# x+y+z=x^2+y^2+z^2=x^3+y^3+z^3 Watch

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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.

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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#2

(Original post by

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.

**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.

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(Original post by

What exactly is the question surrounding this equation?

**RDKGames**)What exactly is the question surrounding this equation?

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#4

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To find all the triples (x,y,z) that satisfies this equation

**cxs**)To find all the triples (x,y,z) that satisfies this equation

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#5

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.

Edit: I suppose it doesn't matter what numbers you use so long as there all positive numbers and one is infinity.

Last edited by Guru Jason; 1 week ago

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#6

(Original post by

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.

**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.

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#7

I think I found a way to generate solutions.

Let x,y,z be the roots of . By Vieta, , and .

Rewriting in terms of , , :

, , .

Apply the condition , and we discover that are the roots of the following function:

.

If you want your equations to equal 4, set d=4 and we obtain the cubic . The roots of this are and you can confirm for yourself that .

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

Let x,y,z be the roots of . By Vieta, , and .

Rewriting in terms of , , :

, , .

Apply the condition , and we discover that are the roots of the following function:

.

**EXAMPLE:**If you want your equations to equal 4, set d=4 and we obtain the cubic . The roots of this are and you can confirm for yourself that .

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.
Last edited by I hate maths!; 1 week ago

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#8

(Original post by

Who said they had to be real numbers. I stand by what I said. kappa

**ghostwalker**)
Unfortunately, infinity is not a real number.

Last edited by Guru Jason; 1 week ago

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(Original post by

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.

**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

Unfortunately, infinity is not a real number.

**ghostwalker**)Unfortunately, infinity is not a real number.

(Original post by

I think I found a way to generate solutions.

Let x,y,z be the roots of . By Vieta, , and .

Rewriting in terms of , , :

, , .

Apply the condition , and we discover that are the roots of the following function:

.

If you want your equations to equal 4, set d=4 and we obtain the cubic . The roots of this are and you can confirm for yourself that .

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

**I hate maths!**)I think I found a way to generate solutions.

Let x,y,z be the roots of . By Vieta, , and .

Rewriting in terms of , , :

, , .

Apply the condition , and we discover that are the roots of the following function:

.

**EXAMPLE:**If you want your equations to equal 4, set d=4 and we obtain the cubic . The roots of this are and you can confirm for yourself that .

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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#10

(Original post by

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.

**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.

Last edited by I hate maths!; 1 week ago

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#11

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

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

. (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 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 .

(FYI the inspiration for the cubic set up I used is STEP 3 2015 Q4 part (ii))

Now the problem becomes finding d such that

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

. (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 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 .

(FYI the inspiration for the cubic set up I used is STEP 3 2015 Q4 part (ii))

Last edited by I hate maths!; 1 week ago

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(Original post by

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

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

. (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 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 .

(FYI the inspiration for the cubic set up I used is STEP 3 2015 Q4 part (ii))

**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

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

. (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 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 .

(FYI the inspiration for the cubic set up I used is STEP 3 2015 Q4 part (ii))

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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#13

(Original post by

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!

**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!

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