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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.

Mathbeginner
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4 Answers4

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

and

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

1123581321
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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?

YiFan Tey
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  • Using Eulers totient function, I would say $\varphi(63)=\varphi(3^{2} \cdot 7) = \varphi( ( 3^{2}-3^{1}) \cdot 7) = \varphi(42)$. Substracting $42$ from $63$ gives $21$, thus I could work in $m^{6} \equiv 1 (\mod 21)$, correct? But I don't see how this makes the problem easier. – Mathbeginner Mar 23 '19 at 13:42
  • @Mathbeginner there's no need to work mod $21$, Euler's theorem gives you mod $63$, which is exactly what you want. (i.e. use $n=63$ and $x=m$ immediately) – YiFan Tey Mar 23 '19 at 15:18
  • But that would be just stating that since we know $m \equiv 1 (\mod 21)$, this implies $m \equiv 1 (\mod 63)$ and that would be the end of it. Is that what you are saying? – Mathbeginner Mar 24 '19 at 08:31
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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, $m^6\equiv 1 \bmod 63$.

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

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

lab bhattacharjee
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  • Downvoting this seems a bit harsh. Though it may well be beyond the OP's knowledge level, it may help to introduce readers to this handy generalization. – Bill Dubuque Mar 23 '19 at 15:30
  • @BillDubuque, Down-voting like up-voting is not controlled here in MSE. Just now received one: https://math.stackexchange.com/questions/771320/evaluate-int-pi-2-pi-2-frac-cosx1ex-dx/771389#771389 .Sometimes I try to learn the flaw: https://math.stackexchange.com/questions/3157905/prove-if-c28-equiv-0-mod-p-then-c3-7c2-8c-is-a-quadratic-residue-mod/3157915#3157915 – lab bhattacharjee Mar 24 '19 at 06:10