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Old 2020-02-15, 20:06   #6
carpetpool
 
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"Sam"
Nov 2016

2·3·53 Posts
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Quote:
Originally Posted by carpetpool View Post
Yes, it should be false in general (Roman B. Popovych's conjecture could be true however).

Assuming the falsehood of the original conjecture, what I was trying to illustrate is that counterexamples should have very specific properties, namely that the only counterexamples (with r prime) should be Carmichael numbers of order(m) where m = (r-1)/2. There is no proof (that I know of), but there is evidence that this should be true.

There should also exist upper bounds on either the Carmichael numbers of order(m) < x (and show that it is 0 for some x), or there is a lower bound such that the smallest Carmichael numbers of order(m) > L.

If these two arguments are true, then there should exist a modified version of the conjecture that is true as we will have conditions for which n must be prime.
I can't seem to get away with it

(x - 1)^(t*n + n) = x^n - 1 mod(n, x^r - 1) where t = 2*r*(n^((r - 1)/2) - 1) if jacobi(n | r) = -1
or
(x - 1)^(t*n + n) = x^n - 1 mod(n, x^r - 1) where t = (n^((r - 1)/2) - 1) if jacobi(n | r) = 1
if
n^2 ≠ 1 mod r and r > 5.

As there are infinitely many values of w such that (x - 1)^w = x^n - 1 mod(n, x^r - 1), proving that w = t*n + n is the smallest such integer (apart from n itself) is false:

n = 61, r = 7

So now I seem to have dug a deeper hole into solving this problem. We need a lower bound B such that
(x - 1)^w ≠ x^n - 1 mod(n, x^r - 1)

if w is any integer (n < w < B).
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