given-that-a-b-mod-n-show-that-a-k-b-k-mod-n-

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

g i v e n t h a t

a \equiv b (m o d n)

s h o w t h a t a^{k} \equiv b^{k} (m o d n)

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

\Rightarrow a - b = q n \Rightarrow a = q n + b \Rightarrow a^{k} = {(q n + b)}^{k}

= \sum_{p = 0}^{k} C_{k}^{p} {(q n)}^{p} b^{k - p} = b^{k} + \sum_{p = 1}^{k} C_{k}^{p} {(q n)}^{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 [n] \Rightarrow

a^{k} \equiv b^{k} [n] .

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

t h a n k y o u

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

y o u a r e w e l c o m e .