The Riemann, generalized Riemann, and Extended Riemann hypotheses use non-conceptually different expressions in which functional equations contain f (s) ↔ f (1 − s) symmetry. Hence, symmetry explains all three of Riemann’s hypotheses (2) (PDF) SIMPLE SYMMETRY PROVES ALL THREE RIEMANN'S HYPOTHESES. Available from: https://www.researchgate.net/publication/381697682_SIMPLE_SYMMETRY_PROVES_ALL_THREE_RIEMANN'S_HYPOTHESES
There's just one Riemann hypothesis, that the zeros of the zeta function ALL have real part equal to 1/2.
The other versions refer to generalizations of what the zeta function can be usefully defined as, when dealing with sets other than the integers.
That the zeta function, if its argument is called s, satisfies a relation that remains invariant under the transformation s to 1-s does not imply that the Riemann hypothesis is true; if that were the case, the hypothesis would have been proven, already. While this symmetry of the functional relation is consistent with the Riemann hypothesis, it would be true, even were the Riemann hypothesis false.
If a function is symmetric about a real number, that doesn't imply that it vanishes at that value: If f(s)=f(1-s), for all s in the interval [0,1), this, simply, implies that f(1/2+it)=f(1/2-it), nothing more.
If it vanishes at that value, that is consistent with it being symmetric about that value: f(1/2+it)=0 and f(1/2-it)=0 are consistent with the previous relation, it doesn't imply them, however.
If a function is antisymmetric about a real number, then it, necessarily, vanishes there: if f(s)=-f(1-s) for all s in the interval [0,1), then f(1/2+it)=-f(1/2-it). So it would imply that the zeta function would vanish for t=0, since, in that case, f(1/2)=-f(1/2)2f(1/2)=0f(1/2)=0; it wouldn't imply that it vanished for t≠0. And a function can vanish at s=1/2 + it, without being antisymmetric about Re s = 1/2. So it wouldn't imply that all the zeroes are of the form 1/2 + it, with t real.
Stam Nicolis
Please check
https://www.researchgate.net/publication/381550752_SIX_ONE-PAGE_PROOFS_OF_THE_RIEMANN_HYPOTHESIS
It's easy to see that the claims made therein are wrong. It would be useful to study a course on number theory, e.g. https://people.maths.ox.ac.uk/greenbj/papers/primenumbers.pdf
Stam Nicolis You have not followed the link to the file, because the first explanation is link to arXiv.org paper. It is co-operation with Stefan Groote.
I have. What Robin's theorem states is that, if and only if the RH is true, then Robin's theorem is true. But Robin's theorem can be false, cf. https://arxiv.org/abs/math/0604314
Stam Nicolis No. I cite, "Robin's theorem states that the truth of the inequality for all n>5040 is equivalent to the Riemann hypothesis (Robin 1984; Havil 2003, p. 207)." https://mathworld.wolfram.com/RobinsTheorem.html
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