13
$\begingroup$

De Gua's theorem is a $3$-dimensional analog of the Pythagorean theorem: The square of the area of the diagonal face of a right-angled tetrahedron is the sum of the squares of the areas of the other three faces.

For certain tetrahedra, this provides a representation of an integer $n$ as the sum of three integer squares.

Let the tetrahedron have vertices at \begin{eqnarray} & (0,0,0)\\ & (a,0,0)\\ & (0,b,0)\\ & (0,0,c) \end{eqnarray} If $a,b,c$ are integers, at least two of which are even, then the squared areas of the three triangles incident to the origin are each integer squares, and so "represent" $n=A^2$, the diagonal-face area squared.

Example. Let $a,b,c$ be $2,3,4$ respectively.


                   
The diagonal face-area squared is \begin{eqnarray} A^2 & = & \left[ (2 \cdot 3)^2 + (3 \cdot 4)^2 + (4 \cdot 2)^2 \right] \,/\, 4\\ & = & (36 + 144 +64) \,/\, 4 \\ & = & 9 + 36 + 16\\ & = & 61\\ A & = & \sqrt{61} \;. \end{eqnarray} So here, $61$ is represented as the sum of three squares: $9+36+16$.

Let $N_T(n)$ be the number of integers $\le n$ that can be represented as a sum of three squares derived from deGua's tetrahedron theorem, as above. Call these tetra-realized. Let $N_L(n)$ be the number of integers $\le n$ that can be represented as a sum of three squares. $N_L$ is determined by Legendre's three-square theorem, which says that $n$ is the sum of three squares except when it is of the form $n=4^a (8 b + 7)$, $a,b \in \mathbb{N}$.

I would like to know how prevalent is tetra-realization:

Q. What is the ratio of $N_T(n)$ to $N_L(n)$ as $n \to \infty$?

I would also be interested in any characterization of the tetra-realizable $n$.

$\endgroup$
  • 1
    $\begingroup$ This has already been answered but the following comment might be of interest. In contrast to the two dimensional case (where the converse of Pythagoras holds), there are infinitely many non right-angled triangles for which the conclusion of Gua's theorem hold. Thus it is not too surprising that the above method catches only a tiny fraction of the examples. Not a proof, of course, but I suspect that it could be solidified to one. $\endgroup$ – dalry Dec 15 '15 at 17:14
17
$\begingroup$

These numbers are not very prevalent and the ratio in question goes to zero. Note first that by Legendre's theorem, a positive proportion of the numbers below $n$ may be expressed as a sum of three squares. Now consider $N_T(n)$. This amounts to counting (with parity restrictions on $x$, $y$, $z$, and all three positive) the number of distinct integers of the form $((xy)^2 + (yz)^2 +(xz)^2)/4$ lying below $n$. So we must have $xy$, $yz$, and $xz$ all lying below $2\sqrt{n}$, which means that $$ xyz = \sqrt{(xy) (yz)(xz)} \le 2\sqrt{2} n^{\frac{3}{4}}. $$ So the total number of possibilities for $(x,y,z)$ is bounded by the number of triples with product at most $X=2\sqrt{2}n^{\frac 34}$, and this is $$ \sum_{xyz\le X} 1 \le \sum_{x,y\le X} \frac{X}{xy} \le X(1+\log X)^2. $$ Thus, even if these choices for $(x,y,z)$ all led to distinct integers of the form $(xy)^2+(yz)^2+(xz)^2$, we still would have no more than $Cn^{\frac 34}(\log n)^2$ integers up to $n$ that may be tetra-realized.

$\endgroup$
  • 1
    $\begingroup$ Nice analysis to bound $xyz$. Thanks! $\endgroup$ – Joseph O'Rourke Dec 14 '15 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.