Can we prove that $(2^{2^{n}}+1)+(2^{2^{n-1}}+1) -1$ have at least $n$ different prime divisors

Question

My point was to prove it using induction. Then for $n=1$ it is true, for n = k it is also true so we assume that $ \color{Red}{ 2^{2^{k}} + 2^{2^{k-1}} + 1}$ has at least k different prime divisors

When we consider $n=k+1$ we have:

$x^4+x^2+1$ where $x=2^{2^{k-1}}$

we also know that:

$x^4+x^2+1 = (x^2+x+1)(x^2-x+1) = \color{Red}{(2^{2^{k}} + 2^{2^{k-1}} + 1)}(2^{2^{k}}-2^{2^{k-1}}+1)$

And my question is: Is it sufficient to show that polynomial $x^2-x+1$ is not divisible by $x^2+x+1$ so $(2^{2^{k}}-2^{2^{k-1}}+1)$ gives us at least one more prime number which implies that this statement is true for all natural numbers?

Answer 1

Assuming $2^{2^k}+2^{2^{k-1}}+1$ has at least $k$ different prime divisors,

for the inductive step it suffices to show that $2^{2^k}-2^{2^{k-1}}+1$ has a different prime factor.

Well, if $p$ divides $2^{2^k}+2^{2^{k-1}}+1$ and $2^{2^k}-2^{2^{k-1}}+1$, then $p$ divides their difference, which is a power of $2$,

so $p$ is a power of $2$, but $2^{2^k}+2^{2^{k-1}}+1$ and $2^{2^k}-2^{2^{k-1}}+1$ are not powers of $2$, so $p=1$.

Since $2^{2^k}-2^{2^{k-1}}+1>1$ for $k\gt 0$, it follows that $2^{2^k}-2^{2^{k-1}}+1$ has different prime factor(s)

from those of $2^{2^k}+2^{2^{k-1}}+1$, so the inductive step works.