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Question Number 124795 by benjo_mathlover last updated on 06/Dec/20

Answered by mathmax by abdo last updated on 17/Dec/20

$$\mathrm{p}\mathrm{prime}\Rightarrow \mathrm{Z}/\mathrm{pZ}\mathrm{is}\mathrm{a}\mathrm{corps}\mathrm{we}\mathrm{have}\mathrm{for}{\mathrm{alla}}_{\mathrm{i}}\in \mathrm{Z}/\mathrm{pZ}$$$${({\mathrm{a}}_1+{\mathrm{a}}_2+...+{\mathrm{a}}_{\mathrm{n}})}^{\mathrm{p}}\equiv {\mathrm{a}}_1^{\mathrm{p}}+{\mathrm{a}}_2^{\mathrm{p}}+....+{\mathrm{a}}_{\mathrm{n}}^{\mathrm{p}}\left[\mathrm{p}\right]\left(\mathrm{we}\mathrm{get}\mathrm{this}\mathrm{by}\mathrm{recurrence}\mathrm{on}\mathrm{n}\right)$$$$\mathrm{for}{\mathrm{a}}_1={\mathrm{a}}_2=....={\mathrm{a}}_{\mathrm{n}}=1\mathrm{we}\mathrm{get}{(1+1+...+1)}^{\mathrm{p}}\equiv 1+1...+1\left[\mathrm{p}\right]\Rightarrow$$$${\mathrm{n}}^{\mathrm{p}}\equiv \mathrm{n}\left[\mathrm{p}\right]$$$$$$

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