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https://en.wikipedia.org/wiki/Wieferich_prime says:

H. S. Vandiver proved that $2^{p−1}\equiv1\pmod{p^3}$ if and only if $$1+\frac 13+\cdots+\frac1{p-2}\equiv0\pmod{p^2}.$$

Is there a similar result for $2^{p−1}\equiv1\pmod{p^4}$?

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    $\begingroup$ There are only two primes known for which $2^{p-1}\equiv1\bmod{p^2}$. It seems peculiar to ask about the congruence modulo $p^4$. $\endgroup$ May 6, 2020 at 13:16
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    $\begingroup$ You are correct and there are no known modulo $p^3$ either. Still, Vandiver proved the statement above in 1917. Someone must have worked on it meanwhile. $\endgroup$
    – Kurtul
    May 6, 2020 at 15:41
  • $\begingroup$ ams.org/journals/mcom/2005-74-250/S0025-5718-04-01666-7/… $\endgroup$
    – Kenta S
    May 12, 2020 at 13:23

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