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Does there exist some odd positive integer $k$ such that, for all integers $n>0$, $2^n+k$ is never prime?
Extension:
If yes, for any $a$, does there always exist some $b$ such that $a^n+b$ is never prime?
If no, do there exist some coprime positive integers $a,b>1$, such that $a^n+b$ is never prime?
$k=78557$ does the job for the first problem.
Proof:
Let $k>0$, $n := 78557+2^k$.
Case 1: $k\equiv 0\pmod 2$. Then $3|n$.
Case 2: $k\equiv 3\pmod 4$. Then $5|n$.
Case 3: $k\equiv 1\pmod{12}$. Then $13|n$.
Case 4: $k\equiv 5\pmod{12}$. Then $7|n$.
Case 5: $k\equiv 9\pmod{36}$. Then $37|n$.
Case 6: $k\equiv 21\pmod{36}$. Then $19|n$.
Case 7: $k\equiv 33\pmod{36}$. Then $73|n$.
Since the distinction covers all possible cases, $n$ is always composite.
Regarding the second part: I believe that an existence of a $b$ such that $a^n+b$ is composite for all $n>0$ is very likely as expessions such as $a^{36}-1$ tend to have "enough" different prime factors for constructing a "covering set" as the set $\{3,5,7,13,19,37,73\}$ above. Then, using the Chinese Remainder Theorem, you should be able to construct a suitable $b$.
