<s>[INST] Find five prime factors of $3^{140 }- 1$ I tried to simplify $3^{140}$ but I couldn't go past $81^{35}$, any help would be greatly appreciated.
 [/INST] $$a^n - b^n = (a-b) ( a^{n-1} + a^{n-2} b + \cdots + b^{n-1})$$
Using $n = 140$, we get that $3 - 1$ is a factor, which gives $2$ as a factor.
Using $n = 35$, we get that $3^4 - 1$ is a factor, which gives $5$ as a factor.
Using $n = 28$, we get that $3^5 - 1$ is a factor, which gives $11$ as a factor.
Using $n = 20$, we get that $3^7-1$ is a factor, which gives $1093$ as a factor.
Using $n = 14$, we get that $3^{10} -1 $ is a factor, which gives $61$ as a factor.
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