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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)
giventhatab(modn)showthatakbk(modn)
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].
ab=qna=qn+bak=(qn+b)k=p=0kCkp(qn)pbkp=bk+p=1kCkp(qn)pbkpakbk=np=1kCpkqpnp1bkp0[n]akbk[n].
Commented by Rio Michael last updated on 01/Nov/19
thank you
thankyou
Commented by mathmax by abdo last updated on 01/Nov/19
you are welcome.
youarewelcome.

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