# A Diophantine equation: $-2a^3 + b^3 + c^3 = 36650$, where $a, b, c > 0$ are all positive integers, and $a, b, c \notin \{2, 9, 15, 16, 33, 34\}$

For a problem that I'm working on, I need to solve this Diophantine equation:-

$-2a^3 + b^3 + c^3 = 36650$, where $a, b, c > 0$ are all DISTINCT positive integers, and $a, b, c \notin$ { 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!

• Oh, I'm so sorry, but I just realized I left out a major constraint; a, b, c are all DISTINCT & NOT IN {2, 16, 34, 9, 33, 15}. Could you help find another solution given the above conditions too? May 21, 2013 at 12:31
• Also, on checking, I realized I'd made a minor error in my equation too. I've edited all that in my question. Sorry for the trouble! May 21, 2013 at 12:57
• @JobinIdiculla - I've checked your original question for 0<a, b, c<1000 and have found no solution so far. May 21, 2013 at 13:05
• @VincentTjeng: Oh, that's rather disheartening to know. But then again, the question's slightly different now (there might still be hope!). Could you share your algorithm or tell me what computing engine you used, so that I can extend the search? May 21, 2013 at 13:10
• I used the following line of code in Mathematica: Do[soln = Solve[{-2*a^3 + b^3 + c^3 == 36650, c == dummyc, 0 < a < 1000, 0 < b < 1000}, {a, b, c}, Integers]; If[Length@soln > 0, Print[soln]], {dummyc, 1, 1000}]. For your modified question, the solutions provided are {{a->11,b->34,c->2}} {{a->2,b->33,c->9}} {{a->11,b->33,c->15}} {{a->15,b->34,c->16}} {{a->2,b->9,c->33},{a->11,b->15,c->33}} {{a->11,b->2,c->34},{a->15,b->16,c->34}} May 21, 2013 at 13:59

For your modified question $a^3 + b^3 + c^3 - 3d = -83449$, there are in fact infinite solutions.

Considered modulo 3,

$$a^3 + b^3 + c^3 \equiv 2 \mod 3$$

Now, we know that modulo 3, any cube is congruent to itself, or $x^3 \equiv x \mod 3$.

Therefore, the above equation reduces to

$$a+b+c \equiv 2 \mod 3$$

One triplet of $(a,b,c)$ that satisfies the above equation is $(1,3,4)$. We can then calculate the value of $d$, which is simply $27847$ in this case.

Rearranging the above equation, since $3d=a^3+b^3+c^3+83449$, $d$ is positive for any positive values of $a,b,c$, allowing us to conclude that there are infinite solutions.

• Your solution is correct indeed, Vincent. But something is bugging me. You see, I was trying to solve the following system of equations:- $40033 + a^3 = d$, $39312 + b^3 = d$, $4104 + c^3 = d$. When you add them together and rearrange, you get $a^3 + b^3 +c^3 - 3d = -83449$, for which you solved and obtained one solution as a=1, b=3, c=4, d=27847 and it indeed satisfies the last equation too. But when you plug in the values of a,b,c and d in the original system of equations, it doesn't seem to satisfy them. What is going on? May 21, 2013 at 15:37
• @JobinIdiculla You might convert your comment to a new question, if that is what you really want to ask. May 21, 2013 at 16:37
• Yes, I shall do that right away, Mark. Sorry I messed it up a lot of times. May 21, 2013 at 16:40

For the similar equations a^3 + b^3 = k + c^3 where k is not +/-4 modulo 9, there is a heuristical argument that for n-digit c, there is a chance of a few percent that there is a solution for each k. In other words, there is likely an infinite number of solutions, but you are lucky to find just one. I suspect you will have the same.

How do you get the heuristics? See what you can find about values modulo 2 or 3. Then let a = 1, 2, 3, …, let b = 1, 2, 3, as long as b^3 <= cube root of a^3 + 36650/2, calculate c as a cube root rounded down. You have a solution if the cube root is an integer, the chances are one in (c+1)^3 - c^3. Add up for all b and n digit a, and you get a not very large number.