Why is $f(n)=n^2+3$, where $f\colon\mathbb{N}\to\mathbb{Z}$, not an onto function? Question: $f_2 :\mathbb{N} \to \mathbb{Z}, f_2(n)=n^2 +3$
Using algebra, making $y=f(n)$, isolating for $n$ and plugging in the expression back, I get $n$. However, the answer key says it is not onto. Why is this? I believe it has something to do with going from the set of Natural numbers to the set of Integers.  
 A: Since $2$ has no pre-image because $f(n) = n^2+3 > 2$. This means you can't find any natural $n$ such that: $f(n) = 2$.
A: 
I believe it has something to do with going from the set of Natural numbers to the set of Integers.

Well, yes and no. Yes, because this is technically correct but no because the reason why is probably not what you had in mind based on some of your comments. 
There do exist onto mappings, numerous ones actually, that go from the set of natural numbers to the set of integers. For example, consider the mapping $f\colon\mathbb{N}\to\mathbb{Z}$ defined by
$$
f(n)=
\begin{cases}
n/2&\text{if $n$ is even},\\
(1-n)/2&\text{if $n$ is odd}.
\end{cases}
$$
This function is onto even though it goes from the set of natural numbers to the set of integers. I won't prove this because the real question at hand concerns your particular function. I have tried to make clear why your function is not onto below. If you have any questions, feel free to ask in a comment.

Detailed explanation of why your function is not onto: Recall what it means for a function to be onto: this is an existence question; that is, if there exists some element in the codomain that is not mapped to by any element in the domain, then your function is not onto. You have the mapping $f\colon\mathbb{N}\to\mathbb{Z}$ defined by $f(n)=n^2+3$. What are the domain and codomain of your mapping? We have
$$
f\colon\underbrace{\mathbb{N}}_{\text{domain}}\to\underbrace{\mathbb{Z}}_{\text{codomain}}
$$
Thus, to show that this mapping is not onto, where this mapping is defined by $f(n)=n^2+3$, we must show that there exists an element in $\mathbb{Z}$ that is not being mapped to by any element in $\mathbb{N}$. Suddenly, it becomes clear that your function is not onto because no element $\eta\in\mathbb{Z}$ is mapped to where $\eta<3$, as BRICS points out. This is simply the quickest or easiest way to see that your function is not onto, but there are other examples where $\eta>3$ also shows that your function is not onto. For example, consider $\eta=5$. We have $\eta=5\in\mathbb{Z}$ but is $5$ ever mapped to? No. This is because $f(1)=4, f(2)=7$, and so on, leaving numerous gaps, any one of which we may point to as an example of your function not being onto.
Does it all make sense now?
A: Assume that $0\in\mathbb N$ (the argument is very similar if we assume that $0\notin\mathbb N$). If $n,m$ are nonnegative integers and $n<m$, then $n^2<m^2$, so $$f(n)=n^2 + 3 < m^2 + 3 = f(m).$$ Since $n^2+3$ is a nonnegative integer, 
$$f(\mathbb N) = \{f(n) : n\in\mathbb N\} $$
is a subset of $\mathbb N$, so by the well-ordering principle, it has a least element. From the monotonicity of $f$ we see that
$$\min f(\mathbb N) = f(0) = 3. $$
It follows that $1\notin f(\mathbb N)$, and hence $f$ is not a surjective (onto) function.
