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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.

achille hui
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Arnold Karn
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    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. – Greg Martin Jun 18 '15 at 06:07

1 Answers1

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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 $10^{20}$.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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Dani Yabano
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  • The three numbers $2^{149^3}+149^{3^2}$, $2^{191^3}+191^{3^2}$, and $2^{29^5}+29^{5^2}$ are still small enough that they can be PRP-tested (i.e. tested if they are a Fermat probable prime, cf. Fermat's little theorem) with software such as PFGW. In the most likely case where they are composite, this will be definitively determined by that test. If they are prime, we will get an indication from PFGW, but a primality proof is not computationally possible for these numbers. I have also added them to factordb.com. – Jeppe Stig Nielsen Jul 31 '20 at 22:02
  • I tested the two smallest ones, 2^(149^3)+149^(3^2) is composite: RES64: [A2BD93E165A25719] respectively 2^(191^3)+191^(3^2) is composite: RES64: [83998DD74952A9AA]. The last one would take more time. The chance of success for such a large number is tiny. I am not doing it. – Jeppe Stig Nielsen Aug 11 '20 at 11:20
  • Anyway, $2^{29^5}+29^{5^2}$ has a relatively small factor so it is not needed to run a PRP test on this number (with over 6 million digits). So it was incorrect you claimed it had no factor up to $10^{20}$. – Jeppe Stig Nielsen May 21 '23 at 16:50