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.

 

[littlemathquark]

Can we say that $x^2+y^2+z^2=1$ and $x^3+y^3+z^3=1$ surfaces tangent to each other at $(1,0,0),(0,1,0),(0,0,1)$ points so there is no infinite common solutions in complex numbers?

 

[PeroK]

Can we say that $x^2+y^2+z^2=1$ and $x^3+y^3+z^3=1$ surfaces tangent to each other at $(1,0,0),(0,1,0),(0,0,1)$ points so there is no infinite common solutions in complex numbers?

I don't see an immediate proof that there are complex solutions, but there are a lot of parameters.

 

[haruspex]

Can we say that $x^2+y^2+z^2=1$ and $x^3+y^3+z^3=1$ surfaces tangent to each other at $(1,0,0),(0,1,0),(0,0,1)$ points so there is no infinite common solutions in complex numbers?

No.
In 2D, consider $x+y=1, x^2+y^2=1$. These intersect at $x=2\pm i\sqrt 2$. Adding a third variable could easily produce a continuous curve of complex solutions.

 

[haruspex]

I don't see an immediate proof that there are complex solutions,

Write each in the form $z^6=P(x, y)$ then eliminate $z$ to obtain a polynomial in $y$.

 

[renormalize]

I continue beating the dead horse that is this problem by exhibiting a solution in terms of trigonometric functions.
Given that $1=a^{2}+b^{2}+c^{2}$, it's natural to interpret $\left(a,b,c\right)$ as the components of a unit-vector in 3D and parametrize those three components in terms of two spherical angles $\theta,\phi\,$:$$a=\sin\theta\cos\phi,\;b=\sin\theta\sin\phi,\;c=\cos\theta\tag{1a,b,c}$$where $0\leq\theta\leq\pi,0\leq\phi<2\pi\,$. The second constraint then becomes an equation interrelating $\theta$ and $\phi\,$:$$1=a^{3}+b^{3}+c^{3}=\cos^{3}\theta+\sin^{3}\theta\left(\cos^{3}\phi+\sin^{3}\phi\right)\tag{2}$$the solution of which will allow us to complete the assigned task of finding:$$x\equiv a+b+c=\cos\theta+\sin\theta\left(\cos\phi+\sin\phi\right)\tag{3}$$An obvious solution of (2) is found by setting $\text{sin}\theta=0\Rightarrow\theta=0$ (since $\text{cos}\pi$ has the wrong sign) and $\phi$ arbitrary. Next, for $\theta\neq 0$ we can divide eq.(2) by $\text{sin}^3\theta$ to get:$$f\left(\theta\right)\equiv\csc^{3}\theta-\cot^{3}\theta=\sin^{3}\phi+\cos^{3}\phi\equiv g\left(\phi\right)\tag{4}$$To solve this, it's important to observe that, almost everywhere, $f\left(\theta\right)>1$ but $g\left(\phi\right)<1$:

That means eq.(4) has only two isolated solution pairs $\theta,\phi\,$, which I enumerate below, along with the $\theta=0$ case:\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \theta & \phi & f\left(\theta\right) & g\left(\phi\right) & a & b & c & x\\
\hline 0 & \text{arb.} & - & - & 0 & 0 & 1 & 1\\
\hline \frac{\pi}{2} & 0 & 1 & 1 & 1 & 0 & 0 & 1\\
\hline \frac{\pi}{2} & \frac{\pi}{2} & 1 & 1 & 0 & 1 & 0 & 1\\
\hline
\end{array}Evidently there are just three possible solutions for $\left(a,b,c\right)$, namely: $\left(1,0,0\right),\left(0,1,0\right),\left(0,0,1\right)$, all of which give $x=a+b+c=1$.
OK, that's all folks!
(Edited on 2/6 to eliminate two spurious solutions arising from having erroneously doubled the range of $\theta$ (duh!).)

 

[bob012345]

I found that permutations of $(-1+\frac{i}{\sqrt{2}},-1-\frac{i}{\sqrt{2}},0)$ work

 

[PeroK]

That also explains why we cannot show algebraically that $a+b+c =1$.

 

[littlemathquark]

I found that permutations of $(-1+\frac{i}{\sqrt{2}},-1-\frac{i}{\sqrt{2}},0)$ work

I found this solution message #22

 

[bob012345]

I found this solution message #22

Sorry, I missed that. But what of your other question of how many complex solutions there are? And what is $a+b+c$ for complex solutions?

 

[haruspex]

Sorry, I missed that. But what of your other question of how many complex solutions there are? And what is $a+b+c$ for complex solutions?

Post #34 shows there is a continuum of complex solutions.

 

[bob012345]

Post #34 shows there is a continuum of complex solutions.

Can you elaborate please? Is it because there are more variables than equations?

 

[haruspex]

Can you elaborate please?

As I wrote, you can reduce it to a sixth degree polynomial in two of the variables. Plug in any value for one and solve for the other. By the Fundamental Theorem of Algebra, there is at least one root in the complex plane.

 

[bob012345]

As I wrote, you can reduce it to a sixth degree polynomial in two of the variables. Plug in any value for one and solve for the other. By the Fundamental Theorem of Algebra, there is at least one root in the complex plane.

Ok, thanks, I see now.

 

[WWGD]

Sorry, I missed that. But what of your other question of how many complex solutions there are? And what is $a+b+c$ for complex solutions?

Since Complex curves of Complex Dimension satisfy the same relation, the dimensions of the Intersection will depend on the dimensions of the ambient space.

 

[bob012345]

I found that there are 18 complex solutions to this problem according to Wolfram Alpha results which are permutations of $(0,-1-\frac{i}{\sqrt{2}},-1+\frac{i}{\sqrt{2}})$, $(\frac{3}{2},-1-\frac{i}{\sqrt(2)},\frac{1}{2}-\sqrt{2}i)$ and $(\frac{3}{2},-1+\frac{i}{\sqrt{2}},\frac{1}{2}+\sqrt{2}i)$ with the sums being $-2, 1+\frac{3i}{\sqrt{2}},1-\frac{3i}{\sqrt{2}}$ in addition to the real solutions each with sum $1$.

 

[frankieliu]

Interesting! Maybe also try expanding (a+b+c)3(a + b + c)^3(a+b+c)3 and comparing it with a3+b3+c3+3(a+b+c)(ab+bc+ca)−3abca^3 + b^3 + c^3 + 3(a + b + c)(ab + bc + ca) - 3abca3+b3+c3+3(a+b+c)(ab+bc+ca)−3abc. That might help isolate abcabcabc more cleanly.

 

[WWGD]

Interesting! Maybe also try expanding (a+b+c)3(a + b + c)^3(a+b+c)3 and comparing it with a3+b3+c3+3(a+b+c)(ab+bc+ca)−3abca^3 + b^3 + c^3 + 3(a + b + c)(ab + bc + ca) - 3abca3+b3+c3+3(a+b+c)(ab+bc+ca)−3abc. That might help isolate abcabcabc more cleanly.

Genesis isolated Abacab.

 

[fresh_42]

Genesis isolated Abacab.

I only get ...