<s>[INST] Number of integer solutions to $Ax + By + Cz = D$ How many integer solutions exits for the equations, $Ax + By + Cz = D$, where 
$B = (A + 1), C = (A + 2)$, and $x, y, z$ are non-negative i.e $x, y, z, >= 0$
I require a general solution which can be implemented using code as well. We would be given the values of $A, D$.
 [/INST] First note that $\gcd(A, A + 1, A + 2) = 1 $. Since this gcd while divide $D$ this Diophantine equation will always have solutions. So first solve $Au + (A+1)v = \gcd(A, A+1) = 1$. Then solve $(1)w + (A+2)s = gcd(1,A+2) = 1$. After finding the integer solutions(values of $u,v,w$ and $s$) using the Extended Euclidean Algorithm we then have
\begin{align}
w + (A +2)s &= 1\\
(Au + (A+1)v)w + (A+2)s &= 1\\
A(uw) + (A+1)(vw) + (A+2)s &= 1\\
A(Duw) + (A+1)(Dvw) + (A +2)(Ds) &= D
\end{align}
Thus we have our solutions $x = Duw$, $y = Dvw$ and $z = Ds$.
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