Losing my hope

Dragos V0d3

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Joined
May 7, 2019
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I have we have to solve this intimidating system in R:
x+y+z=3​
x^3+y^3+z^3=3​
x^5+y^5+z^5=3​
I tried unsuccessfully to define a polynomial with solutions x y z but i dont have enough relationships to tie something.​
Can u guide me to a ideea ,especialy about x^5+y^5+z^5​
 

HallsofIvy

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Jan 27, 2012
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You recognize that x= y= z= 1 is a solution, don't you? Would you be surprised that it was the only solution?
 

Dragos V0d3

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May 7, 2019
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You recognize that x= y= z= 1 is a solution, don't you? Would you be surprised that it was the only solution?
I can"t find any method to demonstrate the unicity of this solution; could u explain your argument?
 

Jomo

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Dec 30, 2014
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I can"t find any method to demonstrate the unicity of this solution; could u explain your argument?
Try to find solutions for the first two equations and show that other than (1,1,1) no other solution satisfies the 3rd equation.
Try cubing some form of the 1st equation.....
 

Jomo

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x=y=z=1 is a solution but there may be many others. If the ONLY solution is (1,1,1) this needs to be shown.

For example: x=y^2 and y=x^3. Clearly x=y=0 is a solution. But so is (1,1) a solution. So the solution is (0,0) and (1,1). MAYBE NOT. In theory there can be other solutions. One you get a solution you must make sure that there are no other solutions. Sometimes it is obvious that there is say one solution while other times it is not. For example if two lines in the plane have two solutions (that is two pts of intersection) then clearly there are an infinty number of solutions. After all, have you ever seen two lines in a plane intersect at exactly 2 pts or exactly 3 pts or exactly 10,002 pts?
 

Jomo

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Did you figure out?
More importantly did you figure it out? What have you been trying? Show us your work so we can guide you. What course is this from? What are you allowed to use? Maybe let z=k and go from there? I suspect that this is not easy at all to show that x=y=z=1 is the only solution
 

JeffM

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Sep 14, 2012
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3,426
I have we have to solve this intimidating system in R:
x+y+z=3​
x^3+y^3+z^3=3​
x^5+y^5+z^5=3​
I tried unsuccessfully to define a polynomial with solutions x y z but i dont have enough relationships to tie something.​
Can u guide me to a ideea ,especialy about x^5+y^5+z^5​
\(\displaystyle \text {DEFINE: } f(x, y, z) = 3 - (x + y + z),\ g(x, y, z) = 3 - (x^3 + y^3 + z^3), \text { and } h(x, y, z) = 3 - (x^5 + y^5 + z^5).\)

The system in the problem above has three equations and three unknowns. In principle, the system DOES have sufficient relationships to generate a positive number of solutions if the equations are independent and consistent. It is easy enough to show that they are independent, but showing that they are consistent is most easily done, perhaps only feasible, by showing that there is at least one solution.

One such solution can be found by symmetry, namely (1, 1, 1), thereby proving consistency. But that does not show that the solution is unique. Moreover, someone might not "see" the symmetry solution. Here is a method to find all the solutions.

Notice first that any solution to any of the equations is a zero of one of the functions that I have defined. Of course, there are an infinite number of zeroes of each function, but a solution to the system must be a point that is a zero of all three functions. The distance between the functions at a solution must be zero. Thus, the set of points where d = 0 will contain the solution if one exists.

The distance between f and g at an arbitrary point (p, q, r) is

\(\displaystyle d = \sqrt{(p^3 - p)^2 + (q^3 - q)^2 + (r^3 - r)^2}.\)

Notice that the addends in the square root are all non-negative. Therefore d = 0 entails that each addend be zero.

\(\displaystyle (p^3 - p)^2 = 0 \implies \implies p = p^3 \implies p = -\ 1, 0, \text { or } 1.\)

\(\displaystyle d = 0 \implies (0,\ 0,\ 0),\ (0,\ 0,\ \pm 1),\ (0,\ \pm 1,\ 0),\ (0,\ \pm 1,\ \pm1),\)

\(\displaystyle (\pm 1,\ 0,\ 0),\ (\pm 1,\ 0,\ \pm1),\ (\pm 1,\ \pm 1,\ 0),\ \text { or } (\pm 1,\ \pm 1,\ \pm1).\)

Only one of those points is a zero either f or g. (1, 1, 1) is a zero of both f and g. Furthermore, (1, 1, 1) is a zero of h. There is a unique solution, namely (1, 1, 1).
 
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