Some one know something about the prime numbers of the form $p = 2^k + 2^n +1$,
I would like to know if is it possible give conditions to $n$ and $k$ for $p$ will be a prime.



primes of the form $2^n + 2^k +1$
Hello, 


For each odd prime $q$, there should be conditions on $n,k$ mod (the order of $2$ mod $q$) that are necessary conditions for $q$ to not divide $2^n+2^k+1$. For $q=3$, the idea is implied in @Charles' comment on @joro's answer to [this question](https://mathoverflow.net/questions/97288/primes-with-more-ones-than-zeroes-in-their-binary-expansion). 


For example, for $q=3$, $n,k$ should not both be even.
For $q=5$, $n,k$ should not both be $1 \bmod 4$, and it should not be the case that one of $n,k$ is $3\bmod 4$ while the other is $0 \bmod 4$.
For $q=7$, it should not be the case that one of $n,k$ is $1 \bmod 3$ while the other is $2 \bmod 3$.
For $q=11$, it should not be the case that one of $n,k$ is $1 \bmod 10$ while the other is $3 \bmod 10$, and it should not be the case that one of $n,k$ is $6 \bmod 10$ while the other is $0 \bmod 10$, and it should not be the case that one of $n,k$ is $7 \bmod 10$ while the other is $8 \bmod 10$, and it should not be the case that one of $n,k$ is $2 \bmod 10$ while the other is $9 \bmod 10$, and it should not be the case that $n,k$ are both $4 \bmod 10$.


In general, if [Artin's conjecture on primitive roots](http://en.wikipedia.org/wiki/Artin's_conjecture_on_primitive_roots) is true, there are infinitely many necessary conditions; one gets necessary conditions for each such $q$ where $2^{(q-1)/2} \equiv -1 \bmod q$, because this translates to $q-2$ values that $(n,k)$ cannot take in $(\mathbb{Z}/(q-1)\mathbb{Z})^2$, one of which is the condition that $n,k$ should not both be $(q-3)/2 \bmod q-1$.


