The Gauss-Legendre theorem on sums of three squares states that $$\{x^2+y^2+z^2:\ x,y,z\in\mathbb Z\}=\mathbb N\setminus\{4^k(8m+7):\ k,m\in\mathbb N\},$$ where $\mathbb N=\{0,1,2,\ldots\}$.

It is easy to see that the set $\{x^3+y^3+z^3:\ x,y,z\in\mathbb Z\}$ does not contain any integer congruent to $4$ or $-4$ modulo $9$. In 1992 Heath-Brown conjectured that any integer $m\not\equiv\pm4\pmod9$ can be written as $x^3+y^3+z^3$ with $x,y,z\in\mathbb Z$. Recently, A. R. Booker [arXiv:1903.04284] found integers $x,y,z$ with $x^3+y^3+z^3=33$.

It is well known that $$\left\{\binom x2+\binom y2 +\binom z2:\ x,y,z\in\mathbb Z\right\}=\mathbb N,$$ which was claimed by Fermat and proved by Gauss.

Here I ask a similar question.

**Question**: Does the set $\{\binom x3+\binom y3+\binom z3:\ x,y,z\in\mathbb Z\}$ contain all integers?

Clearly, $\binom{-x}3=-\binom{x+2}3.$ Via Mathematica I found that the only integers among $0,\ldots,2000$ not in the set $$\left\{\binom x3+\binom y3+\binom z3:\ x,y,z\in\{-600,\ldots,600\}\right\}$$ are $$522,\,523,\,622,\,633,\,642,\,843,\ 863,\,918,\,1013,\,1458,\,1523,\,1878,\,1983.\tag{$*$}$$ For example, $$183=\binom{549}3+\binom{-525}3+\binom{-266}3$$ and $$423=\binom{426}3+\binom{-416}3+\binom{-161}3.$$

In my opinion, the question might have a positive answer. For the number $633$ in $(*)$, I have found the representation $$633=\binom{712}3+\binom{-706}3+\binom{-181}3.$$ Maybe some of you could express the numbers in $(*)$ other than $633$ as $\binom x3+\binom y3+\binom z3$ with $x,y,z\in\mathbb Z$.

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