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Prove:

$(p-2)^{(2p-1)^{n}} \equiv (p-1)^{(2p+1)^{n}} -1($mod $ p(p-1)) $ ,

where $p$ is a prime number and $n$ is a natural number. This congruence mod operation problem was formulated by Leon Rosenfeld in 1923.

Attempt: The only idea I had was using Fermat's Little Theorem, which states that for any prime number $p$ and any integer $a$ not divisible by $p$, $a^{(p-1)}$ is congruent to $1 \pmod p$.

I tried using this for $p-2$ and $p-1$ but did not manage to really solve or prove anything so far. I also thought about Wilson's Theorem or even Euler's, but I cannot see how to use them to help me simplify this problem.

Bill Dubuque
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Mogovan Jonathan
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    Starting point: since $\gcd(p,p-1)=1$, that congruence modulo $p(p-1)$ holds if and only if it holds modulo $p$ and it also holds modulo $p-1$. Modulo $p-1$ shouldn't be too hard; then using Fermat's little theorem is a promising strategy for the modulo $p$ one. – Greg Martin Mar 08 '23 at 16:32
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    I was about to edit a full answer but I find it preferable to let the OP write it. The previous hints are sufficient. – Anne Bauval Mar 08 '23 at 16:34
  • Using little Fermat and mod order reduction shows they are congruent mod $p$ & mod $p-1$ hence also modulo their lcm = product = $p(p-1)$ by CCRT – Bill Dubuque Mar 08 '23 at 19:00
  • Where did the reference to Rosenfled come from? Simple exercises like this are easily derived and are common in elementary number theory textbooks. They are not usually attributed to anyone since their discovery and proofs are so straightforward. – Bill Dubuque Mar 08 '23 at 19:38

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