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Motivated by this question.

Let $f \in \mathbb{Q}[x]$ or$f \in \mathbb{Z}[x]$ .

Consider the sequence $f(x),f(f(x)), \ldots f^n(x)$.

If some $f^k(x)$ is reducible, the rest iterates will be reducible too.

This happens for $g(x) = x^2 - x - 1$.

$$g(g(g(x))) = (x^{4} - 3 x^{3} + 4 x - 1) \cdot (x^{4} - x^{3} - 3 x^{2} + x + 1)$$

Fermat numbers are related to similar quadratic map and if it happens to be reducible for some $k$, this will mean there are infinitely many Fermat composites.

Given $f$, is it possible to decide if some $f^k(x)$ is reducible?

Motivated by this question.

Let $f \in \mathbb{Q}[x]$ or$f \in \mathbb{Z}[x]$ .

Consider the sequence $f(x),f(f(x)), \ldots f^n(x)$.

If some $f^k(x)$ is reducible, the rest iterates will be reducible too.

This happens for $g(x) = x^2 - x - 1$.

$$g(g(g(x))) = (x^{4} - 3 x^{3} + 4 x - 1) \cdot (x^{4} - x^{3} - 3 x^{2} + x + 1)$$

Fermat numbers are related to similar quadratic map and if it happens to be reducible for some $k$, this will mean there are infinitely many Fermat composites.

Given $f$, is it possible to decide if some $f^k(x)$ is reducible?

Motivated by this question.

Let $f \in \mathbb{Q}[x]$ be irreducible polynomial.

Consider the sequence $f(x),f(f(x)), \ldots f^n(x)$.

If some $f^k(x)$ is reducible, the rest iterates will be reducible too.

This happens for $g(x) = x^2 - x - 1$.

$$g(g(g(x))) = (x^{4} - 3 x^{3} + 4 x - 1) \cdot (x^{4} - x^{3} - 3 x^{2} + x + 1)$$

Fermat numbers are related to similar quadratic map and if it happens to be reducible for some $k$, this will mean there are infinitely many Fermat composites.

Given $f$, is it possible to decide if some $f^k(x)$ is reducible?

Motivated by this question.

Let $f \in \mathbb{Q}[x]$ or$f \in \mathbb{Z}[x]$ .

Consider the sequence $f(x),f(f(x)), \ldots f^n(x)$.

If some $f^k(x)$ is reducible, the rest iterates will be reducible too.

This happens for $g(x) = x^2 - x - 1$.

$$g(g(g(x))) = (x^{4} - 3 x^{3} + 4 x - 1) \cdot (x^{4} - x^{3} - 3 x^{2} + x + 1)$$

Fermat numbers are related to similar quadratic map and if it happens to be reducible for some $k$, this will mean there are infinitely many Fermat composites.

Given $f$, is it possible to decide if some $f^k(x)$ is reducible?