Tuesday, 10 April 2018

modular arithmetic - Prove by induction that if aequivbpmodm then anequivbnpmodm

The base case is pretty straightforward. But I'm stuck on the inductive step.



As the base case holds, assume for when n=k holds, show the k+1 case holds true.




Inductive Hypothesis: akbk(modm), then



ak+1bk+1(modm)ak+1bk+1=m(k),kZ.



I think I'm missing some steps, I'm not sure how to manipulate what I have to shows it holds.

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