From: bdsm@poincare.WPI.EDU (Billy Don McConnell)
Newsgroups: sci.math
Subject: 3 "Simple" Quartics...
Date: 23 Jan 1993 02:51:23 GMT
Organization: Worcester Polytechnic Institute
Distribution: world
NNTP-Posting-Host: poincare.wpi.edu
Originator: bdsm@poincare.WPI.EDU


I have spent an obscene amount of time over the past 5 years or so looking for the
solutions to the following three (all but identical) Quartic Equations:

		   4        3      2               2    2
		3 H  - 4 S H  + ( S  - 2 B + 2 C )H  - A   =  0

		   4        3      2               2    2
		3 J  - 4 S J  + ( S  - 2 C + 2 A )J  - B   =  0

		   4        3      2               2    2
		3 K  - 4 S K  + ( S  - 2 A + 2 B )K  - C   =  0,


		where   S = X + Y + Z + R

	                A = ( X - R ) ( Y - Z ) 
                                                   Important(?) Fact:
                        B = ( Y - R ) ( Z - X )       
                                                     A + B + C = 0
	                C = ( Z - R ) ( X - Y ) 

(Why X, Y, Z, and then "R"?  Well, that's a long story. Let's just say that this
 problem relates to an extension of the  x, y, r  Triangle from Trigonometry.)

Note:  The use of S, A, B, and C may not be "convenient" for finding the
       Solutions. (That is, maybe the quantity "X+Y+Z+R" isn't itself very
       important.)  But their use makes writing the Equations easier.

Now, Quartics are fairly easily handled by a method one can find in plenty of 
references (I use the _CRC_Standard_Mathematical_Tables_ myself); that is, they
are easily handled this way when the coefficients are constants, not complicated
expressions in X, Y, Z, and R.

Maple V (Release 2) handles the symbol manipulation pretty well, giving the 
Solutions to these Equations in just a few minutes.  However, the Solutions are in
an inCREDibly complicated form, and Maple doesn't seem to know how to simplify
them in any _real_ sense of "simplification". (My quest these 5 years has been to 
simplify these Solutions.) 

I am _convinced_ that the Solutions have a *NICE* representation, but I don't know
what it is.

When I first loosed Maple on the problem, I was discouraged, because even when I
substituted numerical values for X,Y,Z, and R, I _still_ couldn't find any *nice* 
representation of the results;  H, J, and K were a messy combination of cube roots
and square roots of pretty big numbers ("big" in a relative sense of course: using
2, 3, 5, and 7 for X, Y, Z, and R resulted in solutions involving numbers of up to
12 digits, including some prime numbers of roughly that size).
 
However, I do know that the answers must satisfy 

			H + J + K = X + Y + Z + R   (*)
 
(In fact, for each Solution for H there is a Solution for J and one for K such 
that  (*)  holds.  That is, the 12 Solutions can be put into four groups, each 
with an H, J, and K, satisfying (*).)

The fact that Maple's (numerical) Solutions _do_ this is actually ENcouraging, 
since H, J, and K have completely different complicated parts, and I know that 
these complicated parts have to cancel each other out.  So there is some under-
lying pattern to the solutions.  There must be some weird factorization of
the cube-and-square-root parts that eludes Maple (not surprisingly, since Maple
doesn't seem to think in terms of "weird" factorizations) and has eluded me lo 
these many years (also not surprisingly, although I have tried every "weird"
factorization I could think of).

So I'm exposing this problem to the Net to see what becomes of it.  Any and all
insights are welcome (special properties of Quartics without a first-power term,
for instance), as are questions about the origins of the problem;  SOLUTIONS are
most welcome of all, of course...or (*sigh*) proofs that *NICE* Solutions don't
exist.  (I would even like to hear from folks who decided to think about the
problem for a while, but then gave it up; tell me what made you stop.)  Please
respond by e-mail.



Note:  Some relationships between X, Y, Z, and R are special to this System.
       For instance, if any two of these quantities are equal, then 0 is a 
       Root of one of the Equations.  That one was pretty obvious. Less obvious, 
       however, is:  if   X + Y + Z = R,  then  H = Y + Z,  J = X + Z, and  
       K = X + Y  are three Roots of the Equations (observe that H + J + K = 
       2 X + 2 Y + 2 Z = X + Y + Z + R, as is required), but the other Roots are
       messy (as far as I can tell...I have to study this case some more).
       
Note:  In the original inception of the problem, X, Y, Z, and R are squares
       of quantities (if that matters).  That is, X = x-squared, Y = y-squared,
       etc. (And I'm really looking for  h-squared, j-squared, and k-squared.)
       This might be helpful in viewing the problem, since for instance A, B, 
       and C (in x-y-z terms) are products of Differences of Squares. (One 
       might observe how, translating the previous Note into x-y-z-h-j-k terms,
       the "less obvious" relationships are _Pythagorean_ in nature.)  

Well, that's about it.  I hope SOMEone out there give this problem a shot... I've
been working on it alone for long enough.


Thanks a bunch!

(Oh, and to all the folks who have responded to my search for History of Math
Programs, thanks a bunch for THAT, too!  I've got some good leads to follow up
on.  My apologies for not thanking all respondants personally.)

Don Mc Connell
bdsm@fermat.wpi.edu
Newsgroups: sci.math