When can $\frac{3}{n}$ not be written as the sum of two reciprocals of natural numbers? 
Show that the set of natural numbers $n$ for which $\frac{3}{n}$ cannot be written as the sum of two reciprocals of natural numbers ($S = \left\{n \mid \frac{3}{n} \neq \frac{1}{p}+\frac{1}{q}\right\}$ for any $p,q \in \mathbb{N}$) is not the union of finitely many arithmetic progressions.

I first wanted to see which values of $n$ all $\frac{3}{n}$ to be written as the sum of two reciprocals of natural numbers. We have $\frac{3}{n} = \frac{1}{p}+\frac{1}{q} \iff 3pq = n(p+q)$ where $n,p,q \neq 0$. What do I do from here?
 A: Some initial observations:  $\frac 32=\frac 11+\frac 12, \frac 33=\frac 12+\frac 12$, so $2$ and $3$ are not in $S$.  Any multiple of a non-member is a non-member because if $\frac 3n=\frac 1p+\frac 1q, \frac 3{kn}=\frac 1{kp}+\frac 1{kq}$
I claim primes that are $1 \bmod 3$ are in $S$ and primes that are $2 \bmod 3$ are not in $S$.  This says any number which has a factor not equal to $1 \bmod 3$ are not in $S$.  I do not resolve numbers like $91=7 \cdot 13$, but we do not need to do so for this problem.
Suppose $n$ is prime and greater than $3$.  From $3pq=n(p+q)$ we see that $n$ must divide one of $p,q$ and we can let it be $q$.  Then write $q=nr$ and multiply the original relation by $q$ getting $3r=\frac {nr}p+1$. $p$ cannot divide $n$ because if both $p,q$ do the right side is at most $\frac 2n$, so $p$ must divide $r$.  Write $r=sp$ and substitute in, getting $3sp=ns+1$  Now we can see $s=1$, so $r=p, 3p=n+1$  To have a solution we must have $n\equiv 2 \pmod 3$ an in that case we have the solution $\frac 3n=\frac 1{\frac {n+1}3}+\frac 1{\frac {n(n+1)}3}$.  An example is $\frac 3{11}=\frac 14+\frac 1{44}$
We can use $\frac 3{10}=\frac 14+\frac 1{20}$ to show that not all numbers $1 \bmod 3$ are in $S$.  Any other arithmetic progression than $1,4,7, \dots$ will eventually hit a number not in $S$, so it will contain only finitely many numbers, and a finite number of them will be finite.
