Re: x^2 + y^2 + z^2 = r^2 + s^2 + t^2

OK, let me take one more shot at this.

Given,

r+s+t = x+y+z
r^2+s^2+t^2 = x^2+y^2+z^2

we can reduce the second of these to an elementary symmetric function:

rs+rt+st = xy+xz+yz

If we take t and z as unknowns, and r,s,x,y as parameters, this gives
us a system of two linear equations in two unknowns:

t-z = x+y-r-s
(r+s)t-(x+y)z = xy-rs

The determinant D of the matrix [1,-1;r+s,-x-y] is r+s-x-y. If D is
non-zero, then there is a unique set of values t,z which solve the
linear equations, which in turn solves the original problem. If D is 0
on the other hand, then the equations are linearly dependent, i.e.
either there are no solutions or there are an infinite number of
solutions. D = 0 implies r+s = x+y, which in turn implies t = z, which
in turn implies rs = xy. The two equations r+s = x+y and rs = xy taken
together imply that x,y is a permutation of r,s, since both pairs are
the roots of the same quadratic: a^2 - (r+s)a + rs = a^2 - (x+y)a + xy.

This gives all the solutions of the original problem.

The same approach can be taken to the equations:

r+s+t+u = w+x+y+z, etc.

which can be reduced to elementary symmetric functions in the same way:

r+s+t+u = w+x+y+z
rs+(r+s)t+(r+s+t)u = wx+(w+x)y+(w+x+y)z
rst+(rs+rt+st)u = wxy+(wx+wy+xy)z

However, it is not possible to choose three of the quantities in these
equations so that the equations are linear, since at least two must be
on one side or the other, and their product will appear in the last two
equations. We can get around this issue by writing, say, W = w+x, X =
wx, which leaves the equations:

r+s+t+u = W+y+z
rs+(r+s)t+(r+s+t)u = X+Wy+(W+y)z
rst+(rs+rt+st)u = Xy+(X+Wy)z

which is now linear in u,W,X (taking r,s,t,y,z as arbitrary rational
parameters). We can therefore solve this system for u,W,X, and we will
obtain a solution of the problem provided that W^2-4X is a square,
which will imply that w and x are rational.

I haven't carried this reasoning any further.

.

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