Parametric solutions of Euler Brick?
Hauke Reddmann
(a quader with integer sides and face diagonals -
one with additional space diagonal is still
sought and conjectured to be nonexistent)
can't answer my questions anyway so I don't
bother to explain it.
Oh. Just did. Anyway. :-)
The Euler parametric solution is known since,
eh, Euler, but are there more? All the review
articles I know are extremely vague.
If you can answer, please give it as p,q,r
with (p^2-1)(q^2-1)(r^2-1)=8pqr, the "generators".
--
Hauke Reddmann <:-EX8 fc3...@uni-hamburg.de
His-Ala-Sec-Lys-Glu Arg-Glu-Asp-Asp-Met-Ala-Asn-Asn
Proginoskes
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). (I have a copy of that
paper at home, which has a parameterization of the rational solutions
to
x^2 + y^2 + z^2 = 1,
and which uses the fact that 0, 1, and 144 are the only perfect squares
in the Fibonacci sequence.)
--- Christopher Heckman
Geoff Harland
>
> 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). (I have a copy of that
> paper at home, which has a parameterization of the rational solutions
> to
>
> x^2 + y^2 + z^2 = 1,
>
> and which uses the fact that 0, 1, and 144 are the only perfect squares
> in the Fibonacci sequence.)
>
> > bother to explain it.
> > Oh. Just did. Anyway. :-)
> > The Euler parametric solution is known since,
> > eh, Euler, but are there more? All the review
> > articles I know are extremely vague.
> > If you can answer, please give it as p,q,r
> > with (p^2-1)(q^2-1)(r^2-1)=8pqr, the "generators".
If it has really been proved that no Euler bricks exist, I would have
thought that Wolfram's MathWorld website would have mentioned that by now.
Yet when I recently visited the webpages at
http://mathworld.wolfram.com/PerfectCuboid.html
and
http://mathworld.wolfram.com/UnsolvedProblems.html
(about 4 to 6 weeks ago, and yet again after reading your above message),
they both mention that a "Perfect Cuboid" has yet to be found. (Wolfram's
MathWorld website refers to a brick with integer edges and face diagonals as
an "Euler Brick", and to a brick which also has an integer space diagonal as
a "Perfect Cuboid". And the second URL listed above also advises that it was
last updated on 12 June 2006.)
So has the proof which you referred to yet to be accepted by the
mathematical community at large, or is Wolfram Research a bit slow on the
uptake regarding this matter?
(If one were to be pedantic, it is not untrue to continue to state that a
"Perfect Cuboid" has yet to be found, if in fact it really has been proved
that none exist. That said, I don't think that I'm being unreasonable in
suggesting that it would still be appropriate for the former webpage to
state that it has been proved that none exist, and for the "Perfect Cuboid"
item to be removed from the list contained within the latter webpage.)
Regards,
Geoff Harland.
g_ha...@optum12net.cos.au
(Transpose m & s in address
provided - then also remove
cuberoot of 10^3 + 9^3 - 1^3.)
Aluminium Holocene Holodeck Zoroaster
(where "one squared" is the body diagonal)..
I've been working on a proof, as well,
for quite some time; hopefully, I won't need
to use that fact about perfect secondpowers!
> I thought nonexistence was proved a couple of years ago, by a student
> from Georgia (Georgia Europe, not Georgia USA). (I have a copy of that
> paper at home, which has a parameterization of the rational solutions
> to
>
> x^2 + y^2 + z^2 = 1^2,
>
> and which uses the fact that 0, 1, and 144 are the only perfect squares
> in the Fibonacci sequence.)
thus:
like, *HOW* does MDT resolve the wave-particle duality, or
give a proper burial for Schroedinger's poor, silly cat?
thus:
I realize that the lack of response could be becuase
pf "political correctness," which may be how it is that
we are headed for Iran and "the" Sudan and a military draft....
anyway, what is the *current* usage for Base One,
if any ... ugit?
> would "unary" be better, or just "unit," considering that...
