p {\displaystyle p} 是質數， r {\displaystyle r} 是正整數且 r ≠ 1 {\displaystyle r\neq 1} ， 考慮等比級數總和 S = 1 + r + r 2 + r 3 + ⋯ + r p {\displaystyle S=1+r+r^{2}+r^{3}+\cdots +r^{p}} ，則 S ≡ 1 + r + r 2 + r 3 + ⋯ + r p ( mod p ) {\displaystyle S\equiv 1+r+r^{2}+r^{3}+\cdots +r^{p}{\pmod {p}}} ≡ 1 × ( r p + 1 − 1 ) r − 1 {\displaystyle \equiv {\frac {1\times (r^{p+1}-1)}{r-1}}} ........等比級數求和公式 ≡ r p + 1 − 1 r − 1 {\displaystyle \equiv {\frac {r^{p+1}-1}{r-1}}} ≡ ( r p × r ) − 1 r − 1 {\displaystyle \equiv {\frac {(r^{p}\times r)-1}{r-1}}} ≡ ( r × r ) − 1 r − 1 {\displaystyle \equiv {\frac {(r\times r)-1}{r-1}}} ........根據費馬小定理， r p ≡ r ( mod p ) {\displaystyle r^{p}\equiv r{\pmod {p}}} ≡ r 2 − 1 r − 1 {\displaystyle \equiv {\frac {r^{2}-1}{r-1}}} ≡ ( r + 1 ) ( r − 1 ) r − 1 {\displaystyle \equiv {\frac {(r+1)(r-1)}{r-1}}} ≡ r + 1 ( mod p ) {\displaystyle \equiv r+1{\pmod {p}}} 所以結論是 1 + r + r 2 + r 3 + ⋯ + r p ≡ r + 1 ( mod p ) {\displaystyle 1+r+r^{2}+r^{3}+\cdots +r^{p}\equiv r+1{\pmod {p}}} 請問以上的推論過程或結論有無問題？
