Re: Parametric solutions of Euler Brick?



Proginoskes wrote:
Proginoskes wrote:
Gerry Myerson wrote:
In article <1163021324.077080.226470@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Proginoskes" <CCHeckman@xxxxxxxxx> wrote:

Hauke Reddmann wrote:
Those who don't know what an Euler brick is
(a quader with integer sides and face diagonals -
one with additional space diagonal is still
sought and conjectured to be nonexistent)

I thought nonexistence was proved a couple of years ago, by a student
from Georgia (Georgia Europe, not Georgia USA).

Actually, Georgia, (Western) Asia.

Actually, it was only last year.

----------

"The Diophantine Rectangular Parallelepiped (A Perfect Cuboid)"

Lasha Margishvili

Supervisor: Dr. Mamuka Meskhishvili

Tbilisi, Georgia, 2005

[A proof that there is no Euler Brick, as summarized by C C Heckman.]

[BEGINNING OF PART TWO]

We now consider cases (a)-(d).

If (a) is true, then k^2 = m^2, so we do not have a solution, because
of Proposition 6.

If (b) is true, then k^2 = n^2, so this is not true.

If (c) is true, then k = a^2 (which cannot happen, since then k = m) or

k = a^2 * sqrt((a^4 t^2 - a^8 + t^4) / (a^4 t^2 + a^8 - t^4))

and since m = a^2 and n = t,

k = m * sqrt((m^2 n^2 - m^4 + n^4) / (m^2 n^2 + m^4 - n^4)).

If (d) is true, then k = a^2 (again; this can't happen, since k = n
follows) or

k = n * sqrt((m^2 n^2 - n^4 + m^4) / (m^2 n^2 + n^4 - m^4)).

By symmetry, we only consider the value of k in case (c).

Now, let d = gcd(m,n), and m_1 = m / d, n_1 = n / d.

Proposition 8: The expression

m * sqrt((m^2 n^2 - m^4 + n^4) / (m^2 n^2 + m^4 - n^4))

is integer iff the equation m_1^2 n_1^2 + m_1^4 - n_1^4 = 1 has a
solution, with
m = m_1 d and n = n_1 d, and gcd(m_1,n_1) = 1.

Proof of Prop 8: Since k is an integer, either the expression under the
square root is a perfect (integer) square, or m "must be divided by the
denominator under square root". Since

gcd(m_1, m_1^2 n_1^2 + m_1^4 - n_1^4) = 1,

d^2 must be a multiple of m_1^2 n_1^2 + m_1^4 - n_1^4; then d^4 is a
multiple of
m^2 n^2 + m^4 - n^4. Then we get

m^2 n^2 + m^4 - n^4 = d^4, so
(**) m_1^2 n_1^2 + m_1^4 - n_1^4 = 1.

If the expression under the square root is a perfect square, then

(***) N^2 = (m_1^2 n_1^2 - m_1^4 + n_1^4)
/ (m_1^2 n_1^2 + m_1^4 - n_1^4)

From (***), we get:

m_1^2 n_1^2 (N^2 - 1) = (n_1^4 - m_1^4)(N^2 + 1)

gcd (m_1,n_1) = 1 implies that gcd(m_1^2 n_1^2, n_1^4 - m_1^4) = 1, so

P = (N^2 + 1)/(m_1^2 n_1^2) = (N^2 - 1)/(n_1^4 - m_1^4)

for some positive integer P.

This gives us a system of Diophantine equations:

N^2 + 1 = P(m_1 n_1)^2
N^2 - 1 = P(n_1^4 - m_1^4)

If we subtract the second expression from the first, we get

2 = P [(m_1 n_1)^2 - n_1^4 + m_1^4].

Hence P = 1 or 2.

If P = 1, then m_1^2 n_1^2 + m_1^4 - n_1^4 = 2. If we view this
equation as a quadratic equation in m_1^2, we have

m_1^2 = (-n_1^2 + sqrt(5 n_1^4 + 8))/2

The last digit of 5 n_1^4 + 8 is 3 or 8 (as n_1^4 ends in 0, 1, 5, or
6), which means it can't be a perfect square.

Hence P = 2. Then (**) is true. (End of proof of Prop 8.)

[Section: Fibonacci Enters]

The solutions to (**) are

m_1^2 = F_(2k-1) and n_1^2 = F_(2k),

where F_s is the s-th Fibonacci number. (Proof: _Mathemtical Gems III_,
No 9, 1985, MAA, "A Second Look at the Fibonacci and Lucas Numbers" /
K. R. S. Sastry, "A Fermat-Fibonacci Collaboration", _Crux
Mathematicorum No 5, Vol 23, 1997.)

Furthermore, the only Fibonacci numbers which are perfect squares are
F_0 = 0, F_1 = 1, F_2 = 1, and F_12 = 144. (J. H. E. Cohn, "Square
Fibonacci Numbers, Etc", _Fibonacci Quarterly_, vol 2, 1964, pp.
109-113. Online link:
http://math.asu.edu/~checkman/SquareFibonacci.html ).

Thus m_1 = 1, n_1 = 1, which means m = d = n, contradicting Proposition
6. Consequently, the Diophantine rectangular parallelepiped does not
exist.

[END OF PART TWO]

-------------------

--- Christopher Heckman

.



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