<s>[INST] Product of its Prime Factors Given that $4095 = 8^4 - 1$ write $4095$ as a product of its prime factors.
I know how I could separate $4095$ into prime factors however I'm not sure how I could use $8^4 - 1$ to help me.
I could perhaps move the $1$ over to get $4096$ and then work from there...
 [/INST] To be more systematic:
\begin{align*}
x^{12}-1 &= (x^6-1)(x^6+1)\\
&=(x^3-1)(x^3+1)(x^2+1)(x^4-x^2+1)\\
&=(x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1)
\end{align*}
(As a polynomial it could be factored further).
Now set $x=2$.  You get
$$4095=1\cdot 7\cdot3 \cdot 3 \cdot 5 \cdot13=3^2\cdot5\cdot 7\cdot 13.$$
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