Saturday, January 4, 2020

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

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