<s>[INST] Factor $10^6-1$ completely I know kind of a very elementary method to factor this number. Consider the following:
$$10^6-1 = (10^3-1)(10^3+1)=9 \times 11 \times (10^2+10+1)(10^2-10+1) = 9 \times 11 \times 111\times 91$$
I would then factor each number individually.
Is there a faster method? The great hint is that this number is a rep unit number = $9\ \times 111111$.
 [/INST] \begin{align}
10^6 - 1 &= 1000000-1\\
&= 999999 \\
&= 3^2 \times 111111 \\
&= 3^2 \times 111\times 1001 \\
&= 3^2 \times 111 \times (1100 - 99)\\
&= 3^2 \times 111 \times 11 \times (100-9)\\
&= 3^2 \times 111 \times 11 \times 91 \\
&= 3^2 \times (3 \times 37)\times 11 \times 7 \times 13
\end{align}
</s>