Equality of sums of powers

[littlemathquark]

Homework Statement

$a,b,c\in\mathbb{R}$ if $a^2+b^2+c^2=a^3+b^3+c^3=1$ find $a+b+c=?$

Relevant Equations

$a,b,c\in\mathbb{R}$ if $a^2+b^2+c^2=a^3+b^3+c^3=1$ find $a+b+c=?$

Tt's easy to see permutations of $(1,0,0)$ are trivial solutions. I can use $a^3+b^3+c^3=3abc+(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$ and I can find $ab+ac+bc$ in terms of $a+b+c$ but how can I achive the same thing for $abc$?

 

[PeroK]

That doesn't look promising. The obvious way to start is:
$$a + b + c = (a + b + c)(a^2 + b^2 + c^2) = \dots$$

 

[PeroK]

Alternatively, is a non-trivial solution possible?

 

[littlemathquark]

Alternatively, is a non-trivial solution possible?

May be in complex numbers I think. But how?

 

[PeroK]

May be in complex numbers I think. But how?

The problem, as stated, is confined to real numbers.

 

[fresh_42]

Homework Statement: $a,b,c\in\mathbb{R}$ if $a^2+b^2+c^2=a^3+b^3+c^3=1$ find $a+b+c=?$
Relevant Equations: $a,b,c\in\mathbb{R}$ if $a^2+b^2+c^2=a^3+b^3+c^3=1$ find $a+b+c=?$

Tt's easy to see permutations of $(1,0,0)$ are trivial solutions. I can use $a^3+b^3+c^3=3abc+(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$ and I can find $ab+ac+bc$ in terms of $a+b+c$ but how can I achive the same thing for $abc$?

I do not know whether this ends up in a tautology, but I have found a way to express $abc$ in terms of $x=a+b+c.$ I first calculated $x-1=x\cdot 1- 1=(a+b+c)(a^2+b^2+c^2)-(a^3+b^3+c^3)$ and then $x^2-1=(a+b+c)^2-(a^2+b^2+c^2).$

 

[bob012345]

Write the two curves as follows;
$$b^2 +c^2=1-a^2$$ and $$b^3+c^3=1-a^3$$
Logically, $0≤a,b,c ≤1$.

For any given value of $a>0$ the first curve is a circle on the origin. The second is the $y=x$ line (using the usual x,y instead of b,c) when $a=1$ and asymptotically approaches a line away from the origin when $a≠1$. Now determine if they can ever intersect or not.

 

[fresh_42]

I think there are only the trivial solutions. I'm currently at $0<|a|,|b|,|c|<1$ with strict inequalities since an equality leads to a trivial solution. I do not see how we can assume that all three must be positive. I only see that $a+b+c$ is roughly between $\pm 2.6.$ I also have a polynomial equation $p(a+b+c)=6abc\neq 0$ but I struggle to find a second one that would allow me to eliminate the product.

 

[PeroK]

All three can't be positive, because for $0 < a < 1$, we have $a^3 < a^2$ etc. If one or two are negative, then those cubes are negative, which means the positive cube(s) must be greater than 1. That's impossible. Hence, there are no solutions except 0, 0, 1.

 

[haruspex]

All three can't be positive, because for $0 < a < 1$, we have $a^3 < a^2$ etc. If one or two are negative, then those cubes are negative, which means the positive cube(s) must be greater than 1. That's impossible. Hence, there are no solutions except 0, 0, 1.

I am not following. How do you rule out one negative and two in (0,1)?

 

[PeroK]

I am not following. How do you rule out one negative and two in (0,1)?

If $a <0$ and $0 < b, c < 1$, then $b^3 + c^3 = 1 - a^3 > 1$. But $b^3 + c^3 < b^2 + c^2 = 1 - a^2 < 1$.

 

[WWGD]

This can be solved with Dimensional Analysis. If there was a Real solution to $x^2+ y^2 +z^2=x^3+ y^3+ z^3=1$, given the two surfaces are in general position, the solution set would have dimension$= 2+2-3=1$( Sum of dimensions of surface minus dimension of $\mathbb R^3$ as the ambient space; this holds for subsets in general position, i.e., Transversal) , a curve. The set of points where $x+y+z=c$ ; $c$ a Real constant, represents a plane; a contradiction.

 

[haruspex]

This can be solved with Dimensional Analysis. If there was a Real solution to $x^2+ y^2 +z^2=x^3+ y^3+ z^3=1$, given the two surfaces are in general position, the solution set would have dimension$= 2+2-3=1$( Sum of dimensions of surface minus dimension of $\mathbb R^3$ as the ambient space; this holds for subsets in general position, i.e., Transversal) , a curve. The set of points where $x+y+z=c$ ; $c$ a Real constant, represents a plane; a contradiction.

