Conference rusure::math

	I would say, since sqrt v is >= 0 for whatever variable v
	you plug in, that

		sqrt v + sqrt w + sqrt x + sqrt y + sqrt z = 0.

	implies v = w = x = y = z = 0.

	Dan

    All variable substitutions below are to simplify expressions, not
    some magic transformation.  The tricks, if any, are deciding which
    radicals to bring together and which side gets the nonradical terms.
    In one step we produce sqrt(wz) on both sides of the equation and
    in another step we produce sqrt(wxyz) on both sides.


    sqrt(v) + sqrt(w) + sqrt(x) + sqrt(y) + sqrt(z) = 0

    Rearranging and squaring both sides:

    [sqrt(v) + sqrt(w) + sqrt(x)]^2 = [-sqrt(y) - sqrt(z)]^2

    v+w+x+ 2*[sqrt(vw) + sqrt(vx) + sqrt(wx)] = y+z+ 2*sqrt(yz)

    let a = [y + z - v - w - x]/2.  Then, rearranging and squaring
    both sides:

    [sqrt(vw) + sqrt(vx)]^2 = [a + sqrt(yz) - sqrt(wx)]^2

    vw+vx+ 2v*sqrt(wx) = a^2+yz+wx + 2a*sqrt(yz) - 2a*sqrt(wx) - 2*sqrt(wxyz)

    (2v+2a)*sqrt(wx) - 2a*sqrt(yz) = a^2+yz+wx-vw-vx - 2*sqrt(wxyz)

    let b = [a^2+yz+wx-vw-vx]/2.  let c = v+a.  Then, rearranging and
    squaring both sides:

    [c*sqrt(wx) - a*sqrt(yz)]^2 = [b - sqrt(wxyz)]^2

    (c^2)wx + (a^2)yz - 2ac*sqrt(wxyz) = b^2 - 2b*sqrt(wxyz) + wxyz

    Rearranging and squaring both sides:

    [2(b-ac)*sqrt(wxyz)]^2 = [b^2 + wxyz - (c^2)wx -(a^2)yz]^2

    and substituting back in for a,b, and c should complete the problem.
    
    - Don

    sqrt(u) + sqrt(v) + sqrt(w) + sqrt(x) + sqrt(y) + sqrt(z) = 0

    looks impossible to rationalize.  I'm not sure how to attempt a
    proof, though.

    - Don

    Re .1:
    
    Not necessarily Dan.  The variable may not be real numbers, they may be
    complex variables.  Or, even more likely, they might be polynomials.
    
    --  Barry

        Even as it is, it is wrong.  Take the simplest case,
        sqrt x + sqrt y = 0.  Follow the directions in .0 and
        bring sqrt y to the other side, sqrt x = - sqrt y.  Now
        square both sides:  x = y or x - y = 0, not x^2 - y^2 = 0
        as stated.  Likewise with three variables, bringing sqrt
        z to one side and squaring leaves a single term z on that
        side, and a second squaring yields a a quadratic term in
        z.  But the "solution" given contains z^4.
        
        Dan

   It seems you have to be free to choose arbitrary branches on the roots
   to obtain any nontrivial solutions, but this is not stipulated clearly.

   Suppose we have a homogenous sum of n terms a + b + c + .. = 0.

   Then there will result 2^n/2 possible equations that are "true in the sense
   of rationalization" by bringing subsets of the terms to one side of the
   equal sign, squaring, and bringing the terms back.

   If you can find a linear superposition of the equations that cancels
   the cross terms without cancelling the other terms you can get a
   rationalization, which will always be a symmetric form in the variables
   (I think this has to be the case.)

   If not, then the equations can be treated the same way, yielding a
   larger set.  There will be a lot of symmetry, but there may be enough
   freedom to do the cases of 6 or more variables.

   This is just an idea, I don't know if it is the right way to think
   about the problem.

   - Jim

	Given the equation x = 0, we rearrange it so that all the terms
in *odd* order in a_k are on the right hand side, then we square both sides
and bring everything across to the left hand side again.

	To be more precise:

	Given n > 1
	Set x = sum(i)(a_i)
	For k = n downto 2
		Define y = x with a_k replaced by -a_k
		let f = �(x+y)
		let g = �(x-y)
		let x = f�-g�
	endfor.

	At the end, x is a polynomial of degree 2^(n-1), every term is of
even order, and it has as a factor every poly of the form a_1 � a_2 + ... � a_n
It's our answer.

	It certainly seems to me to be "elementary", in so far as the meaning of
this term was loosely waved at in the basenote.   What's going on?

Regards,
Andrew.

   Sigh... that's the first thing I tried, but I made an algebraic
   error which mislead me into thinking it couldn't work.

   - Jim

re .0:

>Given the equation
> 
>sqrt x + sqrt y = 0
> 
>we can "rationalize it" by bringing sqrt y to the right side
>and then squaring both sides.  You get x^2-y^2=0.

Do you mean squaring both sides twice ?

Squaring once gives:

sqrt(x) = - sqrt(y)
x = y

When you raise equations to powers like this don't you sometimes introduce
extraneous roots ?

Also, you treating sqrt as a function (single valued with positive range)
or a relation ?