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.
 A: $$a^n - b^n = (a-b) ( a^{n-1} + a^{n-2} b + \cdots + b^{n-1})$$
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.
A: $$3^{140}-1=(3^{70}+1)(3^{70}-1)=(3^{70}+1)(3^{35}+1)(3^{35}-1)$$
and from $a-b\mid a^7-b^7$, the number
$ 3^{35}\pm1$ is a multiple of $3^5\pm1$. These are of feasible size and give us $2, 11, 61$ as prime factors. Also, $3^{35}\pm1$ is a multiple of $3^7\pm1$, which are still of manageable size and give us $547$ and $1093$.
Alternatively, note that $3^{p-1}\equiv 1\pmod p$ for $p\ne 3$, hence looking for small primes with $p-1\mid 140$ is helpful. This way you find $p=2, 5, 11, 29, 71$ very easily.
A: He asks for primes. So any prime $p\neq 3$ for which $p-1$ divides $140$ is such a prime.
This includes $2,5,11,29,71$. The are not the only prime factors, but they are the first five that we can find.
For example, $4^3\equiv 3\pmod {61}$, so $3^{20}=4^{60}\equiv 1\pmod {61}$. So $61$ is another prime factor.
A: Hint:  you have a difference of squares, a difference of fifth powers, and a difference of seventh powers.
