Suppose k is an integer such that $6k + 1$, $12k + 1$, $18k + 1$ are primes. Show that $n = (6k + 1)(12k + 1)(18k + 1)$ is a Carmichael number. [duplicate]
Let $6k+1=p, 12k+1=q, 18k+1=r$, so $n=pqr$. Consider $a$ s.t. $(a, n)=1$. Note that $n-1=1296k^3+396k^2+36k$ is divisible by $6k=p-1, 12k=q-1, 18k=r-1$. Then $a^{n-1} \equiv (a^{p-1})^{\frac{n-1}{p-1}} \equiv 1 \pmod{p}$ and similarly $a^{n-1} \equiv 1 \pmod{q}$, $a^{n-1} \equiv 1 \pmod{r}$ so $a^{n-1} \equiv 1 \pmod{n}$. Thus $n$ is a Carmichael number.