What's the easiest way to factor $5^{10} - 1$?

Question

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.

Answer 1

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 $0$.

Answer 2

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.

Answer 3

You could use the fact that $x^n-1=(x-1)(1+x+x^2+x^3+...+x^{n-1})$

so here $5^{10}-1=(5-1)(1+5+5^2+5^3+...+5^9)$

Answer 4

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 $$ 4(5(5(5(5(5(5(5(5(5+1)+1)+1)+1)+1)+1)+1)+1)+1)+1). $$