<s>[INST] Can $11^{13}-1$ be divided exactly by 6? Can $11^{13}-1$ be divided exactly by 6?
My solution:
$$11^2 \equiv 1 \pmod 6$$
$$11^{12} \equiv 1 \pmod 6$$
$$11^{13} \equiv 5 \pmod 6 $$
Hence, $(11^{13}-\mathbf{5})$ can be divided exactly by 6. However, according to the solution on my book, （$11^{13}-\mathbf{1}$）can be divide exactly by 6. What's wrong?
 [/INST] It is already not divisible by $ 3 $; notice that
\begin{align}
11^{13} - 1 &\equiv (-1)^{13} - 1 \\
            &\equiv -1 - 1 \\
            &\equiv -2 \\
            &\equiv 1 \\
            &\not\equiv 0 \, (\text{mod} \, 3).
\end{align}
Note $ 11 \equiv 2 \equiv -1 \, (\text{mod} \, 3) $.
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