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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...

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4  
$x^4-1=(x^2)^2-1^2$. – mathlove Mar 25 at 18:43
up vote 8 down vote accepted

$$8^4-1=(8^2+1)(8^2-1)=5\cdot13\cdot7\cdot3^2$$

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I understand that you have worked out the difference of two squares however how did you conclude with $5, 13, 7, 3^2$ – PerfectNutter Mar 25 at 18:46
4  
@PerfectNutter These are small numbers: $\;8^2+1=65=5\cdot13\;,\;\;8^2-1=63=7\cdot9=7\cdot3^2\;$, etc. – Joanpemo Mar 25 at 18:47

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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