I have heard about Euler bricks, and i have noticed that there are an strong textinfinite number of euler bricks(as for 4D bricks we are still pending even one!) in existence. I get there may be infinite bricks, but this leaves an issue: how does one go about calculating a euler brick?
Could someone actually use this formula to calculate a perfect euler brick that has been eluded for possibly millennia? Thanks in advance.
You ask, “could someone actually use a formula to calculate a perfect Euler brick?”
IMHO, it seems unlikely, for this is such an obvious way. As far as I recall, some parametric solutions have been proven to not lead to a perfect cuboid. Massive brute force has also failed, so unless you have a brilliant new idea and a few years to spare…
Why has the Perfect cuboid problem not been solved yet?
You ask, “how does one go about calculating a Euler brick?”
All variables are positive integers, throughout.
The equations to solve are: $$x^2+y^2=u^2$$ $$y^2+z^2=v^2$$ $$x^2+z^2=w^2$$
Clearly the system is homogenous, so primitive solutions can easily generate others. Looking $\mod4$, only one of $x,y,z$ is odd, and the other two $\equiv0\pmod4$. Also, the shape of the brick does not change with the way it’s viewed, so the order of the values of $x,y,z$ is irrelevant. IMHO, the important variable are $x,y,z$, not $(u,v,w)$.
Without loss of generality define $x\equiv1\pmod2$ and $y<z$
One method is to produce Pythagorean triples, within a range of values of $x$, and then find those that fit. Start with the standard formula. $x^2+m^2=n^2$, $$x=k(a^2-b^2)$$ $$m=2kab$$ $$n=k(a^2+b^2)$$
Where $k,a,b$ are positive integers, $a+b\equiv1\pmod2$ and $\gcd(a,b)=1$
Use this to produce a list of $(x,m)$ pairs, using odd $k$, then sort the list to $(x,m)$ sequence.
Often, but not when $x$ is prime, a given $x$ will have more than one value for $m$.
When $m_i^2+m_j^2=r^2$ for some $i,j,r$ and $i<j$, we have a solution.
For example, with $x=85$, using $(a,b,k)=(11,6,1),(3,2,17),(9,8,5),(43,42,1)$ we have $$m=132,204,720,3612$$
As $132^2+720^2=732^2$ we equate $(y,z)$ to these values and obtain,
$$(x,y,z),(u,v,w)=(85,132,720),(157,732,725)$$
Example 2. When $x=6325$, $$m=528,5100,5520,5796,6240,13860,15180,31692,32760,34500,37548,72600,78936,159960,165252,173880,363660,800100,869676,1818432,4000560,20002812)$$
$$(x,y,z),(u,v,w)=(6325,528,5796),(6347,5820,8579)$$
$$(x,y,z),(u,v,w)=(6325,5520,5796),(8395,8004,8579)$$
However, that last one is $23$ times the smaller solution, $(275,240,252)$.
Please let me know if you need more details.
https://sites.google.com/site/tpiezas/0021
http://www.math.harvard.edu/~knill/various/eulercuboid/lecture.pdf
Update 15 May 2017.
Bye one, get one free.
If $(x,y,z)$ is a solution, then $(xy,xz,yz)$ is also a solution.
For example $(117,44,240)$ gives $(429,880,2340)$, when reduced to a primitive solution.
Among others, Euler and Saunderson devised ways to generate an infinitude of euler bricks, eg Saunderson starts with a pythagorean triple $(u,v,w)$ and sets $a=u|4v^2-w^2|, b=v|4u^2-w^2|, c=4uvw$. However these methods do not generate all (primitive) Euler bricks, so we cannot use them to prove the nonexistence of a perfect cube.
In principle all (primitive) euler bricks can be found like this: A 'candidate euler brick' is constructed from two primitive pythagorean triples $(k,l,m)$ and $(r,s,t)$ having odd $k$ and $r$. Set $a=\text{lcm}(k,r)$ and scale the triples up by a factor $(a/k)$ and $(a/r)$ respectively, to define $b=la/k$ and $c=sa/r$. Two of the three face diagonals are integers, because of the triples. If $b^2+c^2$ is a square, the third is integral as well and the candidate is an euler brick.
I tested over a billion pairs of triples, (all pairs of the 50765 primitive triples $(m^2-n^2, 2mn, m^2+n^2)$ with coprime $1\le n \lt m \le 500)$, and found 1013 primitive euler bricks. (Of which 315 had an odd side $< 10^6$, improving a bit on M. Kraitchik, 1945)
Generating Euler Bricks is hard with just a formula, so I made a Python code to generate them for you. Of course, this will generate already known bricks, and will take a massive amount of time. For with each for loop, you have to cube the range and that is the number of sets it must test. Some improvement could be made, and I am working on that, but it is taking a little while because there aren't really any known conformities that a Euler Brick follows. Here is the code pasted below. You can change the number in the range part of the for loops. This will cause more bricks to be calculated. You can run this code in something like PyCharm, or an online compiler like RepL.it
import math a = 1 b = 1 c = 1 for a in range(1, 800): for b in range(1, 800): for c in range(1, 800): x = math.sqrt(a ** 2 + b ** 2) y = math.sqrt((a ** 2) + (c ** 2)) z = math.sqrt((b ** 2) + (c ** 2)) if x == int(x): if y == int(y): if z == int(z): values = [a, b, c] print(values)
That code is fairly simple, but I think there could possibly be some improvement. Instead of testing to see if each number is an integer, you can instead test them at the same time and speed up the process probably saving around 300 ms per test. I don't know quite how to do this but I am working on it. You could also change the range values (the bolded text) so that you are only calculating past numbers that we don't know if they're Euler Bricks. So you could change those numbers to something like 10^13 so that it is something different. This will allow you to generate new Euler Bricks that haven't been discovered, although it will take longer to generate individual Euler Bricks. I would recommend running this and finding 1 Euler Brick then changing the range values, because the more numbers it compares, the laggier it will get. I am just now realizing that the code is not in line correctly so I'm not really sure what to do about that. I will instead also post a link to it on this thing. https://replit.com/@JacksonVliet/Euler-Brick#main.py
