, `(\alpha+1)^(2k)-1` is the root `\forallkgeq0`

So far: If `\alpha` is a root of `P(x)` then so is `(\alpha+1)^(2k)-1` for all `kgeq1`

Therefore, `\alpha=0` ; `\alpha=-1` ; `\alpha=-2`

Then, `P(x)=x^l (x+1)^m (x+2)^n` which `A=1`

            `P(x-1)=(x-1)^l x^m (x+1)^n`

            `P(x+1)=(x+1)^l (x+2)^m (x+3)^n`

            `P(x^2-1)=(x^2-1)^l (x^2)m (x^2+1)^n`
Hence, `P(x)=x^l` and `P(x)=0`