Proving $a^{n-1} \equiv 1 \mod n $ when n is not prime.

Question

Let $t∈\Bbb N$ and let $x=6t+1, \:y=12t+1$ and $z=18t+1$. $x, y$ and $z$ are all primes and let $n=xyz$. Prove that

$a ^{n-1} \equiv 1 \pmod n\;$ whenever $ a∈\Bbb Z$ and $\gcd(a,n) = 1$.

I have started using the Euler-Fermat Theorem since $\gcd(a,n)=1$. Giving $\phi (n)=(1296t^3) $ But unsure how to proceed?

Answer 1

$\!\bmod \color{#0a0}{z\!-\!1}\!=\!\color{#c00}{18t}\!:\,\ n\!-\!1 = \overbrace{xyz\!-\!1}^{\large\color{#0a0}{z\ \ \equiv\ \ 1}}\!\equiv xy\!-\!1\equiv (6t\!+\!1)(12t\!+\!1)-1 = \color{#c00}{18t}(4t)\!+\!\color{#c00}{18t}\equiv 0$

Thus $\,z\!-\!1\mid n\!-\!1\,$ so by $\,\color{#d0f}{\rm Fermat}\,\bmod z\!:\,\ a^{\large n-1}\equiv (\color{#e0f}{a^{\large z-1}})^{\Large \frac{n-1}{z-1}}\equiv \color{#e0f}{ 1}^{\Large \frac{n-1}{z-1}}\equiv 1$

The same holds for the primes $\,x,y\,$ (see below) so it holds true mod their lcm = product $= n$.

Remark $ $ The proof needs only that the coef's $\, 6,12,18\,$ obey $\,b\mid cd,\,c\!+\!d\,$ for all permutations.