Count the number of integer solutions

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The discussion focuses on counting the integer lattice points satisfying the inequality n + Σ|x_k| ≤ N. It notes that if n > N, the number of solutions is zero, and otherwise, the problem simplifies to finding solutions for Σ|x_k| ≤ N'. The function f(N', n) represents the number of solutions, which can be recursively defined as f(N', n) = f(N', n-1) + 2Σf(N'-k, n-1). The approach involves generating functions and polynomial heights, although the latter is not directly related to the solution. The thread emphasizes the importance of understanding the recursive relationship in finding the integer solutions.
benorin
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Count the number of integer solutions of (rather, # of integer lattice points such that)

n+\sum_{k=1}^{n} \left| x_{k}\right| \leq N

Not homework, so no rush. I have worked it through before with a prof., but he's so brilliant I didn't understand much of anything he said :redface: . His solution involed a generating function, something like the coefficient of the blah term in the expansion of blahblah.

Note that the LHS is the height of a polynomial with integer coefficients, although it has nothing to do with the solution.
 
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If n > N, the answer is 0. Otherwise, the problem reduces to finding the lattice point solutions of:

\sum _{k=1} ^n |x_k| \leq N'

where N' > 0. Let f(N',n) denote the number of solutions. Then we want to find f(N',n) and this is just:

f(N',n) = f(N',n-1) + 2\sum _{k=1} ^{N'} f(N'-k,n-1)

We get this because \sum _{k=1} ^n |x_k| = |x_1| + \sum_{k=2} ^n |x_k| and when |x1| > 0, then x1 has two possible values whereas when |x1| = 0 then x1 = 0 only. It's a start.
 

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