let $f(x)$ be rational non constant polynomial and $f\circ f(x)=3f(x)^4-1$ then find the $f(x)$

Question

let $f(x)$ be rational non constant polynomial and $f\circ f(x)=3f(x)^4-1$ then find the $f(x)$ .

My Try :

$$f(f(x))=3f(x)^4-1$$

Let $f(x)=ax^n+g(x)$ so :

$$a(ax^n+g(x))^n+g(ax^n+g(x))=3(ax^n+g(x))^4-1$$

$$a^2x^{n^2}+h(x)+k(x)=3ax^{4n}+l(x)-1$$

$$3ax^{4n}-a^2x^{n^2}-1=h(x)+k(x)-l(x)$$

Now what ?

Answer 1

The highest degree terms on each side of what you have are:

$a_n^{n+1}x^{n^2} = 3a_n^4x^{4n}$

$n^2 = 4n\\ n = 4\\ a_4^5 = 3a_4^4\\ a_4 = 3$

But, there is a more direct approach.

$f(f(x)) = 3(f(x))^4 - 1$

Replace the $f(x)$ with $u$ everywhere we see it.

$f(u) = 3u^4 - 1$

Answer 2

Note that $3f(x)^4 - 1 = g\circ f(x)$, where $g(x) = 3x^4 - 1$. So $f = g$, at least, is a solution.

Now, if $f$ is a solution, we need to show that $f(x) = g(x)$. But note that $f \circ f(x) = g\circ f(x)$ for all $x$, so $(g - f) \circ f(x) = 0$, so $g - f$ is zero on the image of $f$, which, since $f$ is non-constant, contains some interval $(a,b)$, hence $g - f$ is everywhere zero, so $f = g$ is the only solution.

Answer 3

Note that, because I think that you supposed deg $g(x)\leq n-1$, from $$ a_n(a_n x^n+g(x))^n+g(a_n x^n+g(x))= 3(a_n x^n+g(x))^4-1$$ the degree of the LHS is $n^2$, while on the other hand the degree of the RHS is $4n$. This is an equality between two polynomials, hence their degree must be the same. Then $n=4$. Moreover the two polynomials must have the same leading coefficient. The coefficient of $x^{16}$ in the LHS is $a_4^{5}$, while the coefficient in the RHS is $3a_4^{4}$, hence $a_4=3$. Now I think that you have to put $f(x)=3x^4+\sum_{i=0}^{3} a_i x^i $ in the equation and compute