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20201010, 09:23  #1 
"Roman V. Makarchuk"
Aug 2020
Ukraine
2×17 Posts 
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 socalled Monada for all n in(1), and for example if x=2, n=32, numeric value of (1) factored as
2^32+1=(a^6a^4a^2+a1)*(a^14+a^13+a^12a^11a^10+a^9+a^8+a^7a^6+a^5a^4+a^3a^2a+1) 
20201010, 09:30  #2 
"Roman V. Makarchuk"
Aug 2020
Ukraine
34_{10} Posts 
a^6a^4a^2+a1=641
use Monade))) (641+1)/3=214 (2141)/3=71 (71+1)/3=24 24/3=8 (8+1)/3=3 3=a so (((a^21)*a*a1)*a+1)*a1  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 20201010 at 09:34 Reason: grammatic error 
20201010, 09:38  #3 
"Roman V. Makarchuk"
Aug 2020
Ukraine
42_{8} Posts 
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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