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Dr Sardonicus · 2020-10-07, 11:51

I considered the algebraic factorization of 2^m + 1. Write m = k*2^t where k is odd. Then we have (2^(2^t))^k + 1. If k > 1 this has 2^(2^t) + 1 as a proper factor. So I looked at

2^1 + 1 = 3, 2^2 + 1 = 5, 2^4 + 1 = 17, 2^6 + 1 (algebraic factor 2^2 + 1, cofactor 13 is prime), 2^12 + 1 (algebraic factor 2^4 + 1 which appears earlier on list; cofactor 241 is prime).

Then 2^24 + 1 = (2^8)^3 + 1.

Hmm. Algebraic factor, 2^8 + 1 = 257, prime, not a factor of any preceding 2^m + 1. I was pretty sure the cofactor would not divide the lcm, so I had a counterexample. It was then just a matter of working out the details.

2020-10-07, 11:51	#4
Dr Sardonicus Feb 2017 Nowhere 2⁴·271 Posts	I considered the algebraic factorization of 2^m + 1. Write m = k2^t where k is odd*. Then we have (2^(2^t))^k + 1. If k > 1 this has 2^(2^t) + 1 as a proper factor. So I looked at 2^1 + 1 = 3, 2^2 + 1 = 5, 2^4 + 1 = 17, 2^6 + 1 (algebraic factor 2^2 + 1, cofactor 13 is prime), 2^12 + 1 (algebraic factor 2^4 + 1 which appears earlier on list; cofactor 241 is prime). Then 2^24 + 1 = (2^8)^3 + 1. Hmm. Algebraic factor, 2^8 + 1 = 257, prime, not a factor of any preceding 2^m + 1. I was pretty sure the cofactor would not divide the lcm, so I had a counterexample. It was then just a matter of working out the details.