How to demonstrate that $2^{2^n - 2} + 1$ is a nonprime number?

Question

This, considering $n ≥ 3$.

I have tried by induction; I suppose that it's true for all n less than or equal to k (and greater than or equal to 3), but then I stride when I go to prove for n = k + 1.

$2^{2^{k+1} - 2 } + 1 = $ $2^{2* 2^k - 2} + 1$

Answer 1

Induction isn’t needed. All you need is the fact that $a+b\mid a^n+b^n$ for odd positive integers $n$.

Observe that

$$2^{2^n-2}+1=4^{2^{n-1}-1}+1=4^{2^{n-1}-1}+1^{2^{n-1}-1}\;.$$

$2^{n-1}-1$ is odd for $n\ge 2$, so $4+1=5$ divides the sum of odd powers, and for $n\ge 3$ the expression is greater than $5$.

Answer 2

$2^n+1$ can only be prime, if $n$ is a power of $2$. If $p$ is an odd prime factor of $n$, $2^\frac{n}{p}+1$ is a proper divisor of $2^n+1$. It is clear that $2^n-2$ hast at least $1$ odd prime factor for $n\ge 3$.

Answer 3

Multiply by $4$ and look at things mod $5$:

$$4(2^{2^n-2}+1)=2^{2^n}+4=2^{4\cdot2^{n-2}}+4\equiv1+4\equiv0\mod5$$

so $5$ is a prime divisor of $2^{2^n-2}+1$. Note that for $n=2$, $2^{2^2-2}+1=5$, but for $n\ge3$, $2^{2^n-2}+1\gt5$, so $5$ is a proper divisor.