# 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.

• Hint: Note that $5x=5^{26}=(5^{13})^2$. – Colescu Jun 8 '17 at 14:42
• You're on the right Aurifeuillean track. Compare with this – Jyrki Lahtonen Jun 8 '17 at 14:50
• gross. it has 5 HUGE prime factors – Saketh Malyala Jun 8 '17 at 15:06
• @JaideepKhare Are you sure $9x^2+2x^2-10x^2=x^2$? – kingW3 Jun 8 '17 at 15:50
• en.wikipedia.org/wiki/Aurifeuillean_factorization , see row $b=5$. – Jack D'Aurizio Jun 8 '17 at 16:52

## 1 Answer

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.