Integral brick disproof from last month?
Ross Presser
#10, "Nafets Azereb" (seems like a likely psuedonym!) provides a very
simple proposed proof of the impossibility of a perfect cuboid (Euler
brick with all edges, all external diagonals and the internal diagonal
all integers). The page indicates the proof was submitted 31 March
2009. Has anyone found a flaw in it?
The principle in the proof -- three right triangles in the diagram
share a hypotenuse; no two distinct Pythagorean triples share a
hypotenuse; therefore at least one side each of two of the triangles
must be noninteger -- seems so simple that I cannot believe it has
never been proposed before in the centuries since Euler.
Gerry Myerson
<0f32a930-9cb8-490e...@w35g2000prg.googlegroups.com>,
Ross Presser <rpre...@gmail.com> wrote:
16^2 + 63^2 = 65^2 = 33^2 + 56^2.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Virgil
<0f32a930-9cb8-490e...@w35g2000prg.googlegroups.com>,
Ross Presser <rpre...@gmail.com> wrote:
It is false that no two *distinct* Pythagorean triples can share a
hypotenuse, unless your *distinct* means the same as my *primitive*.
A Pythagorean triple is called *primitive* when the greatest common
positive factor of its sides is 1, so that 3,4,5 is primitive but 6,8,10
is not.
Without the restriction of primitiveness, there can be as many differing
Pythagorean triples as one likes all with the same hypotenuse.
E.G.:
C = 25 has [7,24,25} and {15,20,25}
C = 65 has {33,56,65}, {39,52,65} and {25,60,65}
For a Given C to have any primitive triple, C must be a product of
primes all congruent to 1 mod 4, and the more such factors, the more
non-primitive solutions, in a complicated sort of way.
Virgil
<0f32a930-9cb8-490e...@w35g2000prg.googlegroups.com>,
Ross Presser <rpre...@gmail.com> wrote:
One can have as many distinct Pythagorean triples as one likes sharing a
common hypotenuse.
There are a number of ways of doing it based on prime factors of the
hypotenuse which are congruent to 1 mod 4. For example 5, 13, 17, 29,
37, and so on.
Each of these factors over the Gaussian integers into two complex
conjugate Gaussian primes, where i^2 = -1 in the following.
For example, 5 = (2 + i)*(2 - i), and 13 = (3 + 2*i)(3 - 2*i)
Note that for 5, (2 + i)^2 = (2 - i)^2 =3 + 4*i and 3^2 + 4*2 = 5^2
For 13, (3 + 2*i)^2 = (3 - 2*i)^2 = 5 + 12*i and 5^2 + 12^2 = 13^2
Now consider [(2+i)*(3-2*i)]^2 = [4 + 7*i]^2 = -33 + 56*i
Gerry Myerson
Virgil <Vir...@gmale.com> wrote:
> In article
> <0f32a930-9cb8-490e...@w35g2000prg.googlegroups.com>,
> Ross Presser <rpre...@gmail.com> wrote:
>
> > Linked from http://unsolvedproblems.org/index_files/Solutions.htm at
> > #10, "Nafets Azereb" (seems like a likely psuedonym!) provides a very
> > simple proposed proof of the impossibility of a perfect cuboid (Euler
> > brick with all edges, all external diagonals and the internal diagonal
> > all integers). The page indicates the proof was submitted 31 March
> > 2009. Has anyone found a flaw in it?
> >
> > The principle in the proof -- three right triangles in the diagram
> > share a hypotenuse; no two distinct Pythagorean triples share a
> > hypotenuse; therefore at least one side each of two of the triangles
> > must be noninteger -- seems so simple that I cannot believe it has
> > never been proposed before in the centuries since Euler.
>
> It is false that no two *distinct* Pythagorean triples can share a
> hypotenuse, unless your *distinct* means the same as my *primitive*.
Primitive or not, it's false.
