TITLE: Does the sequence $f_1=x^2+1$ , $f_{n+1}=(f_n)^2+1$ contain only irreducible polynomials?
QUESTION [11 upvotes]: Consider the sequence
$$f_1=x^2+1$$
$$f_{n+1}=(f_n)^2+1$$
of polynomials over the integers.

Is $f_n$ irreducible over $\mathbb Q[x]$ for all $n\ge 1$ ?

With PARI/GP, I found out that upto degree $4\ 096$, all the polynomials are irreducible over $\mathbb Q[x]$. Obviously , the polynomials don't have real roots and are all even. Any ideas ?

REPLY [5 votes]: This is true.
From Ayad, McQuillan, Irreducibility of the iterates of a quadratic polynomial over a field (here), a polynomial is said to be stable over $K$ if all its iterates are irreducible over $K$.
Let $f(X)=X^2-lX+m$ and $d=l^2-4m$ its discriminant.

Theorem 3: If $d=0 \pmod 4$ and $d\neq0 \pmod{16}$, then $f$ is stable over $\mathbb Q$.

In our case, $l=0$, $m=1$, hence $d=-4$ and the theorem applies.