Two related proofs I'm having difficulty with concerning biconditionality and divisibility I've been teaching myself proofs through Velleman's wonderful book (new edition here).  I'm currently working on the Questions 25 and 26, page 135.  I'll go in order, explaining my thought process to see if I'm even in the realm of an accurate approach.
Question 25
Definition.  An integer $a$ divides an integer $b$ if there is an integer $c$ such that $b = a \cdot c$.
Proposition. Prove that for all integers $a$ and $b$ there is an integer $c$ such that $a\ |\ c$ and $b\ |\ c$. 
Proof.  Let $a$ be an arbitrary integer.  Let $b$ be an arbitrary integer.  Then there should be an integer $c$ such that $a\ |\ c$ and $b\ |\ c$.  Since $\ a \in \mathbb{Z}$ and $\ b \in \mathbb{Z}$, $\ a\cdot b \in \mathbb{Z}$.  Let $c = a \cdot b$, then $c \in \mathbb{Z}$.  This means $a\ |\ c$ by definition of divisibility.  Similarly, $c = b \cdot a$.  Thus, $b\ |\ c$ by definition of divisibility.  Since both $a\ |\ c$ and $b\ |\ c$, then we've shown there is an integer which both $a$ and $b$ divide.  This is true for all integers $a$ and $b$.$\ \Box$

Commentary
I'm not sure about this one.  Logically I wrote it as, $\forall a,b \in \mathbb{Z}\ \exists c \in \mathbb{Z}(a\ |\ c\ \land\ b\ |\ c)$.  Is my approach accurate or was I supposed to interpret this as a biconditional?

Question 26
// This one is a 2-part question, but I'll only post the 1st part since
// the other would follow easily from this one's approach.

Prove that for every integer $n$, $15\ |\ n$ iff $3\ |\ n$ and $5\ |\ n$.  I translated this as $\forall n \in \mathbb{Z}\big(15\ |\ n \iff (3\ |\ n \land 5\ |\ n\ )\big)$.

Commentary
I'm kind of lost on this one.  Existential proofs always confuse me for some reason.  I know we have to prove in both directions.  So my approach thus far is to attempt a direct proof on $\big((15\ |\ n \to (3\ |\ n \land 5\ |\ n)\big)$ to satisfy the $(\to)$ direction.  Then I would attempt a direct proof of the converse of this conditional, $\big((3\ |\ n \land 5\ |\ n) \to 15\ |\ n)$ to satisfy the $(\gets)$ direction.
Let's start with the $(\to)$ direction for illustrative purposes.  I'm just going to lay down the scratch work logically since that might help clarify where I'm coming from.
This means $\exists k \in \mathbb{Z}(n = 15k)$. 
Now I need to prove the conjunction $(3\ |\ n \land 5\ |\ n)$
\begin{array}{c|c}
\text{Givens} & \text{Goal} \\ \hline
\text{asm. } \underbrace{15\ |\ n}_{\exists k \in \mathbb{Z}(n = 15k)} & \underbrace{3\ |\ n}_{\exists k \in \mathbb{Z}(n = 3k)} \\ \hline
\text{EI. } \underbrace{k = k_0}_{\text{can k be an actual int?}} & \uparrow \\ 
\end{array}
This is where I get stuck.  I know I should use existential instantiation, I'm just not sure what I'm allowed to use.  In the proofs I've done so far in the book I'm dealing with abstract sets and indexed sets.  So I typically invoke some generic $k_0$, which has the properties of $k$ but is a specific, though unknown element, so must remain a bit abstract.  
My question is, since I'm dealing with an actual integer here, since the domain of discourse was said to be $k \in \mathbb{Z}$, could I simply plug in an integer for k that would make the whole statement true?  So for example,
Since I know that 30 is divisible by 15, 3 and 5, I know my goal should be $n = 30$.  So I "adjust" $k$ accordingly in each case to get that result.  This would mean, Let $k = 2$, then $30 = 15 \cdot 2$.  So by definition of divisibility, $15\ |\ 30$.  
Then I'd do the same for $5\ |\ n$.  Let $j = 6$.  Then $30 = 5 \cdot 6$, or in other words, $5\ |\ 30$. 
My next step would be to say that since I've shown there is such an $n$ such that both $3\ |\ n$ and $5\ |\ n$, then $15\ |\ n$ implies $3\ |\ n \land 5\ |\ n$, and this applies for all $n \in \mathbb{Z}$ since $n$ was arbitrary.

