Re: Diophantine Approach...

On Thu, 14 Jun 2007 08:53:04 -0700, vision_1967@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
.

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