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Question Number 72718 by Rio Michael last updated on 01/Nov/19

$$giventhat$$$$a\equiv b\left(modn\right)$$$$showthat{a}^k\equiv b^k\left(modn\right)$$

Commented by prof Abdo imad last updated on 01/Nov/19

$$\Rightarrow a-b=qn\Rightarrow a=qn+b\Rightarrow a^k={(qn+b)}^k$$$$=\sum _{p=0}^k{C}_k^p{\left(qn\right)}^p{b}^{k-p}=b^k+\sum _{p=1}^k{C}_k^p{\left(qn\right)}^p{b}^{k-p}$$$$a^k-b^k=n\sum _{p=1}^k{C}_p^k{q}^p{n}^{p-1}{b}^{k-p}\equiv 0\left[n\right]\Rightarrow$$$$a^k\equiv b^k\left[n\right].$$$$$$

Commented by Rio Michael last updated on 01/Nov/19

$$thankyou$$

Commented by mathmax by abdo last updated on 01/Nov/19

$$youarewelcome.$$

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