Help with this Diophantine equation Note:  This question was posted in error, and should be closed as no longer relevant.  The correct question is posted at Help with this system of Diophantine equations (End of note)
For a research problem that I'm working on, I need to solve this Diophantine equation:-
$a^3+b^3+c^3-3d=-83449$, where $a,b,c,d>0$ are all DISTINCT positive integers and$ a,b,c∉ ${$2,9,15,16,33,34$}.
How does one go about solving this? Is brute-force the only possible way? Or could there be a case that no integer solutions exist for this equation?
Also, are there any online computing engines, that allow me to set constraints, and solve Diophantine equations of this sort?
Any and all help is appreciated! Thanks!
 A: For $0 < a \le3966887  $ solutions are $(9419, 10418, 8146),(69167, 10776, 87090)$ and (added) $(3966887, 2434179, 4797573)$.
Here is an idea for searching. Loop $a$ from $1$ to certain bound.
You have to solve $x^3 + y^3 = C + 2 a^3 = N$. This is easy to solve
if $N$ can be factored since $x^3+y^3$ factors nicely.
Added to the edited question
You have to solve $ a^3+b^3+c^3 + 83449 = 3 d $
Just pick "random" $a,b,c$ such the the lhs is divisible by $3$ like
$(300,301,304)$ and $d=27482938$
Here is a pari/gp script which found the solutions.
 {
 jobin1()=
 th=thueinit(x^3+1,1);
 C=36650;
 for(a=1,10^5,
 A=C+2*a^3;
 v=thue(th,A);
 if(v == [],next);
 print([a,v]);
 );
 }
 jobin1()

A: Sorry for a long comment. Joro has given a nice answer already.
Writing $36650=b^3+c^3-a^3-a^3$, the question is related to the problem 
which numbers can be represented by the sum of $4$ signed cubes. A result of Demjanenko 
says that all numbers not of the form $9n \pm 4$ are representable as a sum of four signed cubes. 
Indeed, all integers $n\le 10^7$ have such a representation, and for
$n$ sufficently large the representation also exists (see the artcle 
Kenji Koyama, On searching for solutions of the Diophantine equation $x^3 + y^3 + 2z^3 = n$
, Math. Comput. 69 (2000).
EDIT: the new equation seems to be $a^3+b^3+c^3=n=3d-83449$.  The conjecture is that this
has solutions if and only if $n$ is not of the form $9k\pm 4$.
A: Let a,b,c satisfy the restrictions given, as well as $1 + a + b + c $ is a multiple of $3$.  Then
$83449 + a^3 + b^3 + c^3$ is also a multiple of 3, and then $d$ can be chosen to be $1/3$ of
the last quantity.
Gerhard "3D Makes It So Easy" Paseman, 2013.05.21