> it or uit?
thus:
Dick Cheeny, Don Rumsfeld and Osama bin Latin form a mission
to Darfur, to prevent a war instead of to start one:
if Darfur is "100% Muslim," then
what's really going on, there?...
is it just aother British Quag for USA soldiers to get bogged
into, with Iran, Iraq, Afghanistan et al ad vomitorium,
under auspices of the UN and NATO?...
why won't the Bruin publish the fact of Islam on the ground,
therein?
thus:
Why doesn't the [UCLA Daily] Bruin report that
Darfur's populace is "100%" Muslim,
according to the DAC's sponsor,
Terry Saunders?...
"99%" was the figure given
by Brian Steidle, when I finally found
him at the Hammer, after everyone else
had left (he, his friend & I were the
very last to leave!)...
What could it possibly mean?
--The Other Side (if it exists)
Aluminium Holocene Holodeck Zoroaster
by just the rectangular edges of the "la boite parfait."
> (where "one squared" is the body diagonal)..
> > x^2 + y^2 + z^2 = 1^2,
thus:
Dick Cheeny, Don Rumsfeld and Osama bin Latin form a mission
to Darfur, to prevent a war instead of to start one:
if Darfur is "100% Muslim," then
what's really going on, there?
is it just aother British Quag for USA soldiers to get bogged
into, with Iran, Iraq, Afghanistan et al ad vomitorium,
under auspices of the UN and NATO?
why won't the Bruin publish the fact of Islam on the ground,
Gerry Myerson
Hauke Reddmann <fc3...@uni-hamburg.de> 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)
> can't answer my questions anyway so I don't
> bother to explain it.
> Oh. Just did. Anyway. :-)
> The Euler parametric solution is known since,
> eh, Euler, but are there more? All the review
> articles I know are extremely vague.
> If you can answer, please give it as p,q,r
> with (p^2-1)(q^2-1)(r^2-1)=8pqr, the "generators".
Is this of any interest?
MR1285758 (95e:11038)
Colman, W. J. A.(4-ELON2)
A perfect cuboid in Gaussian integers.
Fibonacci Quart. 32 (1994), no. 3, 266--268.
11D09
The author gives a parametric family of cuboids with two rational
integer sides, three rational integer face diagonals and rational
integer internal diagonal. The third side will, in general, be
irrational or complex. A perfect cuboid in Gaussian integers is
obtained by a suitable choice of the parameters.
Reviewed by Ákos Pintér
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Gerry Myerson
"Proginoskes" <CCHe...@gmail.com> 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).
News to me.
> (I have a copy of that
> paper at home, which has a parameterization of the rational solutions
> to
>
> x^2 + y^2 + z^2 = 1,
>
> and which uses the fact that 0, 1, and 144 are the only perfect squares
> in the Fibonacci sequence.)
I hope you'll find that paper & supply some more details.
Gerry Myerson
> 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)
> can't answer my questions anyway so I don't
> bother to explain it.
> Oh. Just did. Anyway. :-)
> The Euler parametric solution is known since,
> eh, Euler, but are there more? All the review
> articles I know are extremely vague.
> If you can answer, please give it as p,q,r
> with (p^2-1)(q^2-1)(r^2-1)=8pqr, the "generators".
Another paper that might be of interest is
Reports of the
Mathematical Institute
University of Leiden
Report MI 2001-12
R.M. van Luijk
On Perfect Cuboids
Abstract. Perfect cuboids are rectangular parallelepipeds of which all
sides, face diagonals and full diagonals are integers. They correspond
to rational points on an explicit algebraic surface. In this paper we
prove that this singular surface is of general type. We also analyse a
surface describing cuboids of which all the lengths above are integral,
except possibly of one of the face diagonals. This is shown to be a
singular K3-surface and we compute its full Néron-Severi group.
It's available at
http://www.math.leidenuniv.nl/reports/2001-12.shtml
Hauke Reddmann
> Is this of any interest?
> MR1285758 (95e:11038)
> Colman, W. J. A.(4-ELON2)
> A perfect cuboid in Gaussian integers.
> Fibonacci Quart. 32 (1994), no. 3, 266--268.
> 11D09
> The author gives a parametric family of cuboids with two rational
> integer sides, three rational integer face diagonals and rational
> integer internal diagonal. The third side will, in general, be
> irrational or complex. A perfect cuboid in Gaussian integers is
> obtained by a suitable choice of the parameters.
