For a 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 $a,b,c>0$ 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!

  • 1
    $\begingroup$ I'm not sure why you put DISTINCT in all caps because it's really unnecessary. If, for example, $a = c$ we would have $40033 = 4104$, which is bad. $\endgroup$ May 21, 2013 at 16:54
  • $\begingroup$ Your post already has the only solutions ruled out. I give an answer below pointing out why no others, motivated by the suggestion of Goos in his answer. $\endgroup$
    – coffeemath
    May 21, 2013 at 19:18

2 Answers 2


The only solutions are $(a,b,c,d)=(2,9,33,40041),\ (15,16,34,43408)$ but you have in your post explicitly ruled these out. If you follow the hint by Goos, you get to $$b^3-a^3=721, \\ (b-a)(b^2+ba+a^2)=7\cdot 103,$$ so that either $b-a=1$ or $b-a=7$ which lead to the two solutions above when the other factor is set to $721$ or $103$ respectively.

Maybe you had found the solutions, but your question is about why there are not any others. If so I hope this answer helps see that.


Here's a first step / hint to get you started.

Since the first two equations both equal $d$, we have that $$ a^3 + 40033 = b^3 + 39312 $$

Subtract $a^3$ and $39312$ from both sides, and then factor each side. What do you get?


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .