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Is $10^n+1$ composite for all $n\in \mathbb{N}$ greater then $2$?

I tried many values of $n$, and $10^n+1$ is composite each time (excpet $n=1,2$).

Is my conjecture correct? Thank you.

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    For odd values of $n$, $11|10^n+1$.2012-12-15

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If $n = 2^l m$ where m is odd, then $\displaystyle (10^{2^l})^m + 1 \equiv 0 \bmod (10^{2^l} + 1)$.

So the interesting question is if $10^n + 1$ is composite when $n$ is a power of $2$.

Unfortunately I don't know what happens when $n$ is a power of $2$.

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    @DanBrumleve: the fact that the series converges predicts that there are only finitely many such primes. The fact that the series (starting at x=11) is very small suggests that there are none.2012-12-15