$\displaystyle 2^{27}+1=(2^9)^3+1^3=(2^9+1)(2^{18}-2^9+1)$$\displaystyle =(2^3+1)(2^6-2^3+1)(2^{18}-2^9+1)$

What next?

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- Jun 18th 2010, 08:14 PMdwsmithCanonical Decomp
$\displaystyle 2^{27}+1=(2^9)^3+1^3=(2^9+1)(2^{18}-2^9+1)$$\displaystyle =(2^3+1)(2^6-2^3+1)(2^{18}-2^9+1)$

What next? - Jun 18th 2010, 08:27 PMroninpro
It looks like you can write $\displaystyle 2^3+1=(2+1)(2^2-2+1)$.

Is this what you are aiming for? - Jun 18th 2010, 08:29 PMdwsmith
No, this section:

$\displaystyle (2^6-2^3+1)(2^{18}-2^9+1)$ - Jun 18th 2010, 08:30 PMroninpro
As polynomials, you can't factor those anymore.

- Jun 18th 2010, 09:00 PMdwsmith
I have broken it down to $\displaystyle 3^3*19*(2^9*7*73+1)$ but it isn't fully decomp.

What to do next? - Jun 19th 2010, 12:01 AMOpalg
$\displaystyle 2^n$ is congruent to $\displaystyle \pm1$ mod 3, according as n is even or odd. So $\displaystyle 2^{18}-2^9+1$ is a multiple of 3. In fact it is equal to 3*87211, and 87211 is prime, as you can check here.