<s>[INST] Prove a bijection between $\mathbb{N}^2$ and $\mathbb{N}$. 
Prove that the function
$$f(m,n)=\frac{1}{2}\left(m^2+2 m n+n^2+m+3 n\right)$$
is a bijection between $\mathbb{N}^2$ and $\mathbb{N}$.

The problem arose in a series problem. I have to show that for each couple $(m,n)$ we get a different natural number and that all natural numbers are got applying $f$.
Below an example of what happens for $m,n$ from $0$ to $6$
$$
\begin{array}{ccccccc}
 0 & 2 & 5 & 9 & 14 & 20 & 27 &\ldots\\
 1 & 4 & 8 & 13 & 19 & 26 & 34 &\ldots\\
 3 & 7 & 12 & 18 & 25 & 33 & 42 &\ldots\\
 6 & 11 & 17 & 24 & 32 & 41 & 51 &\ldots\\
 10 & 16 & 23 & 31 & 40 & 50 & 61 &\ldots\\
 15 & 22 & 30 & 39 & 49 & 60 & 72 &\ldots\\
 21 & 29 & 38 & 48 & 59 & 71 & 84 &\ldots\\
\ldots\\
\end{array}
$$
 [/INST] Here is a rigorous proof, but first we rewrite $f$:
$$f(m,n) = \frac12(m^2+2mn+n^2 + m + 3n) = \frac12((m+n)(m+n+1)+2n)$$
$\Large \textbf{Injectivity}$
Suppose we have $f(m,n) = f(a,b)$. Then $(m+n)(m+n+1)+2n = (a+b)(a+b+1)+2b$.
First, suppose $m+n\ne a+b$. WLOG suppose $m+n > a+b$. Then:
\begin{align}(m+n)(m+n+1)+2n &\ge (a+b+1)(a+b+2)\\&=(a+b)(a+b+1)+2a+2b+2 \\&>(a+b)(a+b+1)+2b \\&= (m+n)(m+n+1)+2n\end{align}
which is a contradiction. Hence $m+n=a+b$.
Using this fact we have $2n=2b$, and hence $(m,n) = (a,b)$.
$\Large \textbf{Surjectivity}$
Your table provides a great insight: $f(m,0)$ are precisely the triangular numbers, and $f(m-1, n+1) = 1+f(m,n)$ for $m > 0$.
We can prove this by:
$$f(m,0) = \frac12(m^2+m) = T_m$$
\begin{align}f(m-1,n+1) &= \frac12((m-1+n+1)(m-1+n+1+1)+2(n+1))\\&=\frac12((m+n)(m+n+1)+2n)+1\\&=f(m,n)+1\end{align}
Now take any $x\in \mathbb N$. We can find a triangular number $T_k = \frac{k(k+1)}2$ such that $T_k \le x < T_{k+1}$.
Intuitively this $k$ would be $m+n$, and we need to shift over by $x-T_k$ numbers.
That is, notice that:
$$f(k-x+T_k, x-T_k) = \frac12((k)(k+1)+2(x-T_k))= T_k+x-T_k=x$$
This shows surjectivity.
</s>