I could have given a perfect Gaussian cuboid twenty years ago
if I had thought that would be of any interest :-)
Answer to CH: Yup, wanna see the proof, if it is understandable
to an amateur. (I doubt I will understand a word of the other
publication mentioned here, but I try :-)
Proginoskes
I'll try to find it then, and post an outline sometime soon.
I do recall there was one lemma which looked iffy, but it involved
something from Number Theory that I wasn't that familar with.
He didn't seem too anxious to get it published for some reason. This is
strange, because his "poster display" included a printout of the
MathWorld Perfect Cuboid page, so he knew its significance.
--- Christopher Heckman
Aluminium Holocene Holodeck Zoroaster
to "have" one side that is complex?
> > The author gives a parametric family of cuboids with two rational
> > integer sides, three rational integer face diagonals and rational
> > integer internal diagonal. The third side will, in general, be
> > irrational or complex. A perfect cuboid in Gaussian integers is
> > obtained by a suitable choice of the parameters.
>
> I could have given a perfect Gaussian cuboid twenty years ago
> if I had thought that would be of any interest :-)
thus:
Aluminium Holocene Holodeck Zoroaster
that is easy to state & prove )using Euler's parameterizations or
equiv.,
I think). to find a sufficient condition to prove nonexistance should
be somewhat harder to do, though true.
teh answer to your question is,
Wolframitism can't do numbertheory worth beans,
as of yet (although you'd *think* that Base One would
be a simple matter to emulate, thereby .-)
> So has the proof which you referred to yet to be accepted by the
> mathematical community at large, or is Wolfram Research a bit slow on the
> uptake regarding this matter?
thus:
Aluminium Holocene Holodeck Zoroaster
a sphere of diameter one is all that you need. or,
was that what the author of the below paper, meant?
> > If you can answer, please give it as p,q,r
> > with (p^2-1)(q^2-1)(r^2-1)=8pqr, the "generators".
> Abstract. Perfect cuboids are rectangular parallelepipeds of which all
> sides, face diagonals and full diagonals are integers. They correspond
> to rational points on an explicit algebraic surface. In this paper we
> prove that this singular surface is of general type. We also analyse a
> surface describing cuboids of which all the lengths above are integral,
> except possibly of one of the face diagonals. This is shown to be a
> singular K3-surface and we compute its full Néron-Severi group.
> http://www.math.leidenuniv.nl/reports/2001-12.shtml
thus:
the fact that you chose a triviaility to correct, is interesting;
magnetic moments of planets is interesting, though!
restating an ill-posed hypothesis could result
in dyscovering the problem with it ... or otherwise, if
there's a good, plausible connection between the metaphors.
> > I agree, I would not want anything to do with people who are not
> > anything short of "scientific and logical". I usually quote sources
> I usually do not like making a post just to correct some of my grammar.
thus:
there is no necessity for a perfect box to exist;
that is easy to state & prove
(using Euler's parameterizations or equiv.,
I think). to find a sufficient condition to prove nonexistance should
be somewhat harder to do, though true....
the answer to your question is,
Wolframitism can't do numbertheory worth beans,
as of yet (although you'd *think* that Base One would
be a simple matter to emulate, thereby .-)
thus:
Proginoskes
Proginoskes wrote:
> Gerry Myerson wrote:
> > In article <1163021324.0...@e3g2000cwe.googlegroups.com>,
> > "Proginoskes" <CCHe...@gmail.com> 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, it was only last year.
> > News to me.
> >
> > > (I have a copy of that
> > > paper at home, which has a parameterization of the rational solutions
> > > to
> > >
> > > x^2 + y^2 + z^2 = 1,
> > >
> > > and which uses the fact that 0, 1, and 144 are the only perfect squares
> > > in the Fibonacci sequence.)
> >
> > I hope you'll find that paper & supply some more details.
>
> I'll try to find it then, and post an outline sometime soon.
First of all, a link:
http://www.mualphatheta.org/Science_Fair/Science_Fair_Winners.html
This shows a picture of Lasha Margishvili, along with his poster
presenting his solution.
Now for the first part of the proof. (It is longer than I remember.) If
anyone can find a nice proof of Proposition 7, then the whole thing
should go through.
