I noticed that my first post in this topic was mathematically not detailled enough and I forgot to mention some assumptions.

(Especially that p must always be prime)

For this reason I redefine my previous post:

WITH

is prime !!!

**A.) let n be prime with n=2^p-1:**
if

is prime then for every

and

the congruence is true.

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**B.) Let n be not prime with n=2^p-1 (But p is prime):**
B.1.) If you choose

when the above congruence is always true.

B.2.) If you choose

and

so that

when the congruence is also always true.

B.3) But for the other combinations of

and

the above congruence

is always false.

(A) (B.1) and (B.2) can easily be proven. But (B.3) is hard to prove.

**Therefore is (B.3) my conjecture.**
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Proof of A:

This is trivial, because A follows directly from the "Fermatsche Satz".

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proof of B.1:

is always true.

This is because 2 has always the order

in

with

prime.

-> for every

you have:

Because:

Now we look at the variable

:

(This is because of the factorisation of the form 2^{p}-1)

->

for every

you have now:

-->

Therefore:

when you choose

and

so that

then you have:

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proof of B.2:

If you choose

and

so that

then

is

with

.

->

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