I was just playing with these numbers and it seems to me that the numbers of the form $10^n+1$, where $n>2$ are composite. I can prove that $10^n+1$ can't be prime unless $n$ is a power of $2$, but I do not know for which values of $n$, these numbers are prime, except for $n=2$. Is it an open problem or is there any result on this? Thank you in advance.

  • $\begingroup$ It's still ambiguous in my opinion. I have changed the wording. $\endgroup$
    – TonyK
    Mar 14, 2014 at 11:09

2 Answers 2


$10^{2^n}+1$ is a generalized Fermat prime $F_{n}(10)$. Landau's fourth problem asks if there are infinitely many generalized Fermat primes $F_n(a)$:

As of 2013, all four problems [including yours] are unresolved.

Here are some more numbers, till $n=20$.


Updated to add: Due to a typo in the original question, it turns out that this proof is already known to the OP. Never mind, I'll leave it here anyway.

We can write $n=2^km$ where $m$ is odd. Then $10^n \equiv (-1)^m \equiv -1$ mod $(10^{2^k}+1)$, so $10^n+1 \equiv 0$ mod $(10^{2^k}+1)$. Thus $10^n+1$ is composite unless $n=2^k$, i.e. $n$ is a power of $2$.

For instance:

  • if $n$ is odd, then $10^n+1 \equiv (-1)^n+1 \equiv 0$ mod $11$
  • if $n \equiv 2$ mod $4$, put $n=4m+2$, then $10^n+1 \equiv 100^{2m+1}+1 \equiv (-1)^{2m+1}+1 \equiv 0$ mod $101$
  • similarly, if $n \equiv 4$ mod $8$, then $10^n+1 \equiv 0$ mod $10001$
  • if $n \equiv 8$ mod $16$, then $10^n+1 \equiv 0$ mod $100000001$

and so on.

If $n$ is a power of $2$, you claim that you can prove that $10^n+1$ is prime, but that's not right. In fact $10^n+1$ is prime for $k=1$ (with $10^n+1 = 101$) and composite for $k=0,2,3,4,5$. I can see no reason to rule out prime values for higher values of $k$.

  • $\begingroup$ I wanted to say that $10^n+1$ being prime implies that $n$ is prime, but mistakenly wrote the opposite. @Hurkyl edited that part to only if. $\endgroup$ Mar 14, 2014 at 11:08

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