Proposition 7: The expression
(m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4)
is a perfect square iff one of the following holds:
(a) P^2 = n^2 m,
(b) P^2 = n m^2,
(c) m = a^2, n = t, P = a^3 (where a and t are integers) OR
(d) m = t, n = a^2, P = a^3 (where a and t are integers).
----------
"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.]
[Section: Rectangular Parallelpiped of the First Class]
Let a, b, c denote the sides of the parallelpiped, and p, q, r the
diagonal, chosen so that
a^2 + b^2 = p^2
a^2 + c^2 = r^2
b^2 + c^2 = q^2
The parameterization of the positive rational solutions to x^2 + y^2 =
1 yields equations:
(1,2) a / p = 2 xi / (1 + xi^2), b / p = (1 - xi^2) / (1 +
xi^2)
(3,4) a / r = 2 zeta / (1 + zeta^2), c / r = (1 - zeta^2) / (1 +
zeta^2)
(5,6) b / q = 2 mu / (1 + mu^2), c / q = (1 - mu^2) / (1 +
mu^2)
for some positive rationals xi, zeta, mu. Multiplying these equations
yields Boltianski's equation:
8 xi mu zeta = (1 - xi^2)(1 - zeta^2)(1 - mu^2)
(also proven in Boltianski's "Pythagorean Tetrahedrons", Journal Kvant
1986, #8).
Proposition 3: If (xi, mu, zeta) are solutions to Boltianski's
equation, and two are rational, then the third is as well.
Proposition 4: If xi, mu, and zeta are rational solutions to B.'s
equation, then there exists an Euler brick. [proof: write
xi = m/n, mu = p/q. Then
a = 4 m n p q
b = 2 p q (n^2 - m^2)
c = (n^2 - m^2) (q^2 - p^2)
is an Euler brick.]
Proposition 5: If (xi, mu, zeta) is a solution to B's equation, then so
is
xi' = (1 - xi) / (1 + xi)
mu' = (1 - mu) / (1 + mu)
zeta' = (1 - zeta) / (1 + zeta).
[Section: Rectangular Parallelpiped of the Second Class]
We also suppose that
a^2 + b^2 + c^2 = L^2
for some integer L.
Proposition: The positive rational solutions to x^2 + y^2 + z^2 = 1 are
parameterized by
x = abs(T1^2 + T2^2 - 1) / (T1^2 + T2^2 + 1)
y = 2 T1 / (T1^2 + T2^2 + 1)
z = 2 T2 / (T1^2 + T2^2 + 1)
where T1 and T2 are positive rational numbers less than 1 such that
T1^2 + T2^2 is not 1.
Thus we may write
(7) a / L = abs(T1^2 + T2^2 - 1) / (T1^2 + T2^2 + 1)
(8) b / L = 2 T1 / (T1^2 + T2^2 + 1)
(9) c / L = 2 T2 / (T1^2 + T2^2 + 1)
[Section: Diophantine Rectangular Parallelpiped Does Not Exist]
Consider all the equations involving a, b, c, xi, mu, zeta, and L
above. WOLOG
T1^2 + T2^2 > 1.
Multiplying the equations (1) and (8), and multiplying the equations
(2) and (7), yields
(10) 4 xi T1 = (1 - xi^2) (T1^2 + T2^2 - 1)
Multiplying the equations (3) and (9), and multiplying the equations
(7) and (4), yields
(11) (1 - zeta^2) T1 = zeta (T1^2 + T2^2 - 1)
Multiplying the equations (5) and (9), and multiplying the equations
(6) and (8), yields
(12) 2 mu T2 = (1 - mu^2) T1
Solve for T2 in (12) and substitute it into (10). Write as a quadratic
equation in T1.
T1^2 [(1 - xi^2)(1 + mu^2)]^2 - T1 16 xi mu^2 - 4 mu^2 (1 -
xi^2) = 0
In order for T1 to be rational, the discriminant must be a perfect
square, so
(B) 16 xi^2 mu^2 + (1 - xi^2)^2 (1 + mu^2)^2
must be a perfect square (rational). Proposition 3 states that
(M) 16 xi^2 mu^2 + (1 - xi^2)^2 (1 - mu^2)^2
must also be a perfect square. Proposition 4 states that there is
another solution, where xi is replaced by
(1 - xi) / (1 + xi). This implies that
(B') (1 - xi^2)^2 mu^2 + xi^2 (1 - mu^2)^2
and
(M') (1 - xi^2)^2 mu^2 + xi^2 (1 + mu^2)^2
must be perfect (rational) squares.
