What's the easiest way to factor $5^{10} - 1$? What's the easiest way to factor $5^{10} - 1$?
I believe $5 - 1$ is a factor based off the binomial theorem. 
From there I do not know. 
We are using congruence's in this class.

Consider $x^{10} - 1.$
The first trivial factorization is $(x^5-1)(x^5+1)$.
Note that $1+x+x^2+x^3+x^4 = \frac{x^5-1}{x-1}$ by the geometric series formula. 
Secondly, replace $x$ with $-x$ to see that $1-x+x^2-x^3+x^4 = \frac{x^5+1}{x+1}.$
$$x^{10} - 1 = (1+x+x^2+x^3+x^4)(1-x+x^2-x^3+x^4)(x-1)(x+1).$$
It should be easy to factor $5^{10} - 1$ into $4\times 6\times 781 \times 521.$
Another trick is to see that $11$ clearly divides $781$ because $7+1 = 8$ (cf. multiplying by $11$ shortcuts).
Simple division gives us $$\boxed{5^{10} - 1 = 2^3 \times 3 \times 11 \times 71 \times 521}.$$
The only thing left is to show that $521$ is prime.
All we have to check is primes less than $\sqrt{521}$ because if there are factors other than $1$ and $521$ it must be less than (or equal to) $\sqrt{521}$. (Note that $23^2 = 529$).
Clearly $2,3,5$ and $ 11$ do not work so now we must check $7, 13, 17$ and $19$.
$7$ does not work because if $7$ divided $521$ it would divide $521-21 = 500$.
$13$ does not work because $521-26 = 495 = 99\times5$. 
$17$ does not work because $521-17*3 = 470 = 47\times10$. 
$19$ does not work because $521-19*9 = 350 = 35\times10$.
I found these terms to subtract by trying to eliminate the $1$ at the end of $571$ by turning it into a $5$ or a $0$. 
