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: a^k \equiv b^k \pmod m, then



a^{k+1} \equiv b^{k+1} \pmod m \iff a^{k+1} - b^{k+1} = m(k), k \in \mathbb{Z}.



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

No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...