Help with this system of Diophantine equations A couple hours ago, I'd posted a Diophantine equation question, but realized that I'd committed a rather preposterous blunder deriving it.
This is the actual question which I'm trying to solve:-
For a research problem that I'm working on, I need to solve the following system of Diophantine equations:-
$ a^3 + 40033 = d$,
$ b^3 + 39312 = d$,
$ c^3 + 4104 = d$, where $a,b,c,d>0$ are all DISTINCT positive integers and $a,b,c∉${$2,9,15,16,33,34$}.
Sorry about the earlier mishap. Any and all help is appreciated! Thanks!
 A: I'm not going to do all the calculations, just get things started.
From $a^3+40033 = d = b^3+39312$, one gets $b^3-a^3=721$, or $(b-a)(b^2+ab+a^2)=7\cdot103$, so $b-a\in 1,7,103,721$.  You can do these cases one at a time.  For example, if $b-a=1$ then you have $(a+1)^2+a(a+1)+a^2=721$, or $a^2+a-240=0$, which has no solutions.  If one of the cases does produce a solution, then you can compute the corresponding $d$ and check it against the equation $d=c^3+4104$.
There may well be a slicker approach, but this should work.
Added 5/22/13:  I just discovered a mildly embarrassing error in my answer.  The equation $a^2+a-240=0$ does have solutions:  The quadratic factors as $(a-15)(a+16)$.  (I had mentally multiplied $4\times240=920$ and knew that a discriminant of $921$ was too close to $900$ to be a perfect square.)  I finally caught my error when I decided to try to make my approach a little slicker:
If you write $b=a+k$, then $b^3-a^3=721$ becomes $k(k^2+3ka+3a^2)=7\cdot103$, so again $k \in 1,7,103,721$.  But now you can immediately rule out $k=103$ and $k=721$, because the quadratic factor $k^2+3ka+a^2$ would obviously produce a number way too large (since $a$ is also required to be positive).  So that leaves $k=1$ and $k=7$, both of which do lead to solutions.  It so happens, though, that they lead to $(a,b,c)=(15,16,34)$ and $(2,9,33)$, respectively, which the OP explicitly disallowed.
A: The first two equations amount to $b^3 - a^3 = 721$. 
Now since for any solution $b^3 - a^3   \ge (a+1)^3 - a^3 = 3a^2 + 3a +1$ this directly gives an upper bound on $a$ namely $15$.
Now checking for which of $a=1, \dots, 15$ one has that $a^3 + 721$ is the third power of an integer, by calculating its third root for example one finds that this is only the case for $a=2$ and $a=15$ (where $b$ would be $9$ and $16$, respectively). 
Both are excluded, so already the first two equations have no solution. 
