Proving a multi variable function bijective I understand the theory of how to prove a mutli variable function bijective, however I somehow can neither prove this function injective or surjective:
$$f: \mathbb N\times \mathbb N \rightarrow \mathbb N, (a, b) \mapsto {(a+b)}^{2} + a$$ 
I tried to start with $f(a, b) = f(a', b')$ but I don't really know how to continue from there.
To prove it surjective I started with: show that for every $n$ there is $(a, b)$ with $f(a, b) = m$ but was not successful there either.
To make matters worse I don't even know yet if this function is injective, surjective, neither or both.
Any help would be appreciated!

The map is certainly not surjective, as it has been pointed out in the comments.
Let us try to prove injectivity. I will rewrite the function $(a+b)^2+a$ as $c^2+a$ where $c=a+b$, and in particular note $c\geq a$ because $b\in\mathbb{N}$. Argue by contradiction, suppose there exist $a, c, \hat{a}, \hat{c}$ with $c\neq \hat{c}$ and $a\neq \hat{a}$ such that
$$
c^2+a=\hat{c}^2+\hat{a},
$$
noting the case $c=0$ and the case $\hat{c}=0$ are excluded because they would imply $a=b=\hat{a}=\hat{b}=0$.
Without loss of generality, let $c>\hat{c}$. Write
$$
c^2=\hat{c}^2+\hat{a}-a,
$$
and we try to prove the LHS is strictly larger than the RHS, deriving a contradiction. We prove it for the "best" case, $a=0$, which then proves it for $a>0$ as well. Thus we write
$$
c^2=\hat{c}^2+\hat{a},
$$
and since $\hat{a}\leq \hat{c}$ the RHS is bounded above by $\hat{c}^2+\hat{c}$. We had $c>\hat{c}$; taking the best case scenario again we try $c=\hat{c}+1$, which yields $c^2=\hat{c}^2+2\hat{c}+1$, strictly larger than the upper bound on the RHS, which yields the contradiction.
