System of three equations and four variables

A_Studen_349q

Well, first, I know that "to solve a system of equations" = "find all its roots or prove that there are no roots", second, teacher told that this system has non-zero roots, third, one of my classmates tried to use some numerical methods and said he obtained positive values.
So, I think that there must be non-zero roots (may be positive).
In any case, I have to solve this problem. So far, I don't know the solution.
Waiting for ideas...

SammyS

ehild has obtained that 4th equation, the one that doesn't have x₁ in it:
Although that gives a fourth equation, it's not independent of the other three.

A_Studen_349q

Agrees.

ehild

The fourth equation has been derived from the first three and the symmetry relation f(a,b)=f(b,a).
You can have three equations of the same type as the original ones either with x2, x3 or x4 as the "leading variable", by combining the original equations and the symmetry condition f(a,b)=f(b,a).

For example, the first two equations can be rewritten as
f(x₄,x₁)-f(x₄,x₃)=f(x₃,x₁)
f(x₄,x₁)-f(x₄,x₂)=f(x₂,x₁)
From these, it follows the third equation with x4:
f(x₄,x₂)-f(x₄,x₃)=f(x₃,x₂).


ehild

jedishrfu

another thing to note is that you can factor the the square root term although that may or may not help:

multiply both side by 2 and pull inside to eliminate the 1/4 factor

then for a given sqrt term notice the x^2 - y^2 pattern to yield (x-y)*(x+y)

sqrt( xi^2 * xj^2 - 4 * (xi - xj)^2 ) = sqrt ( ( xi*xj - 2 * (xi -xj)) * ( xi*xj + 2 * (xi -xj) )

gabbagabbahey

I really don't see how you can claim this. As a specific example, how can one show that [itex]f(x_1, x_2)-f(x_1, x_3)=f(x_3, x_2)[/itex]?

Still, [itex]f(x_i, x_j)=0[/itex] is certainly a permissible solution, but leads to only to trivial solutions, even when restricted to [itex]i \neq j[/itex], and I see no reason that it is the only solution.

Edit: Not all solutions of [itex]f(x_i, x_j)=0[/itex] are trivial; for example [itex](x_1, x_2, x_3, x_4)=(1, 2, 0, 0)[/itex]

ehild

I can not prove directly that f(a,b)-f(a,c) = f(c,b) true for any three from x1, x2, x3, x4. It is a "lemma", based on the symmetry of the function f(a,b)=f(b,a). But it might be false. If it is true then f(xi,xj)=0 for i,j {1,2,3,4} i≠j.


x3=x4=0 means negative values under square roots, so the equations are not defined over the real numbers. Plugging in your values, you get √(-1) on both sides, but it can be either i or -i. You can not say that (1, 2, 0, 0) is a solution.

ehild

gabbagabbahey

I agree that the symmetry relation allows you to derive another equation, but I don't see it leading to f(a,b)-f(a,c) = f(c,b) for all permutations (where a≠b≠c). I think exactly half the permutations satisfy this equation, and the other half satisfy f(a,b)-f(a,c) = -f(c,b).

Yes, that was a silly error on my part!

A_Studen_349q

So, any new ideas?