$A_3=\{x\in \mathbb Z: (x+2)/5\in \mathbb Z\}$, $A_7=\{2,3,4,6,8,9,12,16,...\}$. What is $A_3\setminus A_7$? $A_3=\{x\in \mathbb Z: (x+2)/5\in\mathbb Z\}$
$A_7=\{2,3,4,6,8,9,12,16,...\}=\{2^a3^b:(a,b\in \mathbb Z)\cap(a,b\ge0)\}$
The solution is going to be $\{...,-12,-7,-2\} \cup (x+2)/5$ is not even, but I don't really know how to write this in more standard set notation.
 A: Certainly negative numbers do not belong in $A_7$ therefore $A_3 - A_7$ contains all the negative numbers in $A_3$ which can be written as $\{5x-2 : x \leq 0\}$.
Apart from that, it remains to be seen which positive numbers in $A_3$ are not of the form $2^a3^b$ and add them in. Now , any positive integer is not $2^a3^b$ for $a,b \geq 0$ if and only if it has a prime factor which is not $2$ or $3$. This can be checked using the fact that every positive integer has a unique prime factorization.
That is, if $P$ denotes the set of primes, then $\mathbb N - A_7= \{x \in \mathbb N : \exists p \in P - \{2,3\}, p | x\}$ (note : $p|x$ means $p$ divides $x$).
Now, it is clear that $A_3 - A_7 = \{5x - 2 : x \leq 0\} \cup \{5x-2 : x \in \mathbb N, \exists p  \in P - \{2,3\} , p | 5x-2\}$ from the negative positive break up.
We may also combine using the logical "or" written as $\vee$ to get 
$$ 
A_3 - A_7 = \{5x- 2 : (x \in \mathbb Z : x \leq 0) \ \vee\  (x \in \mathbb N : \exists p \in P - \{2,3\} , p | 5x-2)\}
$$