> A Pythagorean triple is called *primitive* when the greatest common
> positive factor of its sides is 1, so that 3,4,5 is primitive but 6,8,10
> is not.
>
> Without the restriction of primitiveness, there can be as many differing
> Pythagorean triples as one likes all with the same hypotenuse.
Even with the restriction.
> E.G.:
>
> C = 25 has [7,24,25} and {15,20,25}
> C = 65 has {33,56,65}, {39,52,65} and {25,60,65}
and 16, 63, 65.
> For a Given C to have any primitive triple, C must be a product of
> primes all congruent to 1 mod 4, and the more such factors, the more
> non-primitive solutions, in a complicated sort of way.
Also the more primitive solutions, and not in such
a complicated way.
Gerry Myerson
Virgil <Vir...@gmale.com> wrote:
(2 + i) ( 3 - 2 i) = what???
Virgil
Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
> In article <Virgil-EC89F2....@bignews.usenetmonster.com>,
> Virgil <Vir...@gmale.com> wrote:
>
> > In article
> > <0f32a930-9cb8-490e...@w35g2000prg.googlegroups.com>,
> > Ross Presser <rpre...@gmail.com> wrote:
> >
> > > Linked from http://unsolvedproblems.org/index_files/Solutions.htm at
> > > #10, "Nafets Azereb" (seems like a likely psuedonym!) provides a very
> > > simple proposed proof of the impossibility of a perfect cuboid (Euler
> > > brick with all edges, all external diagonals and the internal diagonal
> > > all integers). The page indicates the proof was submitted 31 March
> > > 2009. Has anyone found a flaw in it?
> > >
> > > The principle in the proof -- three right triangles in the diagram
> > > share a hypotenuse; no two distinct Pythagorean triples share a
> > > hypotenuse; therefore at least one side each of two of the triangles
> > > must be noninteger -- seems so simple that I cannot believe it has
> > > never been proposed before in the centuries since Euler.
> >
> > It is false that no two *distinct* Pythagorean triples can share a
> > hypotenuse, unless your *distinct* means the same as my *primitive*.
>
> Primitive or not, it's false.
For non-primitive, the examples below show different Pythagorean triples
sharing a hypotenuse.
>
> > A Pythagorean triple is called *primitive* when the greatest common
> > positive factor of its sides is 1, so that 3,4,5 is primitive but 6,8,10
> > is not.
> >
> > Without the restriction of primitiveness, there can be as many differing
> > Pythagorean triples as one likes all with the same hypotenuse.
>
> Even with the restriction.
>
> > E.G.:
> >
> > C = 25 has [7,24,25} and {15,20,25}
> > C = 65 has {33,56,65}, {39,52,65} and {25,60,65}
>
> and 16, 63, 65.
>
> > For a Given C to have any primitive triple, C must be a product of
> > primes all congruent to 1 mod 4, and the more such factors, the more
> > non-primitive solutions, in a complicated sort of way.
>
> Also the more primitive solutions, and not in such
> a complicated way.
The number of *primitive* solutions for a given hypotenuse, C, appears
to be 2^(n-1) where n is the number of distinct prime factors of C which
Ross Presser
proposed solution).
bleuprint
Do there exist analogous problems where a solution exists only with VERY big numbers?
bleuprint
Martin Musatov
These integers do exist and are well associated with Musatov's proof
P==NP established prior to this discussion. I believe how Musatov put
it was, "Sometimes the most difficult problems have the most
solutions". Some numbers we do not simply refer to by names but ideas.
ge...@math.mq.edu.au
wrote:
> The integral brick is one of many problems with integer numbers, where up to now no solution is found for even VERY big numbers.
> Do there exist analogous problems where a solution exists only with VERY big numbers?
> bleuprint
x^2 - n y^2 = 1 has solutions in positive integers x and y
for every positive squarefree integer n, but even for some
moderate sized n the smallest solution is enormous. This is
called Pell's equation.