So is my reasoning wrong here or right?  I think it's the use of EI that's throwing me off.  Can I use a specific number, rather than a generic, yet specific, element in the integers? 
 A: In regards to question 25, you seem to have done it all correctly and your notation is clean. It was correct to interpret that as merely implication and not a bi-conditional.
Now to answer your question 26 and give guidance. I would like to mention that understanding how to prove things can take time, I know it did for me.
Suppose we wish to prove for two logical statements ($A \text{ and } B$): 
$$A \iff B$$
This requires use to show 
$$(A \implies B) \land (B \implies A)$$
To prove $A \implies B\ $  you assume $A$ is true and show that B logically follows. This may involve using an object with only the properties specified by A. Note that whether or not any objects actually exist that have the properties of $A$ does not matter because we assume we have an abstract one that does. For example:
$$\text{People with blue eyes do not need air} \implies \text{People with blue eyes do not need to breath}$$
However, note that we cannot choose a specific example of blue eyed people to prove it for everyone, we must rely on the fact that we only know that a person's eyes are blue. Here is an example as to why we cannot chose a specific individual.
$$\text{(For every P in Animals)P is a person} \implies \text{P is Rich}$$
Suppose I choose P to be Bill Gates. He is an animal and a person, but clearly not representative of the general population.
Now this may cause a bit of confusion, but you can show that the statement above is FALSE with a specific counter example since it is claiming it is true FOR ALL. For example, choose P to be the poorest person on the planet. However, do not worry too much about this for your question at hand.
Now more concretely for your problem, we will begin by proving $$(\forall n \in \mathbb{Z})\ ((15\mid n) \implies (3\mid n) \land (5\mid n))$$
We cannot choose an arbitrary number from $\mathbb{Z}$ that satisfies the conditions since it may have other special properties. Instead all we get is a generic $n$ such that $15 \mid n$. This $n$ does not need to exist, but we are showing that if one did then this is what follows.
By definition
$$(15 \mid n) \implies (\exists k\in\mathbb(Z))\ (n = k\cdot 15)$$
So
$$n = k \cdot 15 = k(3\cdot 5) = 3(5k) = 5(3k)$$
Since both $5k$ and $3k$ are integers, we have by definition
$$n = 3(5k) \implies 3 \mid n$$
and
$$n = 5(3k) \implies 5 \mid n$$
So we have shown $(\forall n \in \mathbb{Z})\ 15\mid n \implies ((3\mid n) \land (5\mid n))$
Now we must show the opposite direction that is:
$$(\forall n \in \mathbb{Z})\ ((3\mid n) \land (5\mid n) \implies 15\mid n)$$
So we suppose $(\exists n \in \mathbb{Z})$ such that $3\mid n$ and $5\mid n$, regardless if such an n exists.
I will trust your book has introduced you to the notion of Greatest Common Divisors and has hopefully shown you a proof of the following Fact. (If it does not, Page 16 of John A. Beachy and William D. Blair: Abstract Algebra, 3rd ed. Waveland Press, 2006, Has such a proof, though his style may not be clear to you yet since he is lax with explicit formal logic at times)
THEOREM: Let $a,b,c\in\mathbb{Z}$ and $b \neq 0$ then 
$$ ((b\mid a)\ \land\ (c\mid a)\ \land (GCD(b,c)=1)) \implies bc \mid a$$
Now, since $3$ and $5$ are prime we have $GCD(3,5) = 1$ 
and taking 
$a = n$, $b = 3$ and $c = 5$ 
we can apply the theorem and get 
$$3\cdot5 \mid n$$
And we are done the proof since we have show $(\forall n \in \mathbb{Z})$ both 
$$((3\mid n) \land (5\mid n) \implies 15\mid n)$$
and
$$((15\mid n) \implies (3\mid n) \land (5\mid n))$$
So 
I hope this helps
