Mfrhtyj

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 $A, D$.

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 $z = Ds$.

So, assuming that $A,, D$ are non negative integers
$$
left{ matrix{
0 le x,y,z,A,D; in mathbb Z hfill cr
Ax + left( {A + 1} right)y + left( {A + 2} right)z = D hfill cr} right.
$$
when $A=0$ we are left with $y+2z=D$ and it is easy to see that the number of solutions is
$$
N_{A = 0} = 1 + leftlfloor {{D over 2}} rightrfloor = leftlceil {{{D + 1} over 2}} rightrceil
$$

Taking then $1 le A$, we can write
$$
left{ matrix{
0 le x,y,k,A,D hfill cr
Ax + left( {A + 1} right)y = D - left( {A + 2} right)k hfill cr} right.
$$
and since $gcd left( {A,A + 1} right) = 1$ we can apply to the Popoviciu's Theorem which reads
$$ bbox[lightyellow] {
eqalign{
& p_{left{ {a,b} right}} (n) = left| {,left{ matrix{
0 le x,y,a,b,n in Z hfill cr
gcd (a,b) = 1 hfill cr
ax + by = n hfill cr} right.;} right| = cr
& = {n over {ab}} - left{ {{{b^{,left( { - 1} right)} n} over a}} right} - left{ {{{a^{,left( { - 1} right)} n} over b}} right} + 1 cr}
} tag{1}$$
where:
$left{ x right} $ denotes the fractional part : $left{ x right} = x - leftlfloor x rightrfloor $

and the exponent in brackets denotes the modular inverse
$$
b^{,left( { - 1} right)} b equiv 1;left( {bmod a} right)quad a^{,left( { - 1} right)} a equiv 1;left( {bmod b} right)
$$

In this respect we have
$$
eqalign{
& A^{,left( { - 1} right)} A equiv 1;left( {bmod A + 1} right)quad Rightarrow quad A^{,left( { - 1} right)} = A cr
& left( {A + 1} right)^{,left( { - 1} right)} left( {A + 1} right) equiv 1;left( {bmod A} right)quad Rightarrow
quad left( {A + 1} right)^{,left( { - 1} right)} = 1 cr}
$$

Therefore the Popoviciu's theorem gives
$$
N_{,1 le A} = sumlimits_{k = 0}^{leftlfloor {D/left( {A + 2} right)} rightrfloor } {,left( {{{D - left( {A + 2} right)k} over {Aleft( {A + 1} right)}}
- left{ {{{D - left( {A + 2} right)k} over A}} right} - left{ {{{Aleft( {D - left( {A + 2} right)k} right)} over {A + 1}}} right} + 1} right)}
$$

The terms above can be simplified to some extent. Let's put
$$d=leftlfloor {D/left( {A + 2} right)} rightrfloor $$

Then
$$
eqalign{
& sumlimits_{k = 0}^d {{{D - left( {A + 2} right)k} over {Aleft( {A + 1} right)}} + 1}
= {{D + Aleft( {A + 1} right)} over {Aleft( {A + 1} right)}}left( {d + 1} right) - {{left( {A + 2} right)} over {Aleft( {A + 1} right)}}{{left( {d + 1} right)d} over 2} = cr
& = {{left( {d + 1} right)} over {Aleft( {A + 1} right)}}left( {D - d + Aleft( {A - d/2 + 1} right)} right) cr}
$$
and
$$
eqalign{
& left{ {{{D - left( {A + 2} right)k} over A}} right} + left{ {{{Aleft( {D - left( {A + 2} right)k} right)} over {A + 1}}} right} = cr
& = left{ {{{D - 2k} over A}} right} + left{ {{{Aleft( {D - k} right)} over {A + 1}}} right} cr}
$$

Finally we obtain
$$ bbox[lightyellow] {
eqalign{
& N_{,1 le A} = {{left( {d + 1} right)} over {Aleft( {A + 1} right)}}left( {D - d + Aleft( {A - d/2 + 1} right)} right) + cr
& - sumlimits_{k = 0}^d {left{ {{{D - 2k} over A}} right} + left{ {{{Aleft( {D - k} right)} over {A + 1}}} right}} cr}
} tag{2}$$