Show that if $m$ in $\mathbb{Z}$ has greatest common divisor $1$ with $21$, then $m^{6}-1$ is divisible by $63$.

Question

Show that if $m$ in $\mathbb{Z}$ has greatest common divisor $1$ with $21$, then $m^{6}-1$ is divisible by $63$. Also, I have to work in $\mathbb{Z}/63 \mathbb{Z}^{*}$, thus the group $63$ modulo $\mathbb{Z}$ under multiplication.

So in other words, I have to prove $m^{6} \equiv 1 (\mod 63)$.

I have no idea how to tie the fact that $m$ has $1$ as gcd with $21$ together with working in the $63$ modulo $\mathbb{63}$ group. I would really appreciate some hints or suggestions guide me in the right direction.

Answer 1

The key point is that $21$ and $63$ have the same prime divisors.

Indeed, since $21 = 3 \cdot 7$ and $63 = 3^2 \cdot 7$, we have $$ \gcd(m,21)=1 \iff \gcd(m,3)=1=\gcd(m,7) \iff \gcd(m,63)=1 $$ Then Euler's theorem gives $$m^6\equiv 1 \bmod 7, \quad m^6\equiv 1 \bmod 9$$ Therefore, $m^6\equiv 1 \bmod 63$.

Answer 2

Carmichael Function $\lambda(63)=[\lambda(9),\lambda(7)]=6$

$(m,21)=1\implies (m,7)=(m,3)=1$

Answer 3

$(m,7)=1\Rightarrow m^{\phi(7)}=m^{6}\equiv 1 \pmod{7}$

and

$\begin{cases} m\equiv 1\pmod{3}\Rightarrow 9|(m-1)(m^2+m+1)(m^3+1)=m^6-1\\ m\equiv 2 \pmod{3} \Rightarrow 9|(m^3+1)|(m^6-1) \end{cases}\Rightarrow m^6\equiv1\pmod{9}$

Answer 4

Hint: The fact that $(m,21)=1$ implies, clearly, that $(m,63)=1$. Hence we can use Euler's Theorem, which says that $x^{\phi(n)}=1$ mod $n$ for all $n$, and $x$ coprime to $n$. Here, $\phi$ is Euler's totient function. Can you proceed further?