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Say a number $k$ is cool when the number $n^k+k^n$ is not a prime for all $n \geq 2$.

Question 1: Is there a classification of cool numbers?

For example $k=2$ or $k=15$ are not cool but $k=4$ is cool. Do you know some other cool numbers?

Note that $n^4+4^n=(n^2-2^{(n+1)/2}n+2^n)(n^2+2^{(n+1)/2}n+2^n)$ for odd $n$ and for even $n$ the number $n^4+4^n$ is even and thus $k=4$ is cool.

Question 2: Call $k$ eventually cool when $n^k+k^n$ is prime only for finitely many $n \geq 2$. Is there a classification of eventually cool numbers?

Do you know eventually cool numbers that are not cool?

Mare
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    It would help if you show the general factors showing that $k=4$ is cool. Maybe, this can be generalized. – Peter Oct 29 '20 at 10:50
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    @Peter Ok, I added it. – Mare Oct 29 '20 at 10:59
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    Honestly, it seems to be that kind of question without a hope to be solved.. – Paolo Leonetti Oct 29 '20 at 11:04
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    Is this a textbook exercise? a contest question? an online challenge? Or is this a question you've devised yourself? That is, is there any reason to believe the question has a reasonable solution? – Blue Oct 29 '20 at 11:06
  • I think, in some cases there is a prime but too large to be detected. – Peter Oct 29 '20 at 11:07
  • $k=6$ is for example a hard case. Or do I overlook a forced factorization ? – Peter Oct 29 '20 at 11:10
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    @Blue This is just a random question that came to my mind. I agree that a complete solution might be surprising, but partial answers are also welcome. – Mare Oct 29 '20 at 11:34
  • @Peter Yes, it seems $k=6$ is the first non-trivial case. – Mare Oct 29 '20 at 11:38
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    @OscarLanzi $24^5+5^{24}$ and probably $1036^5+5^{1036}$ are primes. – Tan Oct 29 '20 at 13:13
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    https://en.wikipedia.org/wiki/Leyland_number and various links there will be of interest. – Barry Cipra Oct 29 '20 at 14:33

2 Answers2

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Certainly all numbers of the form $4^{2m+1}$ are cool. When $n = 2h+1$ is odd render

$n^{4^{2m+1}}+(4^{2m+1})^n=a^4+4b^4$

$a=n^{4^{2m}}$

$b=2^{h(2m+1) + m}$

And apply the (Sophie Germain) factorization noted in the question for the specific case $k=4$.

Oscar Lanzi
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If $GCD(n,k) = g \ne 1$, $n^k + k^n$ is never prime because

$$n^k + k^n = g\bigg({n^k + k^n \over g}\bigg)$$

vvg
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