Let $n$ and $m$ be positive integers. Find the number of integral solutions to the inequality $x_1 + x_2 + \ldots + x_m \leq n$ such that $x_i ≥ 2i$ for all $i\in \{1, 2, \ldots, m\}$.
My approach is as follows. By using generating functions, I can find the number of integral solutions for each equation $x_1 + x_2 + \ldots + x_m = k$, where $k\leq n$. But when adding them up to get a simplified form, that would be so complicated. Is there any other technique to do this problem?
Any approach is highly appreciated.
Two comments. First, to change the inequality to an equality, you can add a "dummy" variable, like $x_1+\dots+x_m+y=n$. This saves summing solutions to other equations, as you had suggested. Second, to account for any lower bound like $x\ge b$, you can subtract $b$ from both sides of the equality and make a new variable, say $x_1'=x_1-b\ge0$. For the lower bounds you posed, this amounts to subtracting $2+4+\dots+ 2m=m(m+1)$ from $n$. Combining these two ideas results in a concise solution.