Now assume that xi = m/k and mu = n/k. (B') and (M') show that the
following are perfect squares:
(1 - m^2 / k^2)^2 n^2 / k^2 + m^2 / k^2 (1 - n^2 / k^2)^2 and
(1 - m^2 / k^2)^2 n^2 / k^2 + m^2 / k^2 (1 + n^2 / k^2)^2
Consequently, (m^2 + n^2)(n^2 m^2 + k^4) and
(m^2 + n^2)(n^2 m^2 + k^4) - 4k^2 m^2 n^2 are prefect squares.
Hence, we have a Pythagorean triple; there exist P and Q such that
(m^2 + n^2)(n^2 m^2 + k^4) = (P^2 + Q^2)^2 and
2 P Q = 2 n m k
Write Q = n m k / P, substitute into the other equation above, rewrite
as a quadratic equation in k^2:
(k^2)^2 [P^4(m^2+n^2)-n^4 m^4] - k^2 (P^4 n^2 m^2)
+ (P^4 n^2 m^2 (m^2 + n^2) - P^8 = 0
Then solve for k^2.
[If anyone is still reading this ... The next Proposition is where I
feel the weak link of the chain of proof is.]
Proposition 7: The expression
(m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4)
is a perfect square iff one of the following holds:
(a) P^2 = n^2 m,
(b) P^2 = n m^2,
(c) m = a^2, n = t, P = a^3 (where a and t are integers) OR
(d) m = t, n = a^2, P = a^3 (where a and t are integers).
[END OF PART ONE]
-------------------
--- Christopher Heckman
Hauke Reddmann
> [END OF PART ONE]
OK, up to here everything looks trivial on the first
glance, the snag invariably will come when trying to reduce
the perfect square condition to something useful :-)
Anxiously waiting for Part 2 (the inconceivable one
that I *couldn't* have done 20 years ago) :-)
Virgil
"Proginoskes" <CCHe...@gmail.com> wrote:
> (Georgia Europe, not Georgia USA).
Is the Georgia that is not the USA in Europe or in Asia?
Proginoskes
Georgia, Eurasia, then. 8-) Wikipedia actually puts it in Western Asia.
--- Christopher Heckman
Proginoskes
> Proginoskes wrote:
> > Gerry Myerson wrote:
> > > In article <1163021324.0...@e3g2000cwe.googlegroups.com>,
> > > "Proginoskes" <CCHe...@gmail.com> 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
Proginoskes
Proginoskes wrote:
> [second half of proof omitted]
Well? Comments?
I guess there's nothing "obviously" wrong (or someone would have said
so right away).
--- Christopher Heckman
Hauke Reddmann
> Well? Comments?
> I guess there's nothing "obviously" wrong (or someone would have said
> so right away).
So what *has* been proved? No Euler Brick if Proposition 7
is true, but we still lack a proof of that?
Proginoskes
Hauke Reddmann wrote:
> Proginoskes <CCHe...@gmail.com> wrote:
>
> > Well? Comments?
>
> > I guess there's nothing "obviously" wrong (or someone would have said
> > so right away).
>
> So what *has* been proved? No Euler Brick if Proposition 7
> is true, but we still lack a proof of that?
One direction is obvious (to me); if the variables have the special
values, you end up with a perfect square. What I can't see is the
reverse ("only if"), that a perfect square leads to one of the cases.
Here is the statement of Proposition 7, for those of you who have
forgotten it:
Proposition 7: The expression
(m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4)
is a perfect square iff one of the following holds:
(a) P^2 = n^2 m,
(b) P^2 = n m^2,
(c) m = a^2, n = t, P = a^3 (where a and t are integers) OR
(d) m = t, n = a^2, P = a^3 (where a and t are integers).
--- Christopher Heckman
Tim Peters
> ...
> One direction is obvious (to me); if the variables have the special
> values, you end up with a perfect square. What I can't see is the
> reverse ("only if"), that a perfect square leads to one of the cases.
