Can anyone explain the relation between modular inverse of p^k and p^(k-1) given M. Any help would be appreciated.
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Can anyone explain the relation between modular inverse of p^k and p^(k-1) given M. Any help would be appreciated.
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if m is prime we have Fermat's little theorem :
$$$ (p^{k}) ^ {m - 1} \equiv 1 mod m $$$
$$$ (p^{k}) ^ {m - 2} \equiv p^{-k} mod m $$$
$$$ (p^{k - 1}) ^ {m - 2} p ^ {m - 2} \equiv p^{-k} mod m $$$
$$$ p^{-k + 1} p ^ {m - 2} \equiv p^{-k} mod m $$$
$$$ p^{-k + 1} p ^ {-1} \equiv p^{-k} mod m $$$
if m is not prime see Euler's theorem
also, check modular-inverce cp algorithms
$$${p^{k-1}}^{-1} \equiv p^{-(k-1)} \equiv p^{-k+1} \equiv {p^k}^{-1} \cdot p$$$