Friday 27 May 2016

arithmetic - Prove that GCD$(n^a - 1,n^b -1)= n^{GCD(a,b)} -1$

I used Euclid's theorem $a=bq +r$.
$n^a -1 = ((n^b)^q )n^r -1$
I don't know how to move forward.

-

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)

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}...