Bonjour,
Depuis l'année dernière je n'arrive pas à prouver une conjecture reliant l'indicatrice d'Euler et les nombres premiers.
Publiée sur math stack exchange mais je n'ai pas eu de réponses bien convaincantes à part la forme que devrait avoir un contre-exemple pour invalider la conjecture. J'ai laissé le texte de base en anglais pour ne pas dénaturer l'essai de démonstration de la personne citée en référence.
Je viens vous demander de l'aide pour savoir comment démontrer ceci car ça a l'air bien complexe et je n'ai compris que le début de la démonstration ci-dessous.
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Conjecture
ϕ denotes the Euler's totient function, a denotes a natural number > 1 and n denotes a natural number such that n = 4k (k an integer ≥ 1).
If (an − 2) + 1 ≡ n − 1 (mod n) then ϕ(an − 2) + 1 is always a prime number.
Example with a = 119 and n = 20: ϕ(11920 − 2) + 1 ≡ 19 (mod 20) then ϕ(11920 − 2) + 1 is always prime.
Investigations
Max Alekseyev studied this conjecture but no proof has been found [1].
A counterexample would have to satisfy the equation:
an = c · pk
with c ∈ {1, 2}, a prime p ≡ 3 (mod 4), and an integer k > 1. Below I will show that k cannot be even and also cannot be a multiple of 3, implying that k ≥ 5 is coprime to 6.
Denoting x := an/4 we rewrite the two equations as
x4 − 2 = c · pk
.
If k is even, then introducing y := pk/2 we obtain the quartic equation
x4 − 2 = c · y2
In the case c = 1, it is easy to establish absence of meaningful solutions via factoring x4−y2=(x2-y)(x2+y) while in the case c = 2 we can solve it with Magma's ‘IntegralQuarticPoints‘ function, showing that there are no solutions in this case either.
If 3 | k, then the equation is reduced to two elliptic curves (indexed by c):
Y2 = cX3 + 2,
where Y := an/2 and X := pk/3
They have the only integral points (easily computed in
Magma or Sage) (X, Y ) = (−1, 1) for c = 1 and (X, Y ) ∈ {(−1, 0),(1, 2),(23, 156)} for c = 2, neither of which gives us a solution to the original equation.
Hence, we have gcd(k, 6) = 1 and thus k ≥ 5.
PS. We may also notice that for c = 2, x must be even the equation takes form
8(x/2)4-pk=1
while for c = 1 it can be written as x4 − pk = 2
That is, pk if it exists would be the smallest of two powerful numbers that differ in 1 ([OEIS
A060355](https://oeis.org/A060355)) or 2 ([OEIS A076445](https://oeis.org/A076445)).
Reference
[1] Max Alekseyev (https://math.stackexchange.com/users/147470/max-alekseyev), Euler's
totient function and primes, URL (version: 2022-06-23): https://math.stackexchange.com/q/4478910