--
GM
bleuprint
> and y
> for every positive squarefree integer n, but even for
> some
> moderate sized n the smallest solution is enormous.
> This is
> called Pell's equation.
> --
> GM
Hi Gerry,
that's what I was looking for, it's already something but for n<13 we need only 2 digits for x and y .
See http://mathworld.wolfram.com/PellEquation.html
The first "nearly" big number for x appears for n=61. But an equation as x²-61.y²=1 can hardly be called a simple formulation, certainly not as simple as the "integer brick" formulation.
Looking for solutions using brute force programming (three loops) is very quickly done w.r.t. the time that has been used for the brick problem to find NO solution.
bleuprint
PS: There are no solutions for n=m²
Gerry Myerson
<32220199.12947.1243168...@nitrogen.mathforum.org>,
bleuprint <patrick....@versateladsl.be> wrote:
> > x^2 - n y^2 = 1 has solutions in positive integers x and y for
> > every positive squarefree integer n, but even for some moderate
> > sized n the smallest solution is enormous. This is called Pell's
> > equation.
>
> Hi Gerry,
> that's what I was looking for, it's already something but for n<13 we need
> only 2 digits for x and y .
>
> See http://mathworld.wolfram.com/PellEquation.html
>
> The first "nearly" big number for x appears for n=61. But an equation as
> x��-61.y��=1 can hardly be called a simple formulation, certainly not as
> simple as the "integer brick" formulation.
Simplicity is in the eye of the beholder.
Did I mention the congruent number problem? Given a positive,
squarefree integer n, find (if it exists) a right triangle with
rational sides and area n. I recommend n = 157; if you give up,
try http://www.math.umd.edu/~eve/cong_num.html
Oh, and stick to keyboard characters or you get weird things like
the copyright symbols instead of x^2.
bleuprint
> things like
> Simplicity is in the eye of the beholder.
I agree and
> Did I mention the congruent number problem? Given a
> positive, squarefree integer n, find (if it exists)
> a right triangle with rational sides and area n.
seen from my eye, also here there is already a simple solution with n=6, a=3, b=4 and c=5.
But back to the brick, I found following formulation somewhat related to the congruent number problem (perhaps it is a known one):
Does there exist an integer Heronian acute triangle with all sides perfect squares.
If not, an integer brick does not exist.
To see the relation with an integer brick, express the product of its volume times its space diagonal in function of the surface diagonals. You get Heron's formula for a triangle with sides equal to the square of the surface diagonals.
bleuprint
PS: In an integer Heronian triangle all sides and the area are integers.
bleuprint
> with all sides perfect squares.
a^4 + n*(b^2) = c^4
bleuprint
Gerry Myerson
<23773502.16519.1243258...@nitrogen.mathforum.org>,
bleuprint <patrick....@versateladsl.be> wrote:
Huh? Take a and c to be your favorite numbers,
n = c^4 - a^4, and b = 1.
Gerry Myerson
<29147225.15442.1243242...@nitrogen.mathforum.org>,
bleuprint <patrick....@versateladsl.be> wrote:
> > Did I mention the congruent number problem? Given a
> > positive, squarefree integer n, find (if it exists)
> > a right triangle with rational sides and area n.
> seen from my eye, also here there is already a simple solution with n=6, a=3,
> b=4 and c=5.
Yes. Now try to find a solution with n = 157.
bleuprint
About my question:
>> Does there exist an integer Heronian acute triangle
Yes there exist such triangles. The smallest I found was one with sides 114^2, 146^2 and 158^2.
So it does not help to disprove the integer brick conjecture (neither does it prove that such a brick exists).
bleuprint
> >> Does there exist an integer Heronian acute
> triangle
> >> with all sides perfect squares?
> Yes there exist such triangles. The smallest I found
> was one with sides 114^2, 146^2 and 158^2.
> So it does not help to disprove the integer brick
> conjecture (neither does it prove that such a brick
> exists).
