Let $n=2^k$ where $k\ge 3$. Let $a$ be any odd natural number.
Prove that $a^{n/4}\equiv 1\pmod{n}$.
My attempt:
$\phi(n)=n/2$
So, by Euler's formula, $a^{n/2}\equiv 1\pmod{n}$.
I don't know how to proceed.
Let $n=2^k$ where $k\ge 3$. Let $a$ be any odd natural number.
Prove that $a^{n/4}\equiv 1\pmod{n}$.
My attempt:
$\phi(n)=n/2$
So, by Euler's formula, $a^{n/2}\equiv 1\pmod{n}$.
I don't know how to proceed.
Since you have posted questions on groups, you might be interested in the following link: . Here, it is stated that for $n=2^k$, the structure of $Z_n^\ast$ is the direct product of the cyclic group of order 2 and the cyclic group of order $2^{k-2}$; it also states that 3 has order $2^{n-2}$ You might try and prove these things for your self or look in practically any book on groups.