Canonical Decomp 2^{27}+1: A Breakdown of the Equation's Components

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Dustinsfl
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[tex]2^{27}+1=(2^9)^3+1^3=(2^9+1)(2^{18}-2^9+1)=(2^3+1)(2^6-2^3+1)(2^{18}-2^9+1)[/tex]

Now what?
 
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Dustinsfl said:
[tex]2^{27}+1=(2^9)^3+1^3=(2^9+1)(2^{18}-2^9+1)=(2^3+1)(2^6-2^3+1)(2^{18}-2^9+1)[/tex]

Now what?

That would really depend a lot on what the question is. Wouldn't it?
 
Dick said:
That would really depend a lot on what the question is. Wouldn't it?

Canonical Decomp.
 
Dick said:
I give up. What's Canonical Decomp?


Canonical Decomp of a [itex]\mathbb{Z}^+[/itex] [itex]n[/itex] is of the form [itex]n=p_{1}^{a_1}*p_{2}^{a_2}\dots p_{k}^{a_k}[/itex], where [itex]p_1,\ p_2, \dots \ p_k[/itex] are distinct primes with [itex]p_1,< p_2, < \dots \ <p_k[/itex] and each exponent is a [itex]\mathbb{Z}^+[/itex]