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.
 A: Note that
$$
5^{10} - 1 = (5^5+1)(5^5-1)
$$
And that
$$
5^5-1 = (5 - 1)(5^4 + 5^3 + 5^2 + 5 + 1)\\
5^5+1 = (5+1)(5^4 - 5^3 + 5^2 - 5 + 1)
$$
That should give you a fairly good start. We cannot factor this further using formal factorization alone.
A: 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 $0$. 
A: You could use the fact that $x^n-1=(x-1)(1+x+x^2+x^3+...+x^{n-1})$ 
so here $5^{10}-1=(5-1)(1+5+5^2+5^3+...+5^9)$
A: Not sure what kind of factorization or congruence information your are looking for, the question isn't very clear. But here is an easy congruence observation about your number:
We know that 
$$
\sum_{k=0}^9 5^k=\frac{5^{10}-1}{5-1}=\frac{5^{10}-1}{4},
$$
thus
$$
4\cdot\sum_{k=0}^9 5^k=5^{10}-1.
$$
It is clear that $\sum_{k=0}^9 5^k\equiv 1\pmod 5$ and so we conclude that $4\sum_{k=0}^9 5^k\equiv 4\pmod 5$.
You could also use a nested parenthesis "factorization" based on the above and get that 
$$
4(5(5(5(5(5(5(5(5(5+1)+1)+1)+1)+1)+1)+1)+1)+1)+1).
$$
