If $q\mid 2^x-1$ and $q\mid 2^y-1$ does $q\mid 2^{\gcd(x,y)} - 1$?

Question

If $q\mid 2^x-1$ and $q\mid 2^y-1$ does $q\mid 2^{\gcd(x,y)} - 1$?

If the answer is yes, please show proof. By factoring $2^x - 1$ to ${2^{\gcd(x,y)}}^\frac{x}{\gcd(x,y)} - 1$ the converse must be true.

Answer 1

The statement $q\mid 2^x - 1$ is equivalent to saying that:

• the element $2$ is invertible in $\Bbb Z/q\Bbb Z$;
• the order of $2$ in the multiplicative group $(\Bbb Z/q\Bbb Z)^\times$ divides $x$.

Hence if $q \mid 2^x - 1$ and $q \mid 2^y - 1$, then the order of $2$ in $(\Bbb Z/q\Bbb Z)^\times$ divides both $x$ and $y$, therefore divides $\gcd(x, y)$.

This shows that $q\mid 2^{\gcd(x, y) - 1}$.