6
$\begingroup$

After this answer euler bricks: way to calculate them? and this question Cuboid nearest to a cube I had some three hundred primitive Euler Brick solutions, complete with their Twin solutions (see below), and wanted to sort them into some logical order.

The equations, with $x,y,z$ as the edges and $u,v,w$ as the face diagonals, are: $$x^2+y^2=u^2$$ $$y^2+z^2=v^2$$ $$x^2+z^2=w^2$$

I used the (IMHO) well known method:

“If $(x,y,z)$ is a solution, then (xy,xz,yz) is also a solution, but applying this to the new solution returns to the original shape.”

I don’t know when or where this method of producing Twins originated, or if it has been proven, but it works for every example I’ve tried. However, the calculations soon increase in magnitude.

Conjecture 1

If $(x_0,y_0,z_0)$ is a primitive Euler Brick solution then the following is also a solution,

$$(X_1,Y_1,Z_1)=(\frac{x_0y_0}{12},\frac{x_0z_0}{12},\frac{y_0z_0}{12})$$

Put $G_1=gcd(X_1,Y_1,Z_1)$ then a new primitive Euler Brick solution is given by,

$$(x_1,y_1,z_1)=(\frac{X_1}{G_1},\frac{Y_1}{G_1},\frac{Z_1}{G_1})$$

Now repeat using $(x_1,y_1,z_1)$ as the generator,

$$(X_2,Y_2,Z_2)=(\frac{x_1y_1}{12},\frac{x_1z_1}{12},\frac{y_1z_1}{12})$$

Put $G_2=gcd(X_2,Y_2,Z_2)$ to produce,

$$(x_2,y_2,z_2)=(\frac{X_2}{G_2},\frac{Y_2}{G_2},\frac{Z_2}{G_2})$$

If all went well, $x_2=x_0$, $y_2=y_0$ and $y_2=y_0$

Definition If $\frac{G_1}{G_2}<1$ then $(x_1,y_1,z_1)$ is the Good twin and $(x_2,y_2,z_2)$ is the Evil Twin.

If $\frac{G_1}{G_2}>1$ then $(x_1,y_1,z_1)$ is the Evil Twin and $(x_2,y_2,z_2)$ is the Good Twin.

Example $$(x_0,y_0,z_0)=(117,44,240)$$ $$(X_1,Y_1,Z_1)=(429,2340,880)$$ $$G_1=gcd(429,2340,880)=1$$ $$(x_1,y_1,z_1)=(429,2340,880)$$ $$(X_2,Y_2,Z_2)=(83655,31460,171600)$$ $$G_2=gcd(83655,31460,171600)=715$$ $$(x_2,y_2,z_2)=(117,44,240)$$

Hence, as $\frac{G_1}{G_2}=\frac{1}{715}$, we find $(117,44,240)$ is the Good Twin and $(429,2340,880)$ is the Evil Twin.

The product $G_1G_2=715$, whichever Twin we start from, and is the lowest I’ve found.

Conjecture 2 The value of $G_1G_2$ is equal for each of a pair of twins.

Conjecture 3 Except for twins, the value of $G_1G_2$ is unique for primitive solutions.

Conjecture 4 There are no Identical Twins, where the original solution is returned as $(x_1,y_1,z_1)$ in any order. The nearest I’ve found is $(72611,9180,206448)$ and its Evil Twin $(12915,290444,36720)$

My questions I would welcome any help proving or disproving any of these conjectures. Also, if there is an easier way to identify Twins, I’d love to know about it.

Please let me know if I’ve failed to make anything clear.

Update 26 May 2017 I generated $82$ solutions using Saunderson’s parametric formula, http://mathworld.wolfram.com/EulerBrick.html and found they were all Good Twins.

I would appreciate any other Euler Brick parametric formula.

$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

I'm very late for this party, but the OP wishes to know more Euler brick formulas. The two procedures below should give infinitely many.

(Note: Saunderson was blind, by the way. Makes you wonder how he did mathematics before Braille.) Anyway, given the Euler brick $(a,b,c)$.

I. Euler's Method

$$(a,b,c) = \left(\frac{p^2 - 1}{2p},\; \frac{q^2 - 1}{2q},\; 1\right)$$

This makes $a^2+c^2$ and $b^2+c^2$ as squares and denominators can always be cleared later. It remains to make $a^2+b^2$ a square as well. One choses $(p,q)$ such that,

$$(p^2 - 1)^2q^2+p^2(q^2 - 1)^2 = t_1^2$$

So a quartic polynomial to be made a square. Given a rational point, this can be birationally treated as an elliptic curve, hence we can generate subsequent points from the initial one. Given Condition I or $x^2+y^2=z^2$ the Pythagorean triple $(x,y,z)$, then,

\begin{align} p_1&=\frac{2y}{z},\quad q_1=\frac{z}{2x}\\[5pt] p_2&=\frac{2y}{z},\quad q_2=\frac{ y(2x + z)}{(x + z)(2x - z)}\end{align}

Example. Let $(x,y,z)=(3,4,5)$, then

$$(a,b,c) = (117, 44, 240)\\ (a,b,c) = (429, 2340, 880)$$

and so on. Of course, the formula for $x^2+y^2=z^2$ is well-known, so we can have used that instead, and then generate infinitely more formulas. (Note: The variables $(p_1, q_1)$ yield Saunderson's parameterization.)

II. Lenhart's Method

Similar to Euler's,

$$(a,b,c) = \left(\frac{p^2 - 1}{2p},\; \frac{2q}{q^2 - 1},\; 1\right)$$

This makes $a^2+c^2$ and $b^2+c^2$ as squares. To make $a^2+b^2$ a square as well, solve,

$$(p^2 - 1)^2(q^2 - 1)^2 + 16p^2q^2 = t_2^2$$

Again, a quartic polynomial to be made a square. Given Condition II or $x^2+y^2=5z^2$, then,

\begin{align} p_1&=\frac{x y}{2z^2},\quad q_1=\frac{x+z}{x-z}\\[5pt] p_2&=\frac{x y}{2z^2},\quad q_2=\frac{(x^2 + z^2)(x^2 - 4z^2)}{(x^2 - z^2)(x^2 - 6z^2)}\end{align}

Example. Let $(x,y,z)=(11,2,5)$, then

$$(a,b,c) = (348, 373, 365)\\ (a,b,c) = (5236700, 683709, 5240587)$$

and so on. The formula for $x^2+y^2=5z^2$ is also known, so likewise we can have used that to generate infinitely more formulas.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you, Tito Piezas III, for your solution, and your interest. I would love to say it answers my question, but I feel more is needed. Well done! $\endgroup$ Commented Apr 30 at 15:44

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.