In a problem set assigned to my class, we were asked to show that if $(u,v,w)$ is a Pythagorean triple, then a cuboid with side lengths $u|4v^2-w^2|$, $v|4u^2-w^2|$, and $4uvw$ generated an Euler brick. I succeeded by replacing $w$ with $\sqrt{u^2+v^2}$, and then replacing $u$ and $v$ with $m^2-n^2$ and $2mn$ respectively, the well-known generator for Pythagorean triples. However, I am unsure how this would be derived. It seems like something easy to check but hard to get. I wasn't able to find any sort of derivation on the web, so does anyone have one?
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$\begingroup$ I read that. There is no derivation on the wiki page. It simply states the parameterization and that's it. $\endgroup$June Richardson– June Richardson2021-01-30 22:30:06 +00:00Commented Jan 30, 2021 at 22:30
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1$\begingroup$ You just seach to look a little deeper. Maybe that helps: math.ubc.ca/~cass/courses/m446-03/pl322/parametrization.html. $\endgroup$garondal– garondal2021-01-30 22:52:03 +00:00Commented Jan 30, 2021 at 22:52
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2 Answers
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The formula regarding the Euler Bricks is from Nicholas Saunderson. You can dig up the original derivation here.
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In your class, it was discussed how $x^2+y^2=z^2$ can be used to generate Euler bricks. You might be interested to know you can also use the similar $x^2+y^2=5z^2$.
I. Euler's method
This uses $x^2+y^2=z^2$ and Saunderson's formula can be derived by this method. The other answer gives a link, but you can see the method in more detail in this MSE post
II. Lenhart's method
One the other hand, this uses $x^2+y^2=5z^2$. Given the Euler brick $(a,b,c)$, first define,
$$(a,b,c) = \left(\frac{p^2 - 1}{2p},\; \frac{2q}{q^2 - 1},\; 1\right)$$
This makes $a^2+c^2$ and $b^2+c^2$ as squares and denominators can be cleared later. To make $a^2+b^2$ a square as well, solve,
$$(p^2 - 1)^2(q^2 - 1)^2 + 16p^2q^2 = t_2^2$$
Thus is a quartic polynomial to be made a square. Let $x^2+y^2=5z^2$, and an initial solution is,
$$p_1=\frac{x y}{2z^2},\quad q_1=\frac{x+z}{x-z}$$
The quartic then is birationally equivalent to an elliptic curve and, from this initial solution, you can find other points $(p_n, q_n)$. But an interesting feature of Lenhart's method is his first formula, by swapping $(x,y)$, yields a pair of bricks with two common sides,
$$(a,b,c) = \Big((x^2-z^2)(y^2-z^2),\; 4xyz^2,\; 2xz(y^2-z^2)\Big)$$ $$(a,b,c) = \Big((x^2-z^2)(y^2-z^2),\; 4xyz^2,\; 2yz(x^2-z^2)\Big)$$
For example, let $(x,y,z) = (2,11,5)$. After removing some factors, then you have the pair of bricks,
$$(a,b,c) = (\color{blue}{1008, 1100,} 960)\\(a,b,c) = (\color{blue}{1008, 1100,} 1155)$$
and so on for infinitely many solutions to $x^2+y^2=5z^2.$
