[Math] Does the set $\{\binom x3+\binom y3+\binom z3:\ x,y,z\in\mathbb Z\}$ contain all integers

binomial-coefficientsnt.number-theory

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 $x,y,z\in\mathbb Z$.

Best Answer

As a supplement to @dario2994's result: I have obtained 

2613: 27874,-17441,-25379
3337: 60083,-20882,-59229
3362: 20543,19711,-25367
4447: 105313,-35617,-103935

and any solution to $\binom{x}{3}+\binom{y}{3}+\binom{z}{3}=n$ where $n=1523,3603,4482$ must have $|x|,|y|,|z|>50000$.

The searching method is pretty simple: we have $$ \binom{x}{3}+\binom{y}{3}=\frac16(x+y-2)(x^2-x y+y^2-x-y), $$ therefore it is easy to find all solutions to $\binom{x}{3}+\binom{y}{3}=n-\binom{z}{3}$ for fixed $n$ and $z$ by a factorization of $6n-z(z-1)(z-2)$, and we only need to do a one-dimensional search with respect to $z$.

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