# Prove that $\frac{(5^{10n})^5-1}{5^{10n}-1}$ is composite for all $n \in \mathbb{N}$

I am aware that by letting $x=5^{10n}$, one gets $\frac{x^5-1}{x-1}$, and that this is equal to the polynomial $x^4+x^3+x^2+x+1$, which is an integer. My current thinking is to somehow show that if $x-1$ has $k$ factors, $x^5-1$ has more than $k+1$ factors.

(If the opposite is true, then the fraction is prime, as the only way to get a prime number from a fraction of two non-primes is for the fraction to be of the form $\frac{jP}{j}$ for some non-prime $j$ and some prime $P$. )

Looking at the table in this link http://mersennewiki.org/index.php/5_Minus_Tables verifies this to be true by comparing $5^{50}-1$ and $5^{10}-1$, for example, but this does not amount to a proof.

I also note that I have read about cyclotomic/Aureifeuillian factorisation, with particular reference to this problem: Prove that $\frac{ 5^{125}-1}{ 5^{25}-1}$ is a composite number, though I do not believe this to be applicable for this problem.

I am also happy to receive an explanation about why this is not solvable, if it happens to be the case.

• – Jack D'Aurizio Sep 13 '16 at 12:52
• Reverse your exponents and factor by difference of squares on top and bottom. Then you can do your cyclotomic thing to both factors and that should give you a nice factorization. – B. Goddard Sep 13 '16 at 12:55
• @JackD'Aurizio The main problem I'm having with the Aurifeullian factorisation is if I let $n=1$, $5^{50}-1=(5^{10}-1)f(5^{10})$, then I need to solve for $5^{10}=5x^2$, as this is the required form of the argument of the cyclotomic polynomial, which comes to $5^{\frac{9}{2}}=x$. Am I doing something wrong? – user345795 Sep 13 '16 at 13:24
• Almost all these numbers seem to be multiples of $101 \times 251$. – Bob Happ Sep 13 '16 at 21:33

Here's what B. Goddard is saying in the comments:

$$\frac{(5^{10n})^5-1}{5^{10n}-1}=\frac{(5^{5n})^{10}-1}{(5^n)^{10}-1}=\frac{(5^{5n})^5-1}{(5^n)^5-1}\cdot\frac{(5^{5n})^5+1}{(5^n)^5+1}.$$

This is a product of two integers $>1$.