BIG SURPRISE AND DON'T IMITATE MY MISTAKE:
In a integer Heronian triangle also the altitudes are integers. Trying to check these on the example above I found that the altitude on 114^2 is equal to 146^2!!!! Thus it must be a right triangle and then we have a^4+b^4=c^4 and this is impossible (FLT). In other words I did not take enough care for the numerical precision of my computer. Checking the last digits showed the flaw.
Thus up to now I found NO integer Heronian triangles with all sides square integers. So there is still hope to disprove the existence of an integer Euler brick by disproving the existence of such Heronian triangle.
bleuprint
A^2=m^2+n^2
B^2=m^2*n^2-k^2
C^2=m^4+k^2
D^2=n^4+k^2
Is there a solution with A,B,C,D,m,n,k being integers>1.
bleuprint
bleuprint
> > > with all sides perfect squares.
> > Probably this problem is the same as finding an
> > integer solution for: a^4 + n*(b^2) = c^4
> n = c^4 - a^4, and b = 1.
Very nice answer but now I see I forgot TWO important conditions for the equation: a^4 + n*(b^2) = c^4
1) b>c>a>1
2) b^2<a^2+b^2
Having only the first condition we can have
n=5,b=4,c=3,a=1 or
n=6,b=20,c=7,a=1 or
n=7,b=5,c=4,a=3 or
n=13,b=60,c=19,a=17 etc...
but I found no ONE suiting both conditions.
bleuprint
PS:
1) Even without the second condition it's proved that there are no integer solutions for n=1 or n=2 (see http://mathworld.wolfram.com/DiophantineEquation4thPowers.html )
2) The second condition comes from the following fact: Suppose there is a Heronian triangle with square sides b^2,c^2,c^2 and b>c>a then its clear that b^2<a^2+b^2.
In such Heronian triangle call a' and b' the projections of sides a and c on side b^2. Then a'+b'=b^2 and
c'^2-a'^2=(c^2)^2-(a^2)^2 or with c'-a'=n we have:
n*(b^2)=c^4-a^4.
3) The Heronian triangle with square sides does not need to be "acute".
Gerry Myerson
<3507736.5413.12444820...@nitrogen.mathforum.org>,
bleuprint <patrick....@versateladsl.be> wrote:
> > > > Does there exist an integer Heronian acute triangle
> > > > with all sides perfect squares.
> > > Probably this problem is the same as finding an
> > > integer solution for: a^4 + n*(b^2) = c^4
> Gerry answered:
> > Huh? Take a and c to be your favorite numbers,
> > n = c^4 - a^4, and b = 1.
>
> Very nice answer but now I see I forgot TWO important conditions for the
> equation: a^4 + n*(b^2) = c^4
> 1) b>c>a>1
> 2) b^2<a^2+b^2
That second condition is just a^2 > 0, which is already taken care of
by the first condition, a > 1. Finding solutions still seems quite easy,
e.g., c = 20, a = 10, b = 100, n = 15.
> Having only the first condition we can have
> n=5,b=4,c=3,a=1 or
> n=6,b=20,c=7,a=1 or
> n=7,b=5,c=4,a=3 or
> n=13,b=60,c=19,a=17 etc...
> but I found no ONE suiting both conditions.
Then you've mistyped your 2nd condition
(or you're incapable of performing grade school arithmetic).
bleuprint
2) b^2 < c^2 + a^2
as could also be seen in the PS #2.
It's in fact the same as condition:
2a) n > c^2 - a^2
bleuprint
Gerry Myerson
<15686803.5811.12446191...@nitrogen.mathforum.org>,
bleuprint <patrick....@versateladsl.be> wrote:
OK, so would you mind putting the whole thing,
correctly, in one post?
bleuprint
> OK, so would you mind putting the whole thing,
> correctly, in one post?