$c$ is not specified, so it is not "a plane" but all planes normal to $x=y=z$.

Even if it were specified, why would it be a contradiction? The intersection of the curve and a plane would in general be discrete points, but possibly none.

What @PeroK's analysis shows is that the set of points from the two nonlinear constraints is not in fact a continuous curve but consists only of the trivial ones found. Geometrically, the two surfaces merely touch at those points.

 

[WWGD]

$c$ is not specified, so it is not "a plane" but all planes normal to $x=y=z$.

Even if it were specified, why would it be a contradiction? The intersection of the curve and a plane would in general be discrete points, but possibly none.

What @PeroK's analysis shows is that the set of points from the two nonlinear constraints is not in fact a continuous curve but consists only of the trivial ones found. Geometrically, the two surfaces merely touch at those points.

I said _ if_ $x+y+z=c$ , with $c \in \mathbb R$ a constant. Because it's not clear just what else what it would equal, but reasonable to assume, given the previous two equations $x^2+y^3+z^2=1=x^3+y^3+z^3$, it's reasonable to assume the latter was an equation as well. Whether it's regular or singular, the intersection is 1-dimensional, thus not a plane.

So my argument holds under such assumption.

 

[haruspex]

So my argument holds under such assumption.

Ok, but that assumption renders your argument irrelevant to the problem at hand, so I don’t see how

This can be solved with Dimensional Analysis.

… and even with the assumption I still do not get how your argument leads to a contradiction anyway.
It seems to me the question you have answered is “given $(x, y, z)$ satisfies the two equations for the sums of their squares and cubes, show that there is no constant $c$ s.t. $x+y+z=c$ for all such $(x, y, z)$".

 

[WWGD]

Ok, but that assumption renders your argument irrelevant to the problem at hand, so I don’t see how

… and even with the assumption I still do not get how your argument leads to a contradiction anyway.
It seems to me the question you have answered is “given $(x, y, z)$ satisfies the two equations for the sums of their squares and cubes, show that there is no constant $c$ s.t. $x+y+z=c$ for all such $(x, y, z)$".

It wasn't clear to me just what the OP was asking. It seemed to me to ask whether there existed #c# with $x+y+z=c$ was a reasonable interptetation of what they were asking.

 

[haruspex]

It wasn't clear to me just what the OP was asking. It seemed to me to ask whether there existed #c# with $x+y+z=c$ was a reasonable interptetation of what they were asking.

It says “find x+y+z=c”, which I (and everyone else, it seems) read to mean "find c, where c=x+y+z", though it be a rather hamfisted way to express that.

 

[WWGD]

It says “find x+y+z=c”, which I (and everyone else, it seems) read to mean "find c, where c=x+y+z", though it be a rather hamfisted way to express that.
[/QUOTE]
And I showed no such Real c exists, because if it did, $x,y,z$ would satisfy the equation of a plane, which cannot happen when two surfaces in general position, such as the ones in this problem, intersect, as their intersection, if Real-valued, would define a curve, as the former is 2-dimensional, while the latter is one-dimensional. Thus the contradiction, and no such Real c exists. So we're not disagreeing, we just used different arguments.

 

[haruspex]

It says “find x+y+z=c”, which I (and everyone else, it seems) read to mean "find c, where c=x+y+z", though it be a rather hamfisted way to express that.

And I showed no such Real c exists

Yet it does. As was shown in post #1, (1,0,0), (0,1,0), (0,0,1) are all solutions of the given equations and all lie in the plane $x+y+z=1$.

 

[WWGD]

Yet it does. As was shown in post #1, (1,0,0), (0,1,0), (0,0,1) are all solutions of the given equations and all lie in the plane $x+y+z=1$.

A discrete, finite subset of the plane. I didn't consider that option. But, yes, a discrete set of solutions I didn't consider. I assumed the intersection, solution sets would be continuous. Then the solution set isn't $S:=\{(x,y,z): x+y+z=1\}$, as, e.g., $(1/3,1/3,1/3)$ . But, yes, I guess I assumed the solutiIn set was continuous, which was not required.

Edit: Tl; Dr: I thought the question was to find $c$ , so that $x+y+z=c$ is the intersection of the two curves, which may not have been what the OP wanted.

 

[haruspex]

A discrete, finite subset of the plane. I didn't consider that option. But, yes, a discrete set of solutions I didn't consider. I assumed the intersection, solution sets would be continuous. Then the solution set isn't $S:=\{(x,y,z): x+y+z=1\}$, as, e.g., $(1/3,1/3,1/3)$ . But, yes, I guess I assumed the solutiIn set was continuous, which was not required.

Ok, but assuming a continuous solution set still does not lead to your answer.

