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Question Number 72718 by Rio Michael last updated on 01/Nov/19
given that    a ≡ b(mod n)   show that a^k  ≡ b^k  (mod n)
$${given}\:{that}\: \\ $$$$\:{a}\:\equiv\:{b}\left({mod}\:{n}\right)\: \\ $$$${show}\:{that}\:{a}^{{k}} \:\equiv\:{b}^{{k}} \:\left({mod}\:{n}\right) \\ $$
Commented by prof Abdo imad last updated on 01/Nov/19
⇒a−b=qn ⇒a=qn+b ⇒a^k =(qn+b)^k   =Σ_(p=0) ^k  C_k ^p  (qn)^p b^(k−p) =b^k  +Σ_(p=1) ^k  C_k ^p (qn)^p  b^(k−p)   a^k −b^k  =n Σ_(p=1) ^k  C_p ^k  q^p n^(p−1) b^(k−p)    ≡0[n] ⇒  a^k ≡b^k [n].
$$\Rightarrow{a}−{b}={qn}\:\Rightarrow{a}={qn}+{b}\:\Rightarrow{a}^{{k}} =\left({qn}+{b}\right)^{{k}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{{k}} \:{C}_{{k}} ^{{p}} \:\left({qn}\right)^{{p}} {b}^{{k}−{p}} ={b}^{{k}} \:+\sum_{{p}=\mathrm{1}} ^{{k}} \:{C}_{{k}} ^{{p}} \left({qn}\right)^{{p}} \:{b}^{{k}−{p}} \\ $$$${a}^{{k}} −{b}^{{k}} \:={n}\:\sum_{{p}=\mathrm{1}} ^{{k}} \:{C}_{{p}} ^{{k}} \:{q}^{{p}} {n}^{{p}−\mathrm{1}} {b}^{{k}−{p}} \:\:\:\equiv\mathrm{0}\left[{n}\right]\:\Rightarrow \\ $$$${a}^{{k}} \equiv{b}^{{k}} \left[{n}\right]. \\ $$$$ \\ $$
Commented by Rio Michael last updated on 01/Nov/19
thank you
$${thank}\:{you} \\ $$
Commented by mathmax by abdo last updated on 01/Nov/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$

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