Re: Diophantine Approach...
- From: vision_1967@xxxxxxxxxxx
- Date: Thu, 14 Jun 2007 11:52:31 -0700
Thanks,
However if this is the case, then I _am_ doomed because I don't have
the luxury of being able to factor all cases. It is what it is. It
just seems to be that if a sphere is essentially a *ball* and I using
a marker place a point on the ball at a lattice position... I should
be able to rotate that ball around any of potentially 3 axes to move
my point to a new lattice position.
By flattening the case (4,3,2) to (4, sqrt(13)) we *know* that this
point lies on the circle r**2=29. Rotating a point around the origin
of a circle yields another point on the circle. (1,0), (-1,0), (0,1),
(0,-1) are just the trivial cases rotated by multipliers of 90
degrees.
If I could accurately represent this on paper, I am sure that I could
by hand rotate the point at (4, sqrt(13)) until it fell on a lattice
point. I feel that the difficulty in the problem lies in my lack of
understanding moreso that whether it can ultimately be done.
In the interest of my understanding however, can someone explain why
they feel this won't work?
Perhaps I can make use of the fact that (0, sqrt(29)) also lies on the
circle; it has 3 companions just as (4,sqrt(13)) has 4 variations. If
I am successful, I will post the algorithm here.
Sherman
On Jun 14, 1:18 pm, quasi <q...@xxxxxxxx> wrote:
On Thu, 14 Jun 2007 08:53:04 -0700, vision_1...@xxxxxxxxxxx wrote:
Hi,
The numbers can easily be more than 50 digits long. However, pursuing
the rotation idea I can see a glimmer of hope to a quicker solution.
In the case where 29=x^2+y^2+z^2 and the initial point (4,3,2)... it
occurred to me after I tried a 3d rotation, that I might be able to
'flatten' this sphere into a circle and do a 2d rotation.
(4,3,2) -> (x, y, 0)
I did the 3d rotation with the result that to 'rotate' Z to 0, I came
up with
Sin(Theta)/Cos(Theta) = -2/3
In the Z rotation I required, rotation about the X axis looks like:
1 0 0 4
0 CosT SinT 3
0 -SinT CosT 2
Thus for Z to rotate to 0,
Znew = 4*0 - 3SinT + 2CosT;
Which meant that -3SinT = 2CosT, which in turn meant that SinT/CosT =
-2/3.
SinT/CosT = TanT
Tan= Opposite/Adjacent, and the hypotenuse = sqrt(Opposite**2 +
Adjacent**2) = sqrt ((-2)**2 + 3**2) = sqrt(13)
So SinT must equal -2/sqrt(13) and CosT must equal 3/sqrt(13)... A
very long way around to determine that where c**2 = a**2 + b**2, 13 =
3**2 + 2**2 All of which suggested that I could have immediately
flattened my sphere into a circle using the point (4, sqrt(13)).
This is where I am currently stuck although having 13 involved
suggests the proper path via 5**2. I am now contemplating a simpler 2-
d rotation using (4,sqrt(13)) rotated to an integer point.
I agree with Robert Israel's point.
I'm sorry, but I really think that any attempt to somehow force a 3
squares solution to help you efficiently find a 2 squares solution is
doomed to failure.
In the end, I'm fairly sure you will be testing just as many cases as
if you tried to resolve the given number to a sum of 2 squares with no
additional information.
quasi
.
- References:
- Re: Diophantine Approach...
- From: Michael Jørgensen
- Re: Diophantine Approach...
- From: vision_1967
- Re: Diophantine Approach...
- From: quasi
- Re: Diophantine Approach...
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