Consider the following question.
Find the generating function for the number of integer solutions to the equation $c_{1}+c_{2}+c_{3}+c_{4}=20$ where $-3\leq c_{1}, -3 \leq c_{2}, -5\leq c_{3}\leq 5, $ and $0 \leq c_{4}$.
The first solution: $$ (c_{1} + 3)+(c_{2}+3)+(c_{3}+5)+c_{4}=20 + 3 + 3 + 5 \Rightarrow c_{1}'+c_{2}'+c_{3}'+c_{4}'=31 $$ where $0\leq c_{1}', 0 \leq c_{2}', 0\leq c_{3}'\leq 10, $ and $0 \leq c_{4}'$. The generation function is $f(x) = (x^{0}+x^{1}+...)^{3}(x^{0}+x^{1}+...+x^{10})$, and the number of integer solutions is the coefficient of $x^{31}$ in $f(x)$.
The second solution (My solution): $$g(x) = (x^{-3}+x^{-2}+...+x^{3})^{2}(x^{-5}+x^{-4}+...+x^{5})(x^{0}+x^{1}+...)$$, and the number of integer solutions is the coefficient of $x^{20}$ in $g(x)$.
My question: In the solution manual book, the presented solution is the first solution, but I think that the second solution is correct, and it is simpler than the first solution. Is my solution correct? What is the difference between these two solutions?
Replacing your $\ (x^{-3}+x^{-2}+\cdots+x^{3})^2\,$ in $\,g(x)\,$ by $\;(x^{-3}+x^{-2}+\cdots)^2\;$ should make both solutions equivalent (with the $x^3$ upperbound you are imposing the artificial constraints $c_1\le3$ and $c_2\le3$) .