Is there an even number $a$ such that $a^{2^{n}}+1$ is prime for infinitely many $n$?

Question

Is there an even number $a$ such that $\{n: a^{2^{n}}+1 \text{ is prime} \}$ is an infinite set?

Let $a$ be even. Is there infinitely many $n$ such that $a^{2^{n}}+1$ is composite?

Answer 1

A remark that might be noteworthy...

• Either the sequence $\{2^{2^{n}}+1\}_{n \in \mathbb{N}}$ contains infinitely many composite numbers or the sequence $\{6^{2^{n}}+1\}_{n \in \mathbb{N}}$ contains infinitely many composite numbers.

Proof. If there are only finitely many composite numbers in the first sequence (the sequence of Fermat numbers), we might assure the existence of an $n_{0} \in \mathbb{N}$ such that $2^{2^{n}}+1$ is a prime number for every $n$ that belongs to $\mathcal{I}:=[n_{0},\infty) \cap \mathbb{N}$.

We claim that, in such a case, $6^{2^{n}}+1$ is a composite number for every $n \in \mathcal{I}$. Indeed, for any given $n \in \mathcal{I}$, the Fermat number $2^{2^{n}}+1$ is a prime and Pepin's test gives us that $$3^{2^{n-1}} \equiv -1 \pmod{2^{2^{n}}+1}.$$ Taking squares on both sides of the previous congruence we get $3^{2^{n}} \equiv 1 \pmod{2^{2^{n}}+1}$; from this information and the fact that $2^{2^{n}} \equiv -1 \pmod{2^{2^{n}}+1}$, we obtain that $$6^{2^{n}} \equiv -1 \pmod{2^{2^{n}}+1}$$ and the validity of our claim follows.