## System of three equations and four variables

 Quote by Ray Vickson How do you know that goal is achievable? RGV
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...

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 Quote by A_Studen_349q Yes, but this "linear combination" is nothing but eqn (1). Why? Well, lets enumerate our three equations of the initial system as (1), (2) and (3). We have 3 eqns and 4 vars while (1) feels lack of x2, (2) feels lack of x3, (3) feels lack of x4. The "linear combination" (let's call it (4)) you told about will feel lack of x1. Why do I say (4) is nothing but (1)? Well, we can just re-designate our variables around to turn any of our eqns into (4). ...
ehild has obtained that 4th equation, the one that doesn't have x1 in it:
 Quote by ehild Well, not all goals can be achieved. Subtracting 2' from 1' and comparing it with 3': $\dots\ =f(x_3 ,x_4)-f(x_2,x_4)=-f(x_2, x_3)$ ... ehild
Although that gives a fourth equation, it's not independent of the other three.

 Quote by 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.
Agrees.

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 Quote by 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.
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
 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) )

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 Quote by ehild You can get similar relation between any pair of the xi-s: Let be a, b, c any different xi-s. f(a,b)-f(a,c)=f(c,b)
I really don't see how you can claim this. As a specific example, how can one show that $f(x_1, x_2)-f(x_1, x_3)=f(x_3, x_2)$?

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

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

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 Quote by gabbagabbahey I really don't see how you can claim this. As a specific example, how can one show that $f(x_1, x_2)-f(x_1, x_3)=f(x_3, x_2)$?
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.

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

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 Quote by 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). If it is true then f(xi,xj)=0 for i,j {1,2,3,4} i≠j.
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).

 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.
Yes, that was a silly error on my part!
 So, any new ideas?