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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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  • $\begingroup$ I don't obtain the same list as you, for instance how do you represent 558 ? perhaps of course my program is wrong. I went further and found representations for some of the integers you list. It seems plausible that your conjecture is correct. $\endgroup$ – Henri Cohen Mar 18 at 11:30
  • $\begingroup$ @Henri Cohen $558=\binom{16}3+\binom{-1}3+\binom{-1}3$. I also noted that $$633=\binom{712}3+\binom{-706}3+\binom{-181}3.$$ $\endgroup$ – Zhi-Wei Sun Mar 18 at 11:55
  • $\begingroup$ My mistake, I was assuming x, y, z different. I went to 1200 and obtained 633 like you, as well as 1458, but I am not able to represent 1983 which is not in your list. $\endgroup$ – Henri Cohen Mar 18 at 13:33
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    $\begingroup$ The question, whether every integer is a sum of three tetrahedrals, was asked on math.stackexchange a year-and-a-half ago, math.stackexchange.com/questions/2472205/… One user claimed to have verified it up to $10,000$. The solution given for $6398$ there is impressively large. $\endgroup$ – Gerry Myerson Mar 19 at 0:52
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    $\begingroup$ @Zhi-WeiSun For $x, y, z \in \mathbb{Z}$, though, the algorithm is easy: $n = \binom{n+2}{3} + 2 \binom{-n + 1}{3} + \binom{n}{3}$. $\endgroup$ – user44191 Mar 19 at 1:40
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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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Here are the representations (in a succinct form) of all the numbers in your list but $1523$.

522: -3132 -2827 3766
523: -1296 -576 1335
622: -3401 679 3394
633: -2446 -1646 2675
642: -2710 -2127 3093
843: -2851 -1596 3011
863: -6866 -1313 6884
918: -6033 479 6034
1013: -6891 1087 6884
1458: -1001 230 999
1878: -1961 -225 1964
1983: -3601 2244 3286

A row

k: x y z

has to be understood as $$k = \binom{x}{3} + \binom{y}{3} + \binom{z}{3}.$$

As a consequence of the script I used to compute those representations, the only integers among $0\dots 5000$ not in the set $$ \left\{\binom{x}{3} + \binom{y}{3} + \binom{z}{3}: x,y,z\in\{-15000,15000\}\right\}$$ are $$1523, 2613, 3337, 3362, 3603, 4447, 4482 .$$

Edit: And finally I obtained a decomposition for $1523$.

1523: -78361 155442 -148496
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I have obtained that

3603: 184939, -110384, -170771
4482: 125663, 56483, -129355

Now $$ \binom{x}{3}+\binom{y}{3}+\binom{z}{3} $$ can represent all integers among $0,\ldots,5000$.

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