Find all positive integer $k$ such that $k^k+1$ is a prime.
I only found $k=1,2,4$ where the values are $2,5,257$.
But I cannot prove why others will not work. Pls help tks.
PS: It is one of my exam qns which it ask for $k^k+1$ is prime less than $10^{20}$
First of all, $k$ must be the form $k=2^{2^m},m\in \mathbb{N}$. If you want to find all primes being the form $k^k+1$ less than $10^{20}$, then $$2^{2^m\cdot2^{2^m}}+1 \leq 10^{20}$$ Hence, $$2^{2^{2^m+m}}\leq 10^{20}$$ $$2^m+m\leq \log_2\log_2(10^{20})<7$$ Finally, we can try $m=0,1,2$ corresponding to $k=2,4,16$. When $k=2,4$,$k^k+1$ is a prime. When $k=16$, note that this is a Fermat number $F_6$ which is not a prime.