Suppose we have an integral brick with integer sides a,b,c, integer face diagonals ab,ac,bc and integer space diagonal abc (ab,... is NOT a*b but ONE length). No pair can be the same as we should have then an irrational length. So let's have
(1) a<b<c then ab<ac<bc<abc
Expressed as a function of the face diagonals we have
(2a) a^2=(ab^2+ac^2-bc^2)/2
(2b) b^2=(ab^2-ac^2+bc^2)/2
(2c) c^2=(-ab^2+ac^2+bc^2)/2
(2d) abc^2=(ab^2+ac^2+bc^2)/2
Then
(3) a*b*c*abc=sqrt[(ab^2+ac^2-bc^2)/2 * (ab^2-ac^2+bc^2)/2 * (-ab^2+ac^2+bc^2)/2 * (ab^2+ac^2+bc^2)/2]
The right side is Heron's formula for the area of a triangle with sides ab^2,ac^2 an bc^2. And as the left side is an integer, so the area of this triangle is also an integer. This triangle is thus an integer heronian triangle.
Let ab" and ac" be the projections of side ab^2 and ac^2 on the biggest side bc^2, and let h" be the altitude on side bc^2. Then we have:
(4a) ac"^2+h"^2=ac^4
(4b) ab"^2+h"^2=ab^4
(4a) minus (4b) with ac"+ab"=bc^2 gives:
(5) (ac"-ab")*bc^2=ac^4-ab^4
and with ac"-ab"=n we have finally:
(6) n*(bc^2)=ac^4-ab^4
with the conditions
i) 1<ab<ac<bc (*)
ii) bc^2<ac^2+ab^2 or using (6) ii') n>ac^2-ab^2
Suppose n is also an integer (**).
There are no integral solutions for (6) with n=1 or n=2 (***).
see http://mathworld.wolfram.com/DiophantineEquation4thPowers.html
Then n cannot be a square as (n*bc)^2=ac^4-ab^4 is also impossible.
That's it so far, but first look to the remarks as there is a BIG problem.
bleuprint
Remarks:
(*) We can have c<ab as well as ab<ac. Then we have two cases:
(7) 1<a<b<c<ab<ac<abc or 1<a<b<ab<c<ac<bc<abc.
In the first case we have
i') 4<ab<ac<bc<abc and ii") 9<n>ac^2-ab^2
In the second case we have
i') 3<ab<ac<bc<abc and ii") 16<n>ac^2-ab^2
Careful considerations on (7) can perhaps restrict more the two conditions i and ii, but the next remark (**) has first to be cleared.
(**) "Suppose n is an integer", here there is a BIG problem. ac" and ab" are certainly rational so n=ac"-ab" must also be rational, but I have no proof that it must be an integer.
(***) There are integer solutions for (6) with (n,bc,ac,ab) as (5,4,3,1);(6,20,7,1);(7,20,8,6);(13,60,19,17) etc... They are OK for condition i but not for ii.
n=6,
bleuprint
> correctly, in one post?
>
> --
> Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for
> email)
See first reply hereafter on 6/11/09.
bleuprint
bleuprint
In a previous reply, Gerry asked:
> OK, so would you mind putting the whole thing,
> correctly, in one post?
I do it here, and use easy to remember variables.
Suppose we have an integral brick with integer sides a,b,c, integer face diagonals ab,ac,bc and integer space diagonal abc (ab,... is NOT a*b,... but ONE length). No pair can be the same as we should have then an irrational length. So let's have
(1) a<b<c then ab<ac<bc<abc
Expressed as a function of the face diagonals we have
(2a) a^2=(ab^2+ac^2-bc^2)/2
(2b) b^2=(ab^2-ac^2+bc^2)/2
(2c) c^2=(-ab^2+ac^2+bc^2)/2
(2d) abc^2=(ab^2+ac^2+bc^2)/2
Then
(3) a*b*c*abc=sqrt[(ab^2+ac^2-bc^2)/2 * (ab^2-ac^2+bc^2)/2 * (-ab^2+ac^2+bc^2)/2 * (ab^2+ac^2+bc^2)/2]
The right side is Heron's formula for the area of a triangle with sides ab^2,ac^2 an bc^2. And as the left side is an integer, so the area of this triangle is also an integer. This triangle is thus an integer heronian triangle.
