# Is every number a sum of $3$ tetrahedral numbers?

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.

• For positive tetrahedral numbers, this is a significant strengthening of Pollock's tetrahedral numbers conjecture: Every positive integer is the sum of at most five tetrahedral numbers. – Eric Towers Oct 14 '17 at 17:18
• This conjecture is equivalent to Every element of $6\mathbb{Z}$ can be represented as $(x^3+y^3+z^3)-(x+y+z)$ and it probably is a very difficult problem. This is an instance of a similar problem: it is not known if $33$ can be represented as the sum of three integer cubes. – Jack D'Aurizio Oct 14 '17 at 17:25
• The question has now been raised on mathoverflow, mathoverflow.net/questions/325659/… – Gerry Myerson Mar 19 at 0:53
• @Jack, $33$ has been done very recently, by Andrew Booker. See, e.g., aperiodical.com/2019/03/… – Gerry Myerson Mar 19 at 0:56

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 $t$.
• Thanks for you answer. I did only verify it to $t=150$. – Tyrell Oct 15 '17 at 14:03