Question Number 124795 by benjo_mathlover last updated on 06/Dec/20
Answered by mathmax by abdo last updated on 17/Dec/20
![p prime ⇒Z/pZ is a corps we have for alla_i ∈ Z/pZ (a_1 +a_2 +...+a_n )^p ≡a_1 ^p +a_2 ^p +....+a_n ^p [p] (we get this by recurrence on n) for a_1 =a_2 =....=a_n =1 we get (1+1+...+1)^p ≡1+1...+1[p] ⇒ n^p ≡n[p]](Q126172.png)
$$\mathrm{p}\:\mathrm{prime}\:\Rightarrow\mathrm{Z}/\mathrm{pZ}\:\mathrm{is}\:\mathrm{a}\:\mathrm{corps}\:\:\mathrm{we}\:\mathrm{have}\:\:\mathrm{for}\:\mathrm{alla}_{\mathrm{i}} \:\in\:\mathrm{Z}/\mathrm{pZ} \\ $$$$\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} +...+\mathrm{a}_{\mathrm{n}} \right)^{\mathrm{p}} \:\:\equiv\mathrm{a}_{\mathrm{1}} ^{\mathrm{p}} \:+\mathrm{a}_{\mathrm{2}} ^{\mathrm{p}} +....+\mathrm{a}_{\mathrm{n}} ^{\mathrm{p}} \left[\mathrm{p}\right]\:\:\left(\mathrm{we}\:\mathrm{get}\:\mathrm{this}\:\mathrm{by}\:\mathrm{recurrence}\:\mathrm{on}\:\mathrm{n}\right) \\ $$$$\mathrm{for}\:\mathrm{a}_{\mathrm{1}} =\mathrm{a}_{\mathrm{2}} =....=\mathrm{a}_{\mathrm{n}} =\mathrm{1}\:\:\mathrm{we}\:\mathrm{get}\:\left(\mathrm{1}+\mathrm{1}+...+\mathrm{1}\right)^{\mathrm{p}} \:\equiv\mathrm{1}+\mathrm{1}...+\mathrm{1}\left[\mathrm{p}\right]\:\Rightarrow \\ $$$$\mathrm{n}^{\mathrm{p}} \:\equiv\mathrm{n}\left[\mathrm{p}\right] \\ $$$$ \\ $$