Let ab" and ac" be the projections of side ab^2 and ac^2 on the biggest side bc^2, and let h" be the altitude on side bc^2. Then we have:
(4a) ac"^2+h"^2=ac^4
(4b) ab"^2+h"^2=ab^4
(4a) minus (4b) with ac"+ab"=bc^2 gives:
(5) (ac"-ab")*bc^2=ac^4-ab^4
and with ac"-ab"=n we have finally:
(6) n*(bc^2)=ac^4-ab^4 (**)
with the conditions
i) 1<ab<ac<bc (*)
ii) bc^2<ac^2+ab^2 or using (6) ii') n>ac^2-ab^2
There are no integer solutions for (6) with n=1 or n=2 (***).
see http://mathworld.wolfram.com/DiophantineEquation4thPowers.html
Thus n cannot be a square integer as (n*bc)^2=ac^4-ab^4 is impossible. As 2*(n*bc)^2=ac^4-ab^4 is also impossible n cannot be twice an integer square.
That's it so far, but first look to the remarks as there is a complication fo n.
bleuprint
Remarks:
(*) We can have c<ab as well as c>ab. Then we have two cases:
(7) 1<a<b<c<ab<ac<abc or 1<a<b<ab<c<ac<bc<abc.
In the first case we have then
i') 4<ab<ac<bc<abc and ii") n>ac^2-ab^2>=11
In the second case we have then
i') 3<ab<c<ac<bc<abc and ii") n>ac^2-ab^2>=20
Careful considerations on (7) can perhaps restrict more the two conditions i and ii, but the next remark (**) has first to be cleared.
(**) It's not sure that n is an integer. As ac"+ab"=bc^2 must be an integer, ac" and ab" are certainly rational. So n=ac"-ab" must also be rational, but I have no proof that n must be an INTEGER. It should be nice to have one.
OwlHoot
wrote:
>
> I do it here, and use easy to remember variables.
>
> Suppose we have an integral brick with integer sides a,b,c, integer face diagonals
> ab,ac,bc and integer space diagonal abc (ab,... is NOT a*b,... but ONE length).
What?! Denoting a single variable by ab, or abc in a problem like
this (or anywhere), is little short of insane!
Also (although I wouldn't want to sound too dogmatic about this),
all those inequalities seem fairly irrelevant. In the end, if you
find rational solutions then the inequalities will authomatically
be taken care of.
Most people consider the perfect cuboid problem in rational form,
namely to make 1 + x^2, 1 + y^2, x^2 + y^2, 1 + x^2 + y^2 all
squares of rational numbers ("perfect squares") for rational
x, y.
This is known to represent geometrically a so-called "surface of
general type", which has at most a finite number of rational
solutions (possibly none).
The weaker system 1 + x^2, 1 + y^2, 1 + x^2 + y^2, where one
face diagonal can be irrational, represents a so-called K3
surface, which can have an infinite number of rational
solutions but has no two-parameter solutions in the
form x, y = x(u, v), y(u, v).
Because a K3 surface also has an infinite number of lines,
this system can't be reduced by birational transforms to
a single curve such as an elliptic curve.
However, there are loads of curves in it, certainly elliptic
curves (although I haven't managed to express it as a family
of conics, or in so-called tubular form).
There's a bit more I could post, but I have one or two ideas
about how to reduce the finding of all perfect cuboids (or
excluding any) to a finite problem. So I'll leave it at
that for now.
Cheers
John R Ramsden
No pair can be the same as we should have then an irrational length.