>
> Here is the statement of Proposition 7, for those of you who have
> forgotten it:
>
> Proposition 7: The expression
>
> (m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4)
>
> is a perfect square iff one of the following holds:
>
> (a) P^2 = n^2 m,
> (b) P^2 = n m^2,
> (c) m = a^2, n = t, P = a^3 (where a and t are integers) OR
> (d) m = t, n = a^2, P = a^3 (where a and t are integers).
I haven't followed this, so don't know whether I'm missing "hidden"
assumptions. Looks like:
m=15 n=20 P=60
makes the first expression a perfect square (104976000000000000 =
324000000^2) but doesn't satisfy any of #a through #d.
It's easy to find others. It seems to get substantially harder if the
conditions:
P^4 > n^4 m^2
and
P^4 > n^2 m^4
are added. Then, e.g.,
m=21 n=52 P=273
is the first I found with all of {m, n, P} > 0.
Proginoskes
Tim Peters wrote:
> [Proginoskes]
> > ...
> > One direction is obvious (to me); if the variables have the special
> > values, you end up with a perfect square. What I can't see is the
> > reverse ("only if"), that a perfect square leads to one of the cases.
> >
> > Here is the statement of Proposition 7, for those of you who have
> > forgotten it:
> >
> > Proposition 7: The expression
> >
> > (m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4)
> >
> > is a perfect square iff one of the following holds:
> >
> > (a) P^2 = n^2 m,
> > (b) P^2 = n m^2,
> > (c) m = a^2, n = t, P = a^3 (where a and t are integers) OR
> > (d) m = t, n = a^2, P = a^3 (where a and t are integers).
>
> I haven't followed this, so don't know whether I'm missing "hidden"
> assumptions. Looks like:
>
> m=15 n=20 P=60
>
> makes the first expression a perfect square (104976000000000000 =
> 324000000^2) but doesn't satisfy any of #a through #d.
A quick look through the paper only suggests that P has to divide
evenly into m*n*k. Your examples do this, even if k=1. I'll have to
take a closer look then. (The existence of such a small counterexample
suggests there's something strange going on.)
--- Christopher Heckman
Tim Peters
>>> ...
>>> One direction is obvious (to me); if the variables have the special
>>> values, you end up with a perfect square. What I can't see is the
>>> reverse ("only if"), that a perfect square leads to one of the cases.
>>>
>>> Here is the statement of Proposition 7, for those of you who have
>>> forgotten it:
>>>
>>> Proposition 7: The expression
>>>
>>> (m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4)
>>>
>>> is a perfect square iff one of the following holds:
>>>
>>> (a) P^2 = n^2 m,
>>> (b) P^2 = n m^2,
>>> (c) m = a^2, n = t, P = a^3 (where a and t are integers) OR
>>> (d) m = t, n = a^2, P = a^3 (where a and t are integers).
[Tim Peters]
>> I haven't followed this, so don't know whether I'm missing "hidden"
>> assumptions. Looks like:
>>
>> m=15 n=20 P=60
>>
>> makes the first expression a perfect square (104976000000000000 =
>> 324000000^2) but doesn't satisfy any of #a through #d.
[Proginoskes]
> A quick look through the paper only suggests that P has to divide
> evenly into m*n*k. Your examples do this, even if k=1.
Not sure I follow. If there's no restriction on k, then it's trivial that k
exist s.t. P | mnk, no matter what m, n, and P (=/= 0) may be.
It's true anyway that /most/ counterexamples I found had P | mn directly.
Here's one where it's more of a stretch:
m = 98 = 7^2 * 2
n = 259 = 7 * 37
P = 1127 = 7^2 * 23
That is, P|mnk here iff 23|k.
> I'll have to take a closer look then. (The existence of such a small
> counterexample suggests there's something strange going on.)
Seems that way. OTOH, while the integers "are small", the "prop 7
expression" grows quickly. In the <98, 259, 1127> example, it's
14593054812133845870592414670025 = 3820085707432995^2
That is, the arithmetic is tedious enough that it's almost tempting to use a
computer ;-)
>> ...
Proginoskes
There is a requirement that there is an integer Q such that P Q = m n
k; this is so that a particular Pythagorean triple can be constructed.
