I have come across a summation of the form;
$$\sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}f(n+\alpha m)g(n-\alpha m)$$
Where $\alpha\in\mathbb{Z}$, and $\alpha\neq 0$. I want to separate these two summations by changing variables. My first idea was to define new variables, $i=n+\alpha m$, $j=n-\alpha m$. Then as $i,j\in\mathbb{Z}$, I thought that;
$$ \sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}f(n+\alpha m)g(n-\alpha m) = \sum_{i=-\infty}^{\infty}\sum_{j=-\infty}^{\infty}f(i)g(j) $$
Apparently this is incorrect however, as in the $\alpha = 1$ case, $i$ and $j$ are always either both even or both odd, leading to additional complications. With this in mind, I really can't see how to proceed.
Any help would be much appreciated, as would any general pointers on changing variables in sums. I understand the process for integrals, but haven't had to do it for a sum before.
Let us assume that the double sum is absolute convergent, then we can reorder the sum: Split the summation over $n$ by looking at resiude classes modulo $\alpha$. In fact, we have $$\sum_{q=0}^{\alpha-1} \sum_{n \in \mathbb{Z}} \sum_{m \in \mathbb{Z}} f(q+\alpha(n+m)) g(q+\alpha(n-m)).$$ Note that $$\mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}_0 \times \mathbb{Z}_0 \cup \mathbb{Z}_1 \times \mathbb{Z}_1, \quad \text{via} \ (n,m) \mapsto (n+m,n-m),$$ where $\mathbb{Z}_i = \{m \in \mathbb{Z} : m \equiv i \ \mathrm{mod} \ 2\}$, is a bijection. Thus the above sum is equal to \begin{align} &= \sum_{q=0}^{\alpha-1} \Big( \sum_{(n,m) \in \mathbb{Z}_0^2} f(q+\alpha n) g(q+\alpha m)+\sum_{(n,m) \in \mathbb{Z}_1^2}f(q+\alpha n) g(q+\alpha m) \Big) \\ &= \sum_{q=0}^{\alpha-1} \sum_{(n,m) \in \mathbb{Z}^2} \Big( f(q+2\alpha n) g(q+2\alpha m)+ f(q+\alpha+2\alpha n) g(q+\alpha+2\alpha m)\Big) \\ &= \sum_{q=0}^{2\alpha-1} \sum_{(n,m) \in \mathbb{Z}^2} g(q+2\alpha m)f(q+2\alpha n) \\ &= \sum_{(m,n) \in \mathbb{Z}^2 , m \equiv n \, \mathrm{mod} \ 2\alpha} g(m) f(n) \end{align}
Update: We can deduce a smilar equality for rational $\alpha \in \mathbb{Q}$. Just write $\alpha =v/w$ with coprime $v \in \mathbb{Z}$ and $w \in \mathbb{N}$, then the sum is equal to $$\sum_{h=0}^{w-1} \sum_{n \in \mathbb{Z}} \sum_{m \in \mathbb{Z}} f(n+ \alpha h + v m) g(n-\alpha h -vm).$$ Hence, we can apply the previous result to conclude that the initial sum is given by $$\sum_{h=0}^{b-1} \sum_{(m,n) \in \mathbb{Z}^2 , m \equiv n \, \mathrm{mod} \ 2v} f(n+\alpha h)g(m-\alpha h).$$