Is the statement true?

Question

While working on composition of functions with itself that I have noticed a periodic behavior for f(x).

$$f(x)=x^2-1$$ $$f(0)=-1$$

$$f(f(x))=f^2(x)=(x^2-1)^2-1=x^4-2x^2=x^2(x^2-2)$$

$$f^2(0)=0$$

$$f(f(f(x)))=f^3(x)=(x^2-1)^2((x^2-1)^2-2)=(x^2-1)^2(x^4-2x^2-1)$$

$$f^3(0)=-1$$

$$f(f(f(f(x))))=f^4(x)=x^4(x^2-2)^2[(x^2-1)^4-2(x^2-1)^2-1]$$

$$f^4(0)=0$$

I calculated that $f^5(0)=-1$

I could not proceed more because it became very complex.

I suggest that

If n is odd then $f^n(0)=-1$

If n is even then $f^n(0)=0$

Can you please help me how I can prove it or disprove the statement above?

Thanks

EDIT:

I have used induction as Stefan mentioned in his answer as hint

I proved the statement

Proof:

$$f^{n+2}(x)=f^{n}(f^{2}(x))$$

$$f^{n+2}(x)=f^{n}(x^2(x^2-2))$$

We got for $x=0$

$$f^{n+2}(0)=f^{n}(0)$$

We know $$f(0)=-1$$

$$f^{3}(0)=f(0)=-1$$

$$f^{5}(0)=f^3(0)=-1$$

Thus If n is odd then $f^n(0)=-1$

We know $$f^2(0)=0$$

$$f^{4}(0)=f^2(0)=0$$

$$f^{6}(0)=f^4(0)=0$$

Thus If n is even then $f^n(0)=0$

Thanks to Stefan for hint

Answer 1

Hint: Use induction in $n$. $ $

Answer 2

Notice that if $p(x)$ is any polynomial of the form $p(x) = x^2g(x) + c$, for polynomial $g$ and constant $c$, then $$ f(p(x)) = (p(x))^2 - 1 = x^4(g(x))^2 + 2cx^2g(x) + c^2 - 1 = x^2h(x) + (c^2-1), $$ for some polynomial $h(x)$. Notice that $f^n(0)$ is just the constant term, so the above reduces the problem to showing that the sequence defined recursively by $c_1 = -1$ and $c_n = c_{n-1}^2-1$ is the same as $-1, 0, -1, 0, -1, \cdots$.

Answer 3

Even in your induction you still seem to be overcomplicating things. Note that $f^{n+1}(x)=f(f^n(x))=(f^n(x))^2-1$. In particular $f^{n+1}(0)=(f^n(0))^2-1$ so we don't really need to know $f^{n+1}(x)$ to find $f^{n+1}(0)$, we just need to know $f^{n}(0)$.

Now $f^{n}(0)=0\implies f^{n+1}(0)=0^2-1=-1$ and $f^{n}(0)=-1\implies f^{n+1}(0)=(-1)^2-1=0$

So given that $f(0)=0$, the iterates of $f$ evaluated at $0$ just alternate between $0$ and $-1$.