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Answers for “17 在模 19 下的逆元是什么,如何快速计算?”

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18 is its own inverse modulo 19 because 18 is congruent to -1 modulo 19, and -1 is always its own multiplicative inverse since (−1)(−1)=1(-1)(-1)=1. Therefore, 18⋅18≡(−1)(−1)≡1(mod19)18 \cdot 18 \equiv (-1)(-1) \equiv 1 \pmod{19}.

Conditions: Working modulo 19.; Using the fact that 18 ≡ -1 (mod 19).