Show that ## a^{12}\equiv 1\pmod {35} ##.

[Math100]

Homework Statement

If $gcd(a, 35)=1$, show that $a^{12}\equiv 1\pmod {35}$.
[Hint: From Fermat's theorem $a^{6}\equiv 1\pmod {7}$ and
$a^{4}\equiv 1\pmod {5}$.]

Relevant Equations

None.

Proof:

Suppose $gcd(a, 35)=1$.
Then $gcd(a, 5)=gcd(a, 7)=1$.
Applying the Fermat's theorem produces:
$a^{6}\equiv 1\pmod {7}$ and $a^{4}\equiv 1\pmod {5}$.
Observe that
\begin{align*}
&a^{6}\equiv 1\pmod {7}\implies (a^{6})^{2}\equiv 1\pmod {7}\implies a^{12}\equiv 1\pmod {7}\\
&a^{4}\equiv 1\pmod {5}\implies (a^{4})^{3}\equiv 1\pmod {5}\implies a^{12}\equiv 1\pmod {5}.\\
\end{align*}
Thus $7\mid (a^{12}-1)$ and $5\mid (a^{12}-1)$.
Since $gcd(7, 5)=1$, it follows that $35\mid (a^{12}-1)\implies a^{12}\equiv 1\pmod {35}$.
Therefore, if $gcd(a, 35)=1$, then $a^{12}\equiv 1\pmod {35}$.

 

[fresh_42]

I have nothing to say about it. It's simply fine.

I am surprised that this is true for so many numbers. But proof is proof.