A theorem of Édouard Lucas related to the Fermat numbers states that :
Any prime divisor $p$ of $F_n=2^{2^n}+1$ is of the form $p=k\cdot 2^{n+2}+1$ whenever $n$ is greater than one.
Does anyone know is there some similar theorem for generalized Fermat numbers: $F_n(a)=a^{2^n}+1$ ?
I've been searching the internet but I couldn't find any similar theorem .
Let $p$ be an odd prime divisor of $a^{2^n}+1$. Then $p$ is of the form $p=k2^{n+1}+1$.
The simplest argument uses a small amount of group theory. The fact that $p$ divides $a^{2^n}+1$ can be rewritten as $$a^{2^n}\equiv -1\pmod{p}.$$ Squaring, we obtain that $a^{2^{n+1}}\equiv 1\pmod p$. It follows that $2^{n+1}$ is the smallest positive integer $e$ such that $a^e\equiv 1\pmod {p}$, so $a$ has order $2^{n+1}$ modulo $p$.
The full multiplicative group of the integers modulo $p$ has order $p-1$. The order of any element divides the order of the group, so $2^{n+1}$ divides $p-1$. Equivalently, $p$ is of the form $k2^{n+1}+1$.
In the case $a=2$, as noted in the post, if $n>1$ then we have the stronger result that $2^{n+2}$ divides $p-1$. That is not true for general $a$. For example, $41$ is a prime divisor of $3^{2^2}+1$, but $2^4$ does not divide $41-1$.
Asking that $a$ be even does not necessarily help. For example, $10^{2^2}=73\times 137$, but $2^4$ divides neither $73-1$ nor $137-1$.
Comment: The small amount of explicit group theory that we used can be dispensed with, if we develop a few of the basic properties of the order of an integer $a$ modulo $p$.
The usual proof that gives $2^{n+2}$ for $a=2$ uses the additional fact that if $n >1$, then $p$ has shape $8k+1$, and therefore the "base" $2$ is a quadratic residue of $p$. The same idea should work, for instance, with $a=18$.