> It's true anyway that /most/ counterexamples I found had P | mn directly.
> Here's one where it's more of a stretch:
>
> m = 98 = 7^2 * 2
> n = 259 = 7 * 37
> P = 1127 = 7^2 * 23
>
> That is, P|mnk here iff 23|k.
>
> > I'll have to take a closer look then. (The existence of such a small
> > counterexample suggests there's something strange going on.)
>
> Seems that way. OTOH, while the integers "are small", the "prop 7
> expression" grows quickly. In the <98, 259, 1127> example, it's
>
> 14593054812133845870592414670025 = 3820085707432995^2
>
> That is, the arithmetic is tedious enough that it's almost tempting to use a
> computer ;-)
I don't know when I'll be able to get back to checking the paper, so
you don't need to find any more counterexamples until I find out how
they affect the other variables in the paper. 8-)
(OTOH, if the paper has to fail, at least it's failing where I
suspected it would. And after only one reading, at that. 8-).)
--- Christopher Heckman
> >> ...
Proginoskes
Proginoskes wrote:
> Tim Peters wrote:
> > [Proginoskes]
> > >>> ...
> > >>> One direction is obvious (to me); if the variables have the special
> > >>> values, you end up with a perfect square. What I can't see is the
> > >>> reverse ("only if"), that a perfect square leads to one of the cases.
> > >>>
> > >>> Here is the statement of Proposition 7, for those of you who have
> > >>> forgotten it:
> > >>>
> > >>> Proposition 7: The expression
> > >>>
> > >>> (m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4)
> > >>>
> > >>> is a perfect square iff one of the following holds:
> > >>>
> > >>> (a) P^2 = n^2 m,
> > >>> (b) P^2 = n m^2,
> > >>> (c) m = a^2, n = t, P = a^3 (where a and t are integers) OR
> > >>> (d) m = t, n = a^2, P = a^3 (where a and t are integers).
> >
> > [Tim Peters]
> > >> I haven't followed this, so don't know whether I'm missing "hidden"
> > >> assumptions. Looks like:
> > >>
> > >> m=15 n=20 P=60
> > >>
> > >> makes the first expression a perfect square (104976000000000000 =
> > >> 324000000^2) but doesn't satisfy any of #a through #d.
> > It's true anyway that /most/ counterexamples I found had P | mn directly.
> > Here's one where it's more of a stretch:
> >
> > m = 98 = 7^2 * 2
> > n = 259 = 7 * 37
> > P = 1127 = 7^2 * 23
> I don't know when I'll be able to get back to checking the paper, so
> you don't need to find any more counterexamples until I find out how
> they affect the other variables in the paper. 8-)
I have finished off the semester (except for one grade I have to
change; I confused the "passed while auditing" letter grade with the
"failed while auditing" letter grade* and need to fix this) and got
back to checking these solutions.
There is an equation elsewhere in the paper:
k^4*(P^4*(m^2+n^2)-n^4*m^4)-2*k^2*P^4*n^2*m^2+P^4*n^2*m^2*(m^2+n^2)-P^8=0
which has to have a positive integer as a solution. If you let (m,n,P)
= (21,52,273), or
(15,20,60), or (98,259,1127), then there are no integer solutions to
this equation, according to Maple.
(In fact, if (m,n,P) = (15,20,60), then you get a quadratic polynomial
in k; Margishvili and Meskhishvili seem not to have considered this a
necessary part of the proof.)
The expression (m^2 + n^2)(P^4 - n^4 m^2)(P^4 - n^2 m^4) actually
factors even further, into:
(m^2 + n^2) (m^2 n - P^2) (m^2 n + P^2) (n^2 m - P^2) (n^2 m + P^2).
--- Christopher Heckman
* Let's see if you can pass this quiz: The grades for "passed while
auditing (a class)" and "failed while auditing" at ASU are W and X.
Which one is which? (No fair googling!) 8-)
Proginoskes
Actually, this should be:
(11) (1 - zeta^2) T2 = zeta (T1^2 + T2^2 - 1)
rarm...@lausd.k12.ca.us
I don't know if trivial cases count for Prop 7 but set m=1, n=0 (or
vice-versa) and you end up with P^8 which is of course a perfect square
for all P.
-Saint Cad