Hunting for the Perfect Euler Brick
TPiezas
There are only about 5 identities solving the system,
a^2+b^2 = x^2
a^2+c^2 = y^2
b^2+c^2 = z^2
Euler found two. Can you find more?
P.S. A perfect Euler brick, of course, would have a^2+b^2+c^2 = t^2 as
well, but none are known.
See Form 8 at http://sites.google.com/site/tpiezas/007
- Titus
Don Stockbauer
> Hello all,
>
> There are only about 5 identities solving the system,
>
> a^2+b^2 = x^2
> a^2+c^2 = y^2
> b^2+c^2 = z^2
>
> Euler found two. Can you find more?
No.
OwlHoot
To find (or more likely rule out the existence of) a perfect Euler
brick I have focussed on trying to find general rational solutions
to: 1 + x^2, 1 + y^2, 1 + x^2 + y^2 = u^2, v^2, w^2
I've shown that it is equivalent to X + Y + Z + T, X.Y.Z.T = 0, 1
However, this doesn't seem to be a birational equivalence. Instead
one can go from the first set to the second and then back to the
first but with different values, which is slightly perplexing.
I think I've also found a general two parameter solution to the
second set, which then implies the same for the original set.
If this solution is correct (a big if, I must admit) that means
the system is not a K3 surface, contrary to what I seem to recall
reading, because a K3 surface has no rational parametrization.
It also means one could equate expressions from two overlapping
(homogenized) solutions to form a set of equations which could
then by taking resultants be reduced to either a polynomial in
one variable or (in the unlikely event that expressions in the
equations had common algebraic factors) a finite set of higher
dimensional varieties.
In the first, by far the most likely, case there could only be
a finite number of perfect Euler bricks and in principle these
could all be found by a finite search.
Cheers
John Ramsden
TPiezas
>
> - Show quoted text -
There are only 5 known identities that make the simultaneous eqns
x^2+y^2, x^2+z^2, y^2+z^2 as squares. To solve the additional
x^2+y^2+z^2 = t^2, two of these need a 16-deg polynomial, while the
other three yield the 8-deg,
v^8+68v^6-122v^4+68v^2+1 = t^2 (eq.1)
Of course, as the parametrizations are not complete, a negative result
to eq.1 does not imply the perfect Euler brick does not exist.
P.S. The fact there are four quadratics to be made squares does not
necessarily lead to a hyper-elliptic curve like eq.1. I know of a
system of four _quartics_ to be made squares which, after given a
constraint, reduced to merely two _quadratics_, hence was solvable by
an elliptic curve. Whether there is such a constraint that simplifies
the system for perfect Euler bricks is not known.
- Titus