Monday, 19 November 2018

elementary number theory - $gcd(b^x - 1, b^y - 1, b^ z- 1,...) = b^{gcd(x, y, z,...)} -1$







Dear friends,




Since $b$, $x$, $y$, $z$, $\ldots$ are integers greater than 1, how can we prove that
$$
\gcd (b ^ x - 1, b ^ y - 1, b ^ z - 1 ,\ldots)= b ^ {\gcd (x, y, z, .. .)} - 1
$$
?



Thank you!



Paulo Argolo

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}...