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Old 2020-10-10, 09:23   #1
RMLabrador
 
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"Roman V. Makarchuk"
Aug 2020
Ukraine

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Default In honor of my old friend, Gottfried Wilhelm Leibniz, or Monadic factorisation of 2^n+1

2^n+1 rewrite as x^n+1 (1), let k*a^n+1/0/-1, a=3 so-called Monada for all n in(1), and for example if x=2, n=32, numeric value of (1) factored as
2^32+1=(a^6-a^4-a^2+a-1)*(a^14+a^13+a^12-a^11-a^10+a^9+a^8+a^7-a^6+a^5-a^4+a^3-a^2-a+1)
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Old 2020-10-10, 09:30   #2
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a^6-a^4-a^2+a-1=641

use Monade)))

(641+1)/3=214
(214-1)/3=71
(71+1)/3=24
24/3=8
(8+1)/3=3
3=a so (((a^2-1)*a*a-1)*a+1)*a-1 - monadic polinomial, and if we do the same for the second part of 2^32+1=641*6700417 we obtain second polynomial

As you can understand, everyone of factor 2^n+1 is divisible by k*3+1/0/-1 (2) till the very end and this is why i named (2) Monade. Good name, i like it!

Last fiddled with by RMLabrador on 2020-10-10 at 09:34 Reason: grammatic error
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Old 2020-10-10, 09:38   #3
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And of course, in the Mother Nature exist an algorithm for get this polynomial not from factors, but direct from value of 2^n+1. I have the wireld one, and now going to learn C++))
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