show that $a^{12} \equiv 1 \mod 32760$ for each $a \in \mathbb{Z}$ such that $\text{gcd}(a, 32760) = 1$

Question

I am trying to show that $a^{12} \equiv 1 \mod 32760$ for each $a \in \mathbb{Z}$ coprime to $32760$. I have found the prime decomposition $32760 = 2^3 \cdot 3^3 \cdot 5 \cdot 7 \cdot 13$ and determined $\phi(32760)$, where $\phi$ is the Euler's totient function. This gives me that $\phi (32760) = 8640$.

Using Euler's congruence, this gives that $a^{8640} \equiv 1 \mod 32760$, but this is still far from the result I want to show... I also know a consequence of Eulers congruence, which states that $e \equiv e' \mod \phi (n)$ implies that $a^e \equiv a^{e'} \mod n$, but I do not see how to use this...

Any hints would be appreciated.

$\textbf{EDIT:}$ I made a mistake in my primefactorization, it is edited now.

Answer 1

For each maximal prime power $\,p^{\large k}$ dividing $\, 32760 = 2^{\large 3} \cdot 3^{\large 2} \cdot 5\cdot 7\cdot 13\,$ we have $\,\phi = \phi(p^k)\mid 12,\ $ so $\,12 = \phi k,\,$ so $\ {\rm mod}\,\ p^{\large k}\!:\,\ a^{\large 12} \equiv (a^{\large \phi})^{\large k}\equiv 1^{\large k}\equiv 1\, $ by Euler's Theorem.

Thus, since all $\,p^{\large k}$ divide $\, a^{\large 12}\!-1,\,$ so too does their lcm = product $= 32760$

Remark $\ $ See Carmichael's Lambda Theorem for the general result.

Answer 2

We have $$32760 = 2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 13$$ By the Chinese remainder theorem, $$U(32760) \cong U(8) \times U(9) \times U(5) \times U(7) \times U(13) \cong C_4 \times C_6 \times C_4 \times C_6 \times C_{12}$$ has exponent $12$.

In terms of congruences, by Euler–Fermat we have

$a^4 \equiv 1 \bmod 8$

$a^6 \equiv 1 \bmod 9$

$a^4 \equiv 1 \bmod 5$

$a^6 \equiv 1 \bmod 7$

$a^{12} \equiv 1 \bmod 13$

and so, $a^{12} \equiv 1 \bmod 8,9,5,7,13$

By the Chinese remainder theorem,

$a^{12} \equiv 1 \bmod 32760 $