Q: Is number of the form $$\displaystyle 2^{M^N}+M^{N^2}$$ always composite for $M,N$ odd primes?

I observed that:

  • If $M=N$ then this number is absolutely a composite, because it satisfies the identity $a+b \mid a^k+b^k$ if k is odd numbers.

  • If $M \equiv 1 \pmod 3$ then this number is always composite, because it divisible by $3$.

    I see that $2^{2c+1} \pmod 3=2$ for all natural numbers $c$, and $p^t \pmod 3=1$ if $p$ is a prime of type $1 \pmod 3$, then it is clear that $2^{M^N}+M^{N^2}$ divisible by $3$ if $M$ prime number of type $1 \pmod 3$.

  • I have checked it exhaustively, and I always found it composites.

  • $\begingroup$ It's possible that it's always composite for no particular reason. The numbers grow so fast (and large numbers are prime less often) that the expected number of primes is finite; maybe there just aren't any. $\endgroup$ – Greg Martin Jun 18 '15 at 6:07

There is no reason there are no primes of the form $2^{M^N}+M^{N^2}$ with $M,N$ odd primes, because what about if $M\equiv2\, (\text{mod}\, 3)$? I have checked numbers of that form, and I found first $3$ numbers with unknown status, and they are $2^{149^3}+149^{3^2}$, $2^{191^3}+191^{3^2}$, and $2^{29^5}+29^{5^2}$. These numbers have no prime factors up to $10^{20}$.


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