Polynomial functional equation

Question

Find all polynomials $f : R \rightarrow R$ such that $f\left(\frac{1}{x+1}\right)=\frac{1}{f(x)-1}$.

Since the functions must be polynomials, I tried using an argument by degrees, but this did not lead me anywhere. Can someone help with some ideas?

Answer 1

Suppose that the degree of $f$ is $d$ then

$$(x+1)^df\left(\frac{1}{x+1}\right)(f(x)-1)=(x+1)^d\tag 1$$

The point is that $(x+1)^df\left(\frac{1}{x+1}\right)$ is a polynomial in $x$. This is then an identity of polynomials.

$f(x)$ is not identically equal to $1$, and so $\deg(f(x)-1)=\deg(f(x))=d$. It then follows that $\deg((x+1)^df\left(\frac{1}{x+1}\right))=0$. Writing $f(x)=\sum_{n=0}^d a_nx^n$ we have $(x+1)^df\left(\frac{1}{x+1}\right)=\sum_{n=0}^da_n(x+1)^{d-n}$ so that we have $a_n=0$ for $n<d$ and $f(x)=ax^d$. Substituting $f(x)=ax^d$ into $(1)$,

$$a^2x^d-a=(x+1)^da(x+1)^{-d}(ax^d-1)=(x+1)^d$$

If $d=0$, then we get $f(x)=a$ for $a$ any solution to $a^2-a-1=0$. Otherwise, set $x=-1$ to get $a(\pm a-1)=0$, so $a\in \{0,1,-1\}$.

$a=0$ doesn't work.

$a=1$ gives $x^d-1=(x+1)^d$, which doesn't work.

$a=-1$ gives $x^d+1=(x+1)^d$ so that $d=1$.

Thus, the only other solution is $f(x)=-x$.