Prove that $\frac{ 5^{125}-1}{ 5^{25}-1}$ is a composite number Prove that   $\dfrac {\left( 5^{125}-1\right)}{\left( 5^{25}-1\right)}$ is composite number using number theory. Do not use calculator or Wolfram alpha or anything like that.
 A: Hint $\ $ The factorization arises by applying a variation of cyclotomic factorization known as Aurifeuillian factorization (aka Aurifeuillean). For the OP we can employ the following one
$$\begin{align}\frac{(5x^2)^5-1}{5x^2-1} =\, (25x^4\!+15x^2+1)^2 - (5x(5x^2\!+1))^2\\[4pt]
\overset{\large x\, =\, 5^{\large 12}}\Longrightarrow\ \ \frac{5^{125}-1}{5^{25}-1}\, =\, (5^{50}+3\cdot 5^{25}+1)^2 - (5^{13}(5^{25}+1))^2\end{align}$$
Remark $\ $ There are many known Aurifeuillian factorizations (e.g. see below for a few more). For more see this answer and see Aurifeuillian Factorization by A. Granville and P. Pleasants.
$$\begin{align} 
\frac{(3x^2)^3+1}{3x^2+1} &=\, (3x^2\!+1)^2-(3x)^2\\[4pt]
\frac{(5x^2)^5-1}{5x^2-1} &=\, (25x^4\!+15x^2+1)^2 - (5x)^2(5x^2\!+1)^2\\[4pt]
\frac{(7x^2)^7+1}{7x^2+1} &=\, (7x^2\!+1)^6-(7x)^2(49 x^4\!+7x^2+1)^2
\end{align}$$
A: This is problem $87$ of Putnam and Beyond. Here is the solution:
$\dfrac{5^{125}-1}{5^{25}-1}=1+a+a^2+a^3+a^4$ where $a=5^{25}$.
We have $1+a+a^2+a^3+a^4+a^5=(a^2+3a+1)^2-(5^{13}(a+1))^2=(a^2+3a+1+5^{13}(a+1))(a^2+3a+1-5^{13}(a+1))$
The reason the second factor is larger than $1$ is $a^2=5^{50}>5^{39}>5^{13}(a+1)$

This problem was also part of the 1992 imo shortlist and was proposed by Korea, here is a link with the same solution but shorter: https://mks.mff.cuni.cz/kalva/short/soln/sh9216.html
A: HINT:
$x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
and $5^{25}=(5^5)^5$, whilst $5^{125}=((5^5)^5)^5$
Does this get you anywhere?
A: Put $x = 525$. Then $\frac{5^{125}-1}{5^{25}-1} = x^4+x^3+x^2+x+1 = (x^2+3^x+1)^2 - 5x(x+1)^2$ which has factor $(x^2+3x+1-513x-513)$ which is obviously $> 1$ and $< x^4+x^3+x^2+x+1$.
