Prove that if $n$ is a multiple of $k$ ($n$ and $k$ are both integers), then so is $n^2$.
I know how to prove this without quantifiers, but am having difficulty working in the quantifiers. I know the problem utilizes universal specification or existential generalization, and was hoping someone could outline how I proceed with the quantifiers. I believe the symbolic statement would be: $$\forall n\ [\exists k\ (n = ck) \land (k \neq 0) \implies (n^2 = c_1k)]$$ where $c$ and $c_1$ are integers.
As you have written it, the statement reads: "For all $n$, there exists a $k$ such that if $k$ is not zero and $n$ is a multiple of $k$, then $n^2$ is a multiple of $k$." However, this is not what you want to prove. You want to prove the implication for all $k$ that are divisors of $n$, not merely that one such $k$ exists. The revised statement (in English) should be: "For all $n$ and $k$, if $n$ is a multiple of $k$, then $n^2$ is a multiple of $k$."
In symbols this is $$\forall\ n,k\ [(\exists\ c\!: (n = ck)) \implies (\exists\ c\!: (n^2 = ck))]$$ where all variables are integers. (Note that your original statement had the variables $c$ and $c_1$ but they were not quantified.)