Can anyone explain the relation between modular inverse of p^k and p^(k-1) given M. Any help would be appreciated.
# | User | Rating |
---|---|---|
1 | tourist | 3857 |
2 | jiangly | 3747 |
3 | orzdevinwang | 3706 |
4 | jqdai0815 | 3682 |
5 | ksun48 | 3591 |
6 | gamegame | 3477 |
7 | Benq | 3468 |
8 | Radewoosh | 3463 |
9 | ecnerwala | 3451 |
10 | heuristica | 3431 |
# | User | Contrib. |
---|---|---|
1 | cry | 165 |
2 | -is-this-fft- | 161 |
3 | Qingyu | 160 |
4 | Dominater069 | 158 |
5 | atcoder_official | 157 |
6 | adamant | 155 |
7 | Um_nik | 151 |
8 | djm03178 | 150 |
8 | luogu_official | 150 |
10 | awoo | 148 |
Can anyone explain the relation between modular inverse of p^k and p^(k-1) given M. Any help would be appreciated.
Name |
---|
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
$$${p^{k-1}}^{-1} \equiv p^{-(k-1)} \equiv p^{-k+1} \equiv {p^k}^{-1} \cdot p$$$