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I'm looking into an area that appears to be a close variant of Finding primes so that $x^p+y^p=z^p$ is unsolvable in the p-adic units. That question is far beyond my knowledge in this area, but I'm wondering if the following is a simpler task.

Are there any solutions to the following set of congruences?

$$\begin{aligned} t^3 &\not\equiv 1 \pmod {p^3} \\ t^{p-1} &\equiv 1 \pmod {p^3} \\ (t+1)^{p-1}&\equiv 1 \pmod {p^3} \\ \end{aligned}$$

I searched numerically and didn't find any solutions for primes $p<2500$. Should I expect to find a solution for some larger prime?

Or is there some approach to prove that all solutions require $t^3 \equiv 1 \pmod {p^3}$, which would imply $p \equiv 1 \pmod 3$?

amWhy
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kayle
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  • One of $t^{p-1}$ or $(t+1)^{p-1}$ must be congruent to $1\bmod 2p^3$ – Roddy MacPhee Oct 23 '21 at 17:02
  • Looks like higher powers share the same characteristic: $$\begin{aligned} t^3 &\not\equiv 1 \pmod {p^x} \ t^{p-1} &\equiv 1 \pmod {p^x} \ (t+1)^{p-1}&\equiv 1 \pmod {p^x} \ \end{aligned}$$ – kayle Nov 01 '21 at 20:54
  • Looking mod 9 might help. If p is 1 more than a multiple of 3 we are taking cubes mod other cubes. – Roddy MacPhee Jun 12 '22 at 19:08

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