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Old 2005-12-30, 21:01   #3
nibble4bits
 
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Nov 2005

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At 1 and 0 you get the identies 1 and 0. 0 can't be divided at all and 1 has only one possible factor including itself and itself. Hehe we'll just say "period" to make more sense.
At 2 total factors there's only primes. (works both ways: A->B and B->A)
The solutions for 2 factors are a kind of 'key' to the higher-factor-count sets in the 3D tree since obviously primes are the simplest factors possible.
If there's an infinite number of primes, is there an infinite number of the 3-factor results? 4-factor? All? See 2nd post to see why there must be.
Yes, there are infinite primes but no telling how long you'll have to wait to find the next one! I left this in the spoiler so those who want to can do the work themselves to find and understand the proof.
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