$33=\left\{a^3+b^3+c^3 \mid (a, b, c) \in \mathbb{Z}\right\}$ I was just wondering, if I had an equation like:
$$33 = \left\{a^3 + b^3 + c^3 \mid (a, b, c) \in \mathbb{Z}\right\}$$
What are the values of $(a, b \land c)$. Is there a way of proving or disproving that such integers of $a$, $b$, and $c$ exist to satisfy this equation? Recently the equation:
$$74 = a^3 + b^3 + c^3$$
Was solved, and there proved to be integers $a$, $b$, and $c$ that satisfied this equation. Could you please help for $33$? Thanks :)
UPDATE: I did not realise this was an open question and that we need hefty computers to solve an equation like this, but thank you people for helping me out. I guess if $(a, b \lor c) \to \pm \infty$ then it is only a matter of trial and error before we find a solution. After looking at Numberphile, I realised how this question was not just any "ordinary problem".
 A: Note in your case, the max value of $a,b,c$ is achieved when both others are zero, and hence is $33^{1/3} \approx 3.2$, so since they must be integers, you have $a,b,c \in \{0,1,2,3\}$.
Now it is easy to find all combinations of these in 4 variables. Without loss of generality, assume $a \ge b \ge c$, and you get the following list of combinations (I used Excel but you can enumerate any way you like):
$$
\begin{array}{ccc|c}
a & b & c & a^3+b^3+c^3 \\
\hline
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 \\
1 & 1 & 0 & 2 \\
1 & 1 & 1 & 3 \\
2 & 0 & 0 & 8 \\
2 & 1 & 0 & 9 \\
2 & 1 & 1 & 10 \\
2 & 2 & 0 & 16 \\
2 & 2 & 1 & 17 \\
2 & 2 & 2 & 24 \\
3 & 0 & 0 & 27 \\
3 & 1 & 0 & 28 \\
3 & 1 & 1 & 29 \\
3 & 2 & 0 & 35 \\
3 & 2 & 1 & 36 \\
3 & 2 & 2 & 43 \\
3 & 3 & 0 & 54 \\
3 & 3 & 1 & 55 \\
3 & 3 & 2 & 62 \\
3 & 3 & 3 & 81 \\
\end{array}
$$
It is clear that no answer results in 33...
UPDATE
As @Ture points out in the comments below, doing this for $74$ shows that there are no solutions to $a^3+b^3+c^3=74$ for non-negative $a,b,c$ either. This shows conclusively that you cannot restrict $a,b,c$ thus, and the result OP quotes likely features at least one of $a,b,c$ less than zero.
A: The set of possible values is tiny, as $4^3>33$ and $2\cdot 3^3>33$. Assuming $a\le b\le c$, you only have to try with
$$\color{green}{1,1,1}\\1,1,2\\\color{green}{1,1,3}\\\color{green}{1,2,2}\\1,2,3\\2,2,2\\\color{green}{2,2,3}.$$
Note that preliminary computation of the parity (odd) further speeds-up the screening.
