Are there integers $a,b,c$ such that $a$ divides $bc$, but $a$ does not divide $b$ and $a$ does not divide $c$? 
Are there integers $a,b,c$ such that $a$ divides $bc$, but $a$ does not divide $b$ and $a$ does not divide $c$?

I am not quite sure what to do with the given information. I know I could easily find an example.
We know that $a$ divides $bc$ so,
$$bc=aq \text{ for some integer } q.$$
And that $a$ does not divide $b$ or $c$ so,
 how is that represented? 
What would be my first step?  
 A: $$4~|~2\times2$$
$$\quad\quad\quad$$
A: $6$ divides $3\times 8$, but $6$ does not divide $3$ and $6$ does not divide $8$.
If a prime number divides the product of two numbers, then it divides one of those two numbers.  That's "Euclid's lemma". (So $6$ is not prime.)
A: The general result on this question is this generalisation of Euclid's lemma:

Gauß's lemma:  If a number divides a product of two factors, a,d is coprime with one of them, it divides the other.

A: oodles. 
We just need $a=mn$ to be composite. ($m > 1; n> 1$  all variables are assumed to be integers.) 
The we can "split the factors" by letting $b = m\cdot \text{(anything not divisible by $n$)}=mk$ and $c = n\cdot \text{(anything not divisible by $m$)}=nj$.
Then $\frac ba = \frac {mk}{mn} = \frac kn \not \in \mathbb Z$ and $\frac ca= \frac {mj}{mn} = \frac jm \not \in \mathbb Z$. But $\frac {bc}{a} = \frac {mnjk}{mn} = jk \in \mathbb Z$.
If $a$ is prime though this will not be possible.  That should be clear right?
If $b = \prod p_i^{m_i}$ and $c = \prod q_i^{n_i}$ where $p_i, q_i$ are prime and $a$, a prime, divides $bc =  \prod p_i^{m_i}\prod q_i^{n_i}$ then $a$ must equal one of the $p_i$ or $q_i$ and must therefore divide either $a$ or $b$.
If $a$ is composite, this can be avoided by either $pq\mid a$ with $p\mid b$ and $q\mid c$ but not vice versa, or by $p^m\mid a$ and $p\mid a$ and $p\mid b$ but not to the full power of $m$.
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To go to your post:
"$bc=aq$ for some integer $q$; what should be my next step?"
$$bc = \frac a{\gcd(a,b)}\gcd(a,b)q ; \text{ let } q = \frac{b}{\gcd(a,b)}\cdot k.$$
$$bc = \frac a{\gcd(a,b)}bk$$
$c = \frac a{\gcd(a,b)}k$; let $k$ be anything that isn't a multiple of $\gcd(b,a)$.
This requires $\gcd(b,a) \ne 1$ and $a\not \mid b$.
Example:  $a = 36$, $b = 48$: $a = 3\cdot 12; b = 4\cdot 12$ so $c = 3\cdot k; 12\not \mid k$, say $k = 7;c =21$.
So $36\mid 21\cdot 48= 7\cdot 36\cdot 4$ but $36\nmid 21$ and $36\nmid 48$.
