Prove that $n=\frac{5^{125}-1}{5^{25}-1}$ is a composite number 
Prove that $n=\dfrac{5^{125}-1}{5^{25}-1}$ is a composite number

My attempt,
Let $x=5^{25}$, so that $5^{125}-1=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
$=(x^4+9x^2+1+6x^3+6x+2x^2-5x^3-10x^2-5x)(x-1)$
$=((x^2+3x+1)^2-5x(x+1)^2)(x-1)$
I'm stuck at this point and don't know how to continue anymore. Hope someone can provide a detailed solution. Thanks a lot.

The polynomial $\Phi_5(x)=x^4+x^3+x^2+x+1$ fulfills an interesting identity.
We have that $4\cdot \Phi_5(x)$ is pretty close to the square of $2x^2+x+2$, and indeed:
$$ 4 \Phi_5(x) = (2x^2+x+2)^2 - 5x^2 \tag{1}$$
as well as:
$$ \Phi_5(x) = (x^2+3x+1)^2 - 5x(x+1)^2 \tag{2} $$
so if $x=5^{2k+1}$, $\Phi_5(x)$ is the difference of two large squares:
$$\begin{eqnarray*} \Phi_5(5^{2k+1}) &=& \left(5^{4k+2}+3\cdot 5^{2k+1}+1\right)^2 - \left(5^{3k+2}+5^{k+1}\right)^2\\&=&\left(5^{4k+2}+5^{3k+2}+3\cdot 5^{2k+1}+5^{k+1}+1\right)\cdot\left(5^{4k+2}-5^{3k+2}+3\cdot 5^{2k+1}-5^{k+1}+1\right) \end{eqnarray*}$$
and $\Phi_5(5^{2k+1})$ cannot be a prime number. See Aurifeuillean factorization.
