cxs
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A friend showed this question to me.
I found that I can add ①③=2*② to get x(x-1)^2+y(y-1)^2+z(z-1)^2=0, so if x,y,z are all negative or positive then the question is easy.
But what about others? I can also prove that there can not be 2 of them negatice, but I can not prove that it can not be the case that x,y positive, z negative w.l.o.g.
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RDKGames
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(Original post by cxs)
A friend showed this question to me.
I found that I can add ①③=2*② to get x(x-1)^2+y(y-1)^2+z(z-1)^2=0, so if x,y,z are all negative or positive then the question is easy.
But what about others? I can also prove that there can not be 2 of them negatice, but I can not prove that it can not be the case that x,y positive, z negative w.l.o.g.
What exactly is the question surrounding this equation?
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cxs
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(Original post by RDKGames)
What exactly is the question surrounding this equation?
To find all the triples (x,y,z) that satisfies this equation
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mnot
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(Original post by cxs)
To find all the triples (x,y,z) that satisfies this equation
are there a finite number? there's probably infinite solutions.
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Guru Jason
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Had to google this to make sure i wasn't being stupid by if x is 0, y is 1 and z is infinity then all of the above hold true I think.

Edit: I suppose it doesn't matter what numbers you use so long as there all positive numbers and one is infinity.
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ghostwalker
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(Original post by Guru Jason)
Had to google this to make sure i wasn't being stupid by if x is 0, y is 1 and z is infinity then all of the above hold true I think.
Unfortunately, infinity is not a real number.
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I hate maths!
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I think I found a way to generate solutions.

Let x,y,z be the roots of f(k)=k^3+ak^2+bk+c. By Vieta, a=-(x+y+z), b=xy+yz+zx and c=-xyz.

Rewriting in terms of S_1=x+y+z, S_2=x^2+y^2+z^2, S_3=x^3+y^3+z^3:

a=-S_1, b=(S_1^2-S_2)/2, c=(-S_1^3+3S_1S_2-2S_3)/6.

Apply the condition x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=d, and we discover that x,y,z are the roots of the following function:

\displaystyle f(k)=k^3-dk^2+\frac{d^2-d}{2}k+\frac{-d^3+3d^2-2d}{6}.

EXAMPLE:

If you want your equations to equal 4, set d=4 and we obtain the cubic f(k)=k^3-4k^2+6k-4. The roots of this are 2,1+i,1-i and you can confirm for yourself that 2+1+i+1-i=2^2+(1+i)^2+(1-i)^2=2^3+(1+i)^3+(1-i)^3=4.

I'm not sure if any of this is useful, but it was fun playing around. I think it confirms there are infinite complex solutions (x,y,z) and maybe you can use this to find how many real solutions if you want.
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Guru Jason
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(Original post by ghostwalker)
Unfortunately, infinity is not a real number.
Who said they had to be real numbers. I stand by what I said. kappa
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cxs
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(Original post by Guru Jason)
Had to google this to make sure i wasn't being stupid by if x is 0, y is 1 and z is infinity then all of the above hold true I think.

Edit: I suppose it doesn't matter what numbers you use so long as there all positive numbers and one is infinity.
(Original post by ghostwalker)
Unfortunately, infinity is not a real number.
Yeah we are asked to solve it in Real number. Sorry that I did not mention it clearly.
(Original post by I hate maths!)
I think I found a way to generate solutions.

Let x,y,z be the roots of f(k)=k^3+ak^2+bk+c. By Vieta, a=-(x+y+z), b=xy+yz+zx and c=-xyz.

Rewriting in terms of S_1=x+y+z, S_2=x^2+y^2+z^2, S_3=x^3+y^3+z^3:

a=-S_1, b=(S_1^2-S_2)/2, c=(-S_1^3+3S_1S_2-2S_3)/6.

Apply the condition x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=d, and we discover that x,y,z are the roots of the following function:

\displaystyle f(k)=k^3-dk^2+\frac{d^2-d}{2}k+\frac{-d^3+3d^2-2d}{6}.

EXAMPLE:

If you want your equations to equal 4, set d=4 and we obtain the cubic f(k)=k^3-4k^2+6k-4. The roots of this are 2,1+i,1-i and you can confirm for yourself that 2+1+i+1-i=2^2+(1+i)^2+(1-i)^2=2^3+(1+i)^3+(1-i)^3=4.

I'm not sure if any of this is useful, but it was fun playing around. I think it confirms there are infinite complex solutions (x,y,z) and maybe you can use this to find how many real solutions if you want.
An excellent idea! Perhaps we can proceed from this to get all the real solutions.
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I hate maths!
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(Original post by cxs)
Yeah we are asked to solve it in Real number. Sorry that I did not mention it clearly.

