Question

What are all the possible solutions, real or complex?

Answer 1

\[a+b+c=3\]\[a^2+b^2+c^2=3\]\[a^3+b^3+c^3=3\]

Answer 2

1,1,1 is the easiest one

Answer 3

Yup yup, definitely good idea to get that one outta the way.

Answer 4

Why did you delete my posts?
Before, you try to prove something, one as to try some symbolic manipulation first.
Please do not censor what people respond to your post.
Thanks

Answer 5

Can one prove that 1,1,1 is the only solution?

Answer 6

Haha sure, but that's no fun. If you want to cheat and get the answer before you try to prove it on your own, what's the point? I'll let other people cheat for themselves if they want.

Answer 7

Let us see your solution.

Answer 8

I wanna see what other people come up with first, I barely just posted the question!

Answer 9

I found the general proof.
Take the sphere x^2 +y^2 + z^2=3.
Find the tangent plane to it at (1,1,1)
You find it equal to x+y+z=3
What it means that the sphere and the plane intersectsat (1,1,1) only

Answer 10

The third cubic equation is not needed

Answer 11

The gradient vector on the sphere at (1,1,1) is
(2,2,2)
So the equation of the tangent plane is
(2,2,2) dot ( x-1,y-1,z-1)=0
2(x-1) + 2(y-1) + 2(z-1)=0
2 x + 2y +2 z=6
x+y+z=3

Answer 12

Aha very interesting. But this only shows that (1,1,1) is the only all real solution! There can still be complex solutions you've left out!

Answer 13

does it have to do with e^i2pi/k form solutions

Answer 14

Yeah good idea, maybe?

Answer 15

i dont like ur tone mister

Answer 16

Haha well I already figured out a solution before I posted this question so I am curious to see how other people work it. I'll share my answer later haha. Work with @agent0smith maybe, he's lurking this question lol.

Answer 17

kay ill get on it, im gonna go to mcdonalds first

Answer 18

I'll leave a hint
\[x^3-3x^2+3x-1=0\]

Answer 19

Alright I'll go ahead and post the solution I discovered!

First, let's prove that there is no complex part by subtracting the complex conjugate of the second equation from itself!

\[a^2+b^2+c^2=3\]
\[\frac{a^2-(a^2)^*}{2i}+\frac{b^2-(b^2)^*}{2i}+\frac{c^2-(c^2)^*}{2i}=0\]
Let's call the imaginary parts of a, b, and c \(\alpha\), \(\beta\) and \(\gamma\) respectively. These are all real numbers.
\[\alpha^2 +\beta^2 +\gamma^2 = 0\]
This can only be 0 when each of these are 0, therefore the imaginary part of all the solutions must be zero! Now from here the proof is done combined with @eliesaab However I'll explain the rest of the proof I found.

What I did was I substituted in \(a=1+x\), \(b=1+y\) and \(c=1+z\) just putting in a deviation from the known root to see if that did anything interesting. It immediately gives us the first equation becoming:
\[x+y+z=0\]
Now for the final stroke, plug this into the second equation to get:
\[2x+x^2+2y+y^2+2z+z^2=0\]
However we know from the first that it simplifies to:
\[x^2+y^2+z^2=0\]
and hey, same reasoning for imaginary. No other solutions exist!

Answer 20

Nicee

Answer 21

Here is the solution using my earlier hint :

Consider the monic polynomial \(P(x)=x^3 - 3 x^2 + mx+n\)
Let \(a,b,c\) be its roots and let \(P_i = a^i + b^i + c^i\).

\(P_0 = 3\)

\(P_1 =a+b+c=3\tag{1}\)

\(P_2=a^2+b^2+c^2 = (a+b+c)(a+b+c) - 2(ab+bc+ca) \tag{2}\)

\(P_3=a^3+b^3+c^3 = (a+b+c)(a^2+b^2+c^2) - (ab+bc+ca)(a+b+c) +3abc \tag{3}\)


From (2),
\(9-2m = 3\implies m=3\)

From (3),
\(3*3-m*3+3(-n) = 3\implies n=-1\)

That means the unique monic polynomial representing the given sum of powers of roots is
\[P(x)=x^3-3x^2+3x-1\]
and the only root is \(x=1\).

Answer 22

Oh, woah that's pretty cool. I have to think about it