Number of pairs satisfying $2\alpha + 3 \beta =k$ How can I prove the following:

Let $\mathbb{W}$ denote the set of non-negative integers. Then what is the cardinality of the set $$\bigl\{ (\alpha,\beta) \in \mathbb{W} \times \mathbb{W} \ | \ 2\alpha + 3\beta =k \bigr\}$$ I think it’s $\left\lfloor\frac{k}{6}\right\rfloor$ if $k \equiv 1 \ (\text{mod} \: 6)$ and $\left\lfloor\frac{k}{6}\right\rfloor + 1$ if $k \not\equiv 1 \ (\text{mod} \ 6)$. 

But I am having trouble doing this. Can anyone provide me an answer, or thoughts on how to go about a solution. 
 A: If $k=6n$, $\beta$ can be $0,2,4,\dots,2n$. 
If $k=6n+1$, $\beta$ can be $1,3,5,\dots,2n-1$. 
If $k=6n+2$, $\beta$ can be $0,2,4,\dots,2n$. 
And so on. 
A: Here’s a straightforward (if slightly ugly) approach.
Suppose that $k=6n$; then $\langle\alpha,\beta\rangle=\langle 0,2n\rangle$ is a solution, and it’s clearly the solution with minimum $\alpha$. Suppose that $\langle a,b\rangle$ is a solution with the smallest possible positive $a$. Then $2a+3b=6n$, so $2a=6n-3b=3(2n-b)$, and $3\mid 2a$. Since $2$ and $3$ are relatively prime, $3\mid a$, and we must have $a\ge 3$. But in fact $2\cdot 3+3(2n-2)=6n=k$, so $\langle 3,2n-2\rangle$ is a solution. By arguing along similar lines you should be able to show that the solutions are precisely the pairs $\langle 3m,2(n-m)\rangle$ such that $0\le m\le n$, so there are $n+1$ of them. And in this case $\left\lfloor\frac{k}6\right\rfloor+1=n+1$, as you conjectured.
Next consider the case $k=6n+1$. Then $k$ is odd, so if $2\alpha+3\beta=k$, $\beta$ must be greater than $0$. Can we find a solution with $\beta=1$? That would require that $2\alpha=6n+1-3=6n-2=2(3n-1)$, which is fine: $\langle\alpha,\beta\rangle=\langle 3n-1,1\rangle$ is a solution with minimal $\beta$. In the first case I got the remaining solutions by changing $\alpha$ by $3$ and $\beta$ by $2$, but in opposite directions. Suppose that in this case I try increasing $\beta$ from $1$ to $d+1$; if $\langle a,d+1\rangle$ is a solution, it must satisfy $2a+3(d+1)=6n+1$, or $3d=6n-2-2a=2(n-1-a)$. As before, this implies that $2\mid d$, so $d\ge 2$. And again we find that the smallest possible value works: $\langle 3(n-1)-1,3\rangle$ is a solution, and you can show without too much trouble that the solutions are the pairs $\langle 3(n-m)-1,2m+1\rangle$ such that $0\le m<n$. There are $n=\left\lfloor\frac{k}6\right\rfloor$ of them, precisely as you conjectured.
The other cases can all be handled similarly.
A: If $k$ is even, then $\beta$ must be even. 
Thus $\beta=2\gamma$, with $0 \leq \gamma \leq \frac{k}{6}$. Each such $\gamma$ leads to an unique solution $\alpha=\frac{k-6\gamma}{2} \,;\, \beta =2\gamma$.
If $k$ is odd, then $\beta$ must be odd. 
Thus $\beta=2\gamma+1 $, with $0 \leq \gamma \leq \frac{k-3}{6}$. Each such $\gamma$ leads to an unique solution $\alpha=\frac{k-6\gamma-3}{2} \,;\, \beta =2\gamma+1$.
