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Consider the number $(2m)^{2^n}+1$ where both $m$ and $n$ are positive integers.

Can it be shown that for any given $n$, there exists an $m$ such that $(2m)^{2^n}+1$ is a prime number.

Edit: It can be shown that there exists an integer k such that (2m)2n can be written k2 for any integers m and n. If the conjecture in the question was true, then it would thus imply there are an infinite number of primes of the form k2+1. This is Landau's fourth problem which I believe still remains unsolved. Therefore I suspect this conjecture also cannot be proved at the current time

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    $\begingroup$ No, he is asking a very important question here. Presumably you have been asked to show this as part of something else, and we need to know what that something else is to know how to answer it properly. $\endgroup$ Aug 10, 2018 at 14:36
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    $\begingroup$ Please provide some context; where did you come across this problem, and what are your thoughts on it? What have you tried and where did you get stuck? $\endgroup$
    – Servaes
    Aug 10, 2018 at 14:39
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    $\begingroup$ An interesting problem (+1) , but a solution is probably out of reach. $\endgroup$
    – Peter
    Aug 11, 2018 at 9:04

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Showing that for every positive integer $n$ there is a positive integer $m$ , such that $(2m)^{(2^n)}+1$ is prime, would not just solve the $k^2+1$-problem. It would also show that there infinite many primes of the form $k^4+1,k^8+1,k^{16}+1$ and so on.

So, a positive answer is very unlikely. But a negative answer is unlikely as well considering that it is not known whether infinite many Fermat-primes exist.

So, this question will probably remain undecided.

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