An excellent idea! Perhaps we can proceed from this to get all the real solutions.
If you want real solutions, you can probably do something with the necessary and sufficient condition that there must be one turning point above the x-axis and one turning point below the x-axis for the cubic to have three distinct real roots, so investigating f'(k) may be fruitful. EDIT: I forgot about repeated real roots, you should examine that as well.
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I hate maths!
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Did you get an answer to this? I couldn't figure out a nice way with A Level knowledge so I had to do a little bit of research into generalised discriminants, which can yield a very nice necessary and sufficient condition for the roots to be real (although it takes some amount of work to get there). Have a read here: https://brilliant.org/wiki/cubic-discriminant/ (there is also an alternative method with linear algebra to find the discriminant of a general polynomial using something called the Sylvester matrix, see here https://socratic.org/questions/what-...al-of-degree-n).

Now the problem becomes finding d such that

\displaystyle d^2 \Big[\frac{d^2-d}{2}\Big]^2-4\Big[\frac{d^2-d}{2}\Big]^3+4d^3\Big[\frac{-d^3+3d^2-2d}{6} \Big]-27\Big[\frac{-d^3+3d^2-2d}{6} \Big]^2-18d \Big[\frac{d^2-d}{2} \Big] \Big[\frac{-d^3+3d^2-2d}{6}\Big] \geqslant 0

I tried simplifying that by hand but mucked up somewhere, but according to Wolfram Alpha it should simplify to

\displaystyle -\frac{1}{6}d^2(d-3)^2(d-2)(d-1) \geqslant 0. (very nice!!)

It's not too bad to do by hand, you can pull out a factor of d^2(d-1) from that horrible expression and you are left with factorising a cubic, but I mucked up somewhere. In any case, it is clear now that there are infinitely many real solutions (x,y,z) from a simple graph sketch of that sextic in d.

You will get solutions of the form (m,m,m) when d=0 or d=3, solutions of the form (m,n,n) when d=1 or d=2, and solutions of the form (m,n,o) when 1 < d < 2 for m \neq n \neq o.

(FYI the inspiration for the cubic set up I used is STEP 3 2015 Q4 part (ii))
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cxs
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(Original post by I hate maths!)
Did you get an answer to this? I couldn't figure out a nice way with A Level knowledge so I had to do a little bit of research into generalised discriminants, which can yield a very nice necessary and sufficient condition for the roots to be real (although it takes some amount of work to get there). Have a read here: https://brilliant.org/wiki/cubic-discriminant/ (there is also an alternative method with linear algebra to find the discriminant of a general polynomial using something called the Sylvester matrix, see here https://socratic.org/questions/what-...al-of-degree-n).

Now the problem becomes finding d such that

\displaystyle d^2 \Big[\frac{d^2-d}{2}\Big]^2-4\Big[\frac{d^2-d}{2}\Big]^3+4d^3\Big[\frac{-d^3+3d^2-2d}{6} \Big]-27\Big[\frac{-d^3+3d^2-2d}{6} \Big]^2-18d \Big[\frac{d^2-d}{2} \Big] \Big[\frac{-d^3+3d^2-2d}{6}\Big] \geqslant 0

I tried simplifying that by hand but mucked up somewhere, but according to Wolfram Alpha it should simplify to

\displaystyle -\frac{1}{6}d^2(d-3)^2(d-2)(d-1) \geqslant 0. (very nice!!)

It's not too bad to do by hand, you can pull out a factor of d^2(d-1) from that horrible expression and you are left with factorising a cubic, but I mucked up somewhere. In any case, it is clear now that there are infinitely many real solutions (x,y,z) from a simple graph sketch of that sextic in d.

You will get solutions of the form (m,m,m) when d=0 or d=3, solutions of the form (m,n,n) when d=1 or d=2, and solutions of the form (m,n,o) when 1 < d < 2 for m \neq n \neq o.

(FYI the inspiration for the cubic set up I used is STEP 3 2015 Q4 part (ii))
Genius! This is a very elegant solution!
The answer is somewhat out of my expectation. I initially thought that the only solution is that {x,y,z} are either 0 or 1. Now it seems that there are infinite solutions that can be derived from your beautiful equation. (And, this answer shows that why I can not proceed with the case "2 positive,1 negative", for I used to think that it can be proved of having no roots... Now it turns out that your method is definitely the right path)
Thx!
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I hate maths!
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(Original post by cxs)
Genius! This is a very elegant solution!
The answer is somewhat out of my expectation. I initially thought that the only solution is that {x,y,z} are either 0 or 1. Now it seems that there are infinite solutions that can be derived from your beautiful equation. (And, this answer shows that why I can not proceed with the case "2 positive,1 negative", for I used to think that it can be proved of having no roots... Now it turns out that your method is definitely the right path)
Thx!
Thank you.
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