Equality of sums of powers

[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 ...