So let's have
> (1) a<b<c then ab<ac<bc<abc
> Expressed as a function of the face diagonals we have
> (2a) a^2=(ab^2+ac^2-bc^2)/2
> (2b) b^2=(ab^2-ac^2+bc^2)/2
> (2c) c^2=(-ab^2+ac^2+bc^2)/2
> (2d) abc^2=(ab^2+ac^2+bc^2)/2
> Then
> (3) a*b*c*abc=sqrt[(ab^2+ac^2-bc^2)/2 * (ab^2-ac^2+bc^2)/2 * (-ab^2+ac^2+bc^2)/2 * (ab^2+ac^2+bc^2)/2]
> The right side is Heron's formula for the area of a triangle with sides ab^2,ac^2 an bc^2. And as the left side is an integer, so the area of this triangle is also an integer. This triangle is thus an integer heronian triangle.
> Let ab" and ac" be the projections of side ab^2 and ac^2 on the biggest side bc^2, and let h" be the altitude on side bc^2. Then we have:
> (4a) ac"^2+h"^2=ac^4
> (4b) ab"^2+h"^2=ab^4
> (4a) minus (4b) with ac"+ab"=bc^2 gives:
> (5) (ac"-ab")*bc^2=ac^4-ab^4
> and with ac"-ab"=n we have finally:
> (6) n*(bc^2)=ac^4-ab^4 (**)
> with the conditions
> i) 1<ab<ac<bc (*)
> ii) bc^2<ac^2+ab^2 or using (6) ii') n>ac^2-ab^2
> There are no integer solutions for (6) with n=1 or n=2 (***).
> seehttp://mathworld.wolfram.com/DiophantineEquation4thPowers.html
Bart Goddard
c1a8b3...@x1g2000prh.googlegroups.com:
> What?! Denoting a single variable by ab, or abc in a problem like
> this (or anywhere), is little short of insane!
I'm not so sure...
B.
--
Cheerfully resisting change since 1959.
James Waldby
> OwlHoot <raven...@googlemail.com> wrote ...:
>
>> What?! Denoting a single variable by ab, or abc in a problem
>> like this (or anywhere), is little short of insane!
>
> I'm not so sure...
I assume you (BG) are unsure that such notation falls
short of insanity, and have the notion that it doesn't
fall short. That could well be; but in many cases,
stupidity is probably a better explanation than insanity.
--
jiw
OwlHoot
> OwlHoot <ravensd...@googlemail.com> wrote in news:677015e7-54b7-4b10-b796-
> c1a8b3c4b...@x1g2000prh.googlegroups.com:
>
> > What?! Denoting a single variable by ab, or abc in a problem like
> > this (or anywhere), is little short of insane!
>
> I'm not so sure...
>
> B.
>
> --
> Cheerfully resisting change since 1959.
Well I was thinking of the confusion that would arise in transforms
and so forth. It's all too easy to make a mistake anyway; but when
you're carting around a variable called "abc" it's a near certainty!
Incidently, the system 1 + x^2, 1 + y^2, 1 + x^2 + y^2 all square
is birationally equivalent to all of the following (reusing x, y
each time):
(x^2 - y^2).(x^2.y^2 - 1) = z^2
x.y, (x^2 - 1)(y^2 - 1) = z^2, t^2
x^2 + 1, x^2 + y^4 = z^2, t^2
x.y, (x + 1).(y + 1) = z^4, t^2
x^2 + y^2 = z^2 + t^2 x.y.z.t = u^2 (in homogenous form)
x^2 - y^2, (x^2 - 1)(y^2 - 1) = z^2, t^2
Interestingly a weaker form of the final pair namely:
(x^2 - y^2)(x^2 - 1)(y^2 - 1) = z^2
is equivalent to:
X + Y = Z + T X.T.Z.T = u^4
which is a weaker form of the preceding pair because in
the latter each of X, Y, Z, T must be the same multiple
of a perfect square to give a solution to the original.
