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.
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.
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 $2$.