Show that $n=(6k+1)(12k+1)(18k+1)$ is an absolute pseudoprime Prove that any integer of the form $n=(6k+1)(12k+1)(18k+1)$ is an absolute pseudoprime if all three factors are prime.
Proof: I am going to use this theorem to show that a number is a pseudoprime. Let n be a square-free number, say $n=p_1...p_r$ where $p_i$ are all distinct primes. If $p_i -1|n-1$ for $i= 1,2,....r$ then n is an absolute pseudoprime. 
so by using this theorem I have $$ (6k+1)-1| n-1$$ $$(12k+1)-1|n-1$$ $$ (18k+1)-1|n-1$$. Now let's work on one equation at a time $$ (6k+1)-1= 6k| n-1$$
now i need to show that $$ 6k| n-1$$ this is where I am having difficulty. Any idea will be appreciated.
 A: FACT 1 : If $p_1,...,p_m$ are $m$ distinct primes and $y\equiv 1 \pmod {p_j}$ for $ j\in \{1,...,m\}$ then $$y\equiv 1 \pmod {\prod_{j=1}^{j=m}p_j} .$$ FACT 2 :If $p$ is prime and $\gcd(p,x)=1$ then $$x^{p-1}\equiv 1 \pmod p, $$ $$\text {and so }  x^v\equiv 1 \pmod p $$ $$\text { whenever } (p-1)|v .$$ Now let $\{6k+1,12k+1,18k+1\}=\{p_1,p_2,p_3\} . $ Observe that  $(p_j-1) |(36k)$ for $j\in \{1,2,3\} ,$ and that $(36k)|(n-1). $ $$ $$ Therefore,  $\gcd(n,x)=1 \implies (\gcd (p_j,x)=1\text { for } j\in \{1,2,3\})$ $\implies (x^{36k}\equiv 1 \pmod {p_j}$ for $j\in \{1,2,3\}$ $\implies (x^{n-1}\equiv 1 \pmod {p_j}  \text{ for } j\in \{1,2,3\})$ $\implies x^{n-1}\equiv 1 \pmod {p_1 p_2 p_3}.$ $$. $$  In general $n$ is a pseudo-prime (a.k.a. Carmichael number) iff $n=\prod_{j=1}^{j=m}(p_j)$ where $\{p_1,...,p_m\}$ is a set of $m$ distinct odd primes with $m\geq 3$ and $LCM \{p_1-1,...,p_m-1\}$ is a divisor of $n-1$.
A: $$n=(6k+1)(12k+1)(18k+1)=6\cdot12\cdot13k^3+(6\cdot12+6\cdot18+12\cdot18)k^2+(6+12+18)k+1;$$
$$n-1=6\cdot12\cdot18k^3+(6\cdot12+6\cdot18+12\cdot18)k^2+(6+12+18)k;$$
$$\frac{n-1}{36k}=2\cdot18k^2+(2+3+6)k+1=36k^2+11k+1,\text{ an integer }.$$
So $n-1$ is divisible by $36k$, which is divisible by $6k$, by $12k$, and by $18k$.