Let's replace the given two equations with abstract equations $f(x,y,z)=0, g(x,y,z)=0$, and say we are asked to find all values of $h(x,y,z)$. (In the actual question it is hinted that there is only one possible value for h, but we can ignore that.)
In general, the two equations each generate a surface, and, except in special cases, the two surfaces will intersect in a curve or not at all.
If they intersect , the answer is the set of values $h$ takes on the curve.
According to the hint, they all produce the same value for $h$ where $h=x+y+z$, so the curve must lie in a plane normal to $x=y=z$. For example, the two surfaces could be a cylinder and a cone, each with axis $x=y=z$.

In the actual problem, the two surfaces only touch at a finite set of points, which do all indeed happen to lie in such a plane.

 

[littlemathquark]

I found this solution:
Permutation of $(1,0,0)$ trivial solutions.

$c^2=1-(a^2+b^2)$ and $c^3=1-(a^3+b^3)$ find $c^6$ for each of them. So we find $2a^6-3(1-b^2)a^4+2(b^3-1)a^3+3(1-b^2)^2a^2+(b^3-1)^2-(1-b^2)^3=0$

If $b=1$ we find $a=0$ trivial solution.

For $b=0$ we find $a^2(2a^4-3a^2-2a+3)=0$ and since $a=1$ value is a solution $a^2(a-1)(2a^3+2a^2-a-3)=0$ thus $a^2(a-1)^2(2a^2+4a+3)=0$

Roots of that equation $a_1=0$, $a_2=1$, $a_3=-1-i/\sqrt2$, $a_4=-1+i/\sqrt2$

So real roots are $a=0$ ve $a=1$

We can see that only $(0,1,0),(1,0,0),(0,0,1)$ are real solutions. So $a+b+c=1$

 

[WWGD]

Ok, but assuming a continuous solution set still does not lead to your answer.

Let's replace the given two equations with abstract equations $f(x,y,z)=0, g(x,y,z)=0$, and say we are asked to find all values of $h(x,y,z)$. (In the actual question it is hinted that there is only one possible value for h, but we can ignore that.)
In general, the two equations each generate a surface, and, except in special cases, the two surfaces will intersect in a curve or not at all.
If they intersect , the answer is the set of values $h$ takes on the curve.
According to the hint, they all produce the same value for $h$ where $h=x+y+z$, so the curve must lie in a plane normal to $x=y=z$. For example, the two surfaces could be a cylinder and a cone, each with axis $x=y=z$.

In the actual problem, the two surfaces only touch at a finite set of points, which do all indeed happen to lie in such a plane.

It does , and this is my last reply here: there is no $c$ so that $x+y+z=c$, a plane, is the solution set to the intersection of the two surfaces. That is the question I answered, and, as such, it's correct. I won't be replying to this anymore, as I have explained myself more than once. I stand by the correctness of my answer.

 

[haruspex]

It does , and this is my last reply here: there is no $c$ so that $x+y+z=c$, a plane, is the solution set to the intersection of the two surfaces. That is the question I answered, and, as such, it's correct. I won't be replying to this anymore, as I have explained myself more than once. I stand by the correctness of my answer.

Fwiw, when I wrote post #21 I had not seen your edit to post #20. It seems that your misreading made it more radically different from the intent than I had realised.

 

[bob012345]

If anyone is interested, here is the 3D plot from Desmos;

https://www.desmos.com/3d/vfds28zvvz

 

[littlemathquark]

In complex numbers, how many solutions are there?

 

[bob012345]

In complex numbers, how many solutions are there?

How does one define complex numbers for three dimensional space?

 

[haruspex]

How does one define complex numbers for three dimensional space?

There is no difficulty having spaces of any number of dimensions in which the underlying field is ℂ. The "complex plane" can be represented as

• a 2D vector space over the reals, together with a rule for multiplying vectors:$(x_1,y_1)\times(x_2,y_2)=(x_1x_2-y_1y_2,x_1y_2+x_2y_1)$ (vector spaces in general have no vector multiplication rule) or
• simply as the field ℂ, or
• as a one dimensional space over the field ℂ with the vector multiplication rule: $( x)\times(y)=(xy)$.

These are all isomorphic.

 

[littlemathquark]

If $a+b+c=t$ , $ab+ac+bc=u$ and $abc=v$ then $6v=t(t^2-3)+2$ but I can't find another equation which eliminate $v$

 

[martinbn]

It does , and this is my last reply here: there is no $c$ so that $x+y+z=c$, a plane, is the solution set to the intersection of the two surfaces.

Just to point out that the question doesn't ask for that. It's the other way around. The intersection of the surfaces needs to be contain in the plane, not that all points in the plane have to be in the intersection. And a curve can be in a plane.