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Dec 18, 2023 at 8:13 history edited Denis Shatrov CC BY-SA 4.0
clarified formulations
Dec 17, 2023 at 21:44 comment added Bogdan Grechuk Ok, I think I found the way around this. We should relax the condition that every prime divisor of p is 1 modulo 8 to the condition that every prime divisor of p that enters its prime factorization in an odd exponent is 1 modulo 8. Other odd prime factors are allowed to be arbitrary. The proof of Theorems works for this more general statement as well, and it suffices for application to our equation.
Dec 17, 2023 at 20:08 comment added Bogdan Grechuk Thank you, this presentation is much better, but I still do not understand whether you applied Theorem 2 to $p'$ or to $p$. For $p'$, I do not see how you get representation $p'=a^2-8b^2$. If you apply it to $p$, how you get a condition that every prime divisor of $p$ is $1$ modulo $8$? $p=p_1 d^2$, and $d$ can have prime factors equal to $-1$ modulo $8$.
Dec 16, 2023 at 21:29 vote accept Bogdan Grechuk
Dec 16, 2023 at 9:47 history edited Denis Shatrov CC BY-SA 4.0
more details
Dec 16, 2023 at 9:42 history edited Denis Shatrov CC BY-SA 4.0
more details
Dec 15, 2023 at 21:04 comment added Denis Shatrov @BogdanGrechuk The proof definitely needs more explanation. What I initially missed is that the $p = u^2 + 32v^2$ representation is not necessarily primitive. I will rewrite the proof tomorrow.
Dec 15, 2023 at 19:00 comment added Bogdan Grechuk Wow, thank you! The proof would be easier to follow if you provide more details in places such as "similar analysis", "modulo p considerations shows", "can be generalized", etc.
Dec 14, 2023 at 15:29 history edited Denis Shatrov CC BY-SA 4.0
completed the proof
Dec 13, 2023 at 21:17 history answered Denis Shatrov CC BY-SA 4.0