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Suppose I want to find the number of solutions in the natural numbers to solve $ax + by + cz = N$ where $a, b, c, N \in \mathbb{Z}$ (not necessarily all positive). How would I set this problem up using generating functions?

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You have infinite number of solutions whenever you have a solution with non-0 terms on the LHS. –  ashley Dec 13 '12 at 22:00
    
Perhaps you meant you had some lower bounds for $x,y,z$ but they were possibly negative? Then change variables to $X = x-x_0$ for example, so the new ones will be non-negative and use standard techniques for non-negative solutions... –  gt6989b Dec 13 '12 at 22:06
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The number of solutions should be the $q^N$ coefficient in the Laurent expansion of $$\left(1+q^a+q^{2a}+\cdots\right)\left(1+q^b+q^{2b}+\cdots\right)\left(1+q^c+{q‌​^{2c}}+\cdots\right)$$ $$=\frac{1}{(1-q^a)(1-q^b)(1-q^c)}.$$ –  anon Dec 13 '12 at 22:08
    
@Cocopuffs No solutions is always an option. But if there are solutions, no lower bound will force there to be an infinite number of them. You just need solve $ax+by+cz=0$ for $x,y,z$ and then add the combination as many times as you like... –  gt6989b Dec 13 '12 at 22:11
    
@anon How does that account for possibly negative $x,y,z$? –  gt6989b Dec 13 '12 at 22:12

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