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Does anyone know whether there are only finitely many of primes of the form $6^{2n}+1$, where $n$ zero or any natural number?

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Any prime you pick that is of the form $6^{2n}+1$ will be finite, unless you're using definitions other than the most conventional ones. Maybe you meant to ask whether there are only finitely many of them? If so, that's the right way to say it. This is actually a fairly frequently occurring way of misunderstanding the terminology. – Michael Hardy Dec 30 '11 at 5:01
Thanks, Michael, for your suggestion. – Vassilis Parassidis Dec 30 '11 at 5:04
Using the admittedly very bad heuristic assumption that $6^{2n}+1$ is randomly distributed, I would conjecture so on the basis of the Prime Number Theorem: the probability that $6^{2n}+1$ is prime is roughly $1/(\log 6^{2n})\sim 1/n$, and the harmonic series diverges giving an infinite expected value for the number of such primes. – Alex Becker Dec 30 '11 at 5:09
After checking the values with wolfram alpha, it seems that it begins with a few primes but then each of the terms seem to be divisible by some previous term in the sequence... I would conjecture there are only finitely many of them, probably because of some simple algebra trick I can't quite think of. – Pedro Dec 30 '11 at 5:17
@AlexBecker Maybe I am misinterpreting your argument, but doesn't that same heurestic tell us that there are infinitely many primes of the form $6^{2n}$? – Alex Youcis Dec 30 '11 at 5:20

I'm pretty sure the in-text question is "no," which means the title question is pretty tough to answer cleanly. But just for the sake of mentioning it, it's easy to check that $n$ would have to be a power of 2 for this quantity to prime: Indeed, if we re-write it as $36^n+1$, then the argument is verbatim the same as for Fermat primes: If $n$ were not a power of two, it would admit a proper factorization $n=ab$ with (say) $b$ odd. Then $36^a+1\mid 36^{ab}+1$, contradicting primeness.

So now you're looking for primes of the form $36^{2^k}+1$, a problem which doesn't, at least to me, appear any more tractable than the (wide open) analogous problem with base 2, the problem of finding the Fermat primes (taking the form $2^{2^k}+1$). In fact, there is some literature on the topic -- mostly computational, as primes of this form are rare but algorithmically interesting -- going under the eminently-Googlable name "generalized Fermat number."

Finally, let me just note that via direct computation (SAGE), $36^2+1$ is prime, and $36^{2^k}+1$ is composited for $2\leq k\leq 6$.

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Checked $36^{2^2}+1$ to $36^{2^{18}}+1$ are all composite using OpenPFGW. – Douglas S. Stones Nov 16 '12 at 4:36

It is not known whether there are infinitely many primes of the form $a^2 + 1.$ Your numbers are the subset with $a = 6^n.$ So it is not known that your set is infinite. The strongest results are in two variables, the well-known result that infinitely many primes are of the form $x^2 + y^2$ (indeed all primes $p \equiv 1 \pmod 4$) and the result of Iwaniec and Friedlander that there are infinitely many primes of the form $x^2 + y^4.$

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This is the fourth of Landau's problems. – Jacopo Notarstefano Dec 30 '11 at 5:56
Landau was smart. – Will Jagy Dec 30 '11 at 6:11
If it is unknown whether there are finitely many or infinitely many primes of the form $a^2 +1$, how does it follow that this is also not known for a subset? It could well be that the subset is better understood and it were known that there are only finitely many primes in it. – user20266 Dec 30 '11 at 11:10
We don't know if there are infinitely many primes of the form $6^{2n}+1$ (since if there were, we would know that there are infinitely many of the form $a^2+1$); but it's [logically] possible to know that there are only finitely many primes of the form $6^{2n}+1$ without knowing if there are finitely or infinitely many of the more general form. – Arturo Magidin Dec 30 '11 at 18:26
@Thomas and Arturo, Cam had already answered and i was not careful about the consequences of the anecdote. – Will Jagy Dec 30 '11 at 21:07

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