Is there a prime which is the form of $10^n + 1$ except $2, 11, 101$?

Question

I confirm there isn't such prime for $n < 64$.

We know if $n$ has an odd prime factor, the result is composite. So the only possibilities are for $n$ a power of two. These grow rapidly enough that heuristic arguments suggest an "expected" number of these primes will be small. — hardmath, Jan 22 '17 at 00:24
These are also a special case of generalized Fermat primes and you can find a little more information about them under that name. — Steven Stadnicki, Jan 22 '17 at 00:25
That $10^{2^{10^{10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}}}}}+1$ is not prime is almost certain but it's likely to be unprovable. — Count Iblis, Jan 22 '17 at 01:03

Ng Chung Tak · Accepted Answer · 2017-01-22T01:31:43.233

5

Note that $(10^{n}+1)|10^{n(2k+1)}+1$ for $n$, $k\in \mathbb{N}$.

According to the Table $\boldsymbol{1}$ (page $24$ or $\frac{30}{55}$ of the pdf) from this journal:

$10^{2^{n}}+1$ has no (known?) prime factors for $n=13$, $14$, $21$, $23$, $24$, $25$, $\ldots $

That means $10^m+1$ is composite continually for $3\le m \le 8195$.

edited Jan 22 '17 at 01:31

answered Jan 22 '17 at 00:56

Ng Chung Tak

@TransfiniteNumbers : n = 2 :) since $10^4+1=73\times 37$ – Adren Nov 15 '19 at 11:10
@Adren Do you mean 10^4+1 = 73 x 137? – Nov 15 '19 at 19:10
@TransfiniteNumbers ooops , yes of course – Adren Nov 16 '19 at 06:43

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