Did you try to describe Everett's interpretation? There observation of one particle doesn't change the state of the other indeed. If everything is local, then no FTL is ever implied. In copenhagen collapse is nonlocal and happens instantaneously across the universe, so FTL actually happens.

>In copenhagen collapse is nonlocal and happens instantaneously across the universe, so FTL actually happens.

I don't know if this is a mathematical deduction or something empirically verified. Technically when you measure one particle the wave function of both collapses, because now you know what the properties of the other particle are.

But it could be like a pair of dice that always land on opposite faces. When the first die lands, the fate of the second die is sealed and it's no longer in a "superpostion" despite the second die still tumbling around in the air. I strongly suspect this is the case, because otherwise, like you said, you could transmit information faster than light, breaking relativity.

That's more or less right, as far as I know. Manually turning one of the dice doesn't manipulate the other one, and tossing your die only tells you what the other die did or will come up if it's ever thrown. It doesn't tell you if it was thrown.

Any superluminal signalling method proposed using entanglement has to explain which assumption of the no-communication theorem it's negating, otherwise it's breaking the known laws of quantum physics. Also, explain why nobody's noticed this effect in the lab before and sold it to high frequency traders and then won the Nobel prize.

wikipedia says: there can be also cases where is still possible to communicate through the quantum channel encoding more than the classical information

What I propose is to switch between deterministic and indeterministic process, the former encoding 0, the latter encoding 1. Maybe indeterministic process counts as more than the classical information. Also since it's indeterministic, it has a small chance of miscommunication. Since communication isn't precisely certain, maybe the theorem treats it as failure.

Wikipedia is referring to superdense coding, which is not superluminal and still needs you to physically send a particle from place to place to transfer the information.

"an entangled state (e.g. a Bell state) is prepared using a Bell circuit or gate by Charlie, a third person. Charlie then sends one of these qubits (in the Bell state) to Alice and the other to Bob. Once Alice obtains her qubit in the entangled state, she applies a certain quantum gate to her qubit depending on which two-bit message (00, 01, 10 or 11) she wants to send to Bob. Her entangled qubit is then sent to Bob who, after applying the appropriate quantum gate and making a measurement, can retrieve the classical two-bit message." (emphasis mine)

Alice physically transmits her qubit to Bob, so it's not superluminal and doesn't break the no-communication theorem.

https://en.m.wikipedia.org/wiki/Superdense_coding

Okay, I found my error: deterministic measurement I proposed is impossible, it works only for factorized state, but entangled state produces second spooky result with uniform distribution, so collapsed photon is indistinguishable from noncollapsed.

Mathematically speaking entanglement is local: in entangled state |11>|22>+|12>|21> the state |11> coexists with |22> and not with |21>, so when the particles fly away and meet again, |11> still meets with |22>, it's sealed at the time of creation of the state.