TITLE: Is every number a sum of $3$ tetrahedral numbers?
QUESTION [7 upvotes]: It is known that every number can be represented by a sum of $3$ triangular numbers. According to Gauss (see formula $35$ in mathworld article)
$$
\text{num}=\Delta+\Delta+\Delta
$$
I did some numerical experiments that suggest the above formula is correct when triangular numbers are replaced by tetrahedral numbers
$$
\Delta=\frac{n(n+1)(n+2)}6
$$
if $n$ is allowed to be negative.
Is this conjecture correct?
I tried to google representation of integers by tetrahedral numbers but didn't find anything.

REPLY [7 votes]: If $\Delta(n)$ for negative $n$ is allowed, then certainly the integers $t=0\dots 10000$ are all possible. The most awkward of these is $t=6398=\Delta(-1121877)+\Delta(1037512)+\Delta(665832)$. The size of the summands might give you an idea of the size of the task of seeking an explicit solution for each total $t$.