Also replacing X, Y, Z, T by X/u, Y/u, -Z/u, - T/u this
pair is equivalent to:
X + Y + Z + T = 0 X.Y.Z.T = 1
Then there are forms resembling more a Weierstrass
equation, such as
x.y.(x.y + 1).(x + y) = z^2
But as I said, no forms equivalent to the original
triplet have a two-parameter solution (although loads
of one parameter solutions).
But even more interestingly, an apparently weaker form of
x.y, (x^2 - 1)(y^2 - 1) = z^2, t^2
namely (reusing x, y, z again):
x.(x^2 - 1) = y.(y^2 - 1).z^2
*is* equivalent to the original. This is apparently an
isogeny.
Likewise, the original is equivalent to:
x.y, (x + y)(x.y + 1) = z^2, t^2
and again the apparently weaker form obtained by
multiplying this pair is *also* equivalent to the
original:
x.y.(x + y).(x.y + 1) = z^2
Cheers
John R Ramsden
OwlHoot
> On Fri, 12 Jun 2009 21:02:20 +0000, Bart Goddard wrote:
>
> >> What?! Denoting a single variable by ab, or abc in a problem
> >> like this (or anywhere), is little short of insane!
>
> > I'm not so sure...
>
> I assume you (BG) are unsure that such notation falls
> short of insanity, and have the notion that it doesn't
> fall short. That could well be; but in many cases,
> stupidity is probably a better explanation than insanity.
Yes, the word "insane" was a figure of speech. I'm sure
blueprint isn't insane or stupid, just a bit misguided
maybe using notation like that ;-)
Cheers
John R Ramsden
Bart Goddard
news:FbSdnRXq0tdOQK_X...@bresnan.com:
Well, one definition of "insanity" (supposedly Einstein's) is
"doing the same thing over and over again and expecting different
results."
How many versions of this post have we seen?
bleuprint
So let's have a look at the situation now,
Gerry is well informed about this kind of problems,
OwlHoot knows a lot about the "classical" approach,
Bleuprint has insane(*) variables (and don't understand the "mathematical violence" of OwlHoot) and
Bart is not sure if it's so insane at all.
Why not try to disprove(**) this problem altogether?
Going to the moon was teamwork as all big achievements in the last half century. If successful it will probably be not the first "internet proof" ( I don't know) but surely it will not be the last one, and in fact it's nicer than computerproofs and it's also amusement.
OwlHoot wrote something like:
1) "the classical approach to this problem is like ..."
but up to now it has neither been proved nor disproved with this "classical" approach.
2) thus if there is a disprove it will probably be with a "non classical approach" ("no solution at all" for the finite possibilities).
So my proposal is to continue in the direction of the first reply that comes in.
To be or not to be that's no question, enjoy it.
bleuprint or patrick if you prefer
PS:
*) The "insane" variables help to keep in touch with the original problem, and they are also inspired by a visual approach to the problem. It's imaginable in 3 dimensions and the Heronian square sided triangle reduces it into 2 dimensions. By the way the tangents of the incircle (it's radius must be an integer) divide each side ab^2 (or ac^2 or bc^2) into two parts a^2 and b^2 (or a^2 and c^2 or b^2 and c^2)). The altitude on the smallest side must also be an integer (look to the parametric equations for the sides of a Heronian triangle)
**) I suggest the disprove option but it can as well go the other way.
Bart Goddard
news:4531476.13380.12449326...@nitrogen.mathforum.org:
> Bart is not sure if it's so insane at all.
Correction: Bart is not sure if it's _short_ of insane.
I've spent a certain amount of time working on a similar
problem (rational sided triangles with rational median
and area) and I'm pretty sure the algebraic geometry
approach that Ramsden describes is the right one. And I'm
pretty sure that messing about with high-school algebra-
type manipulations won't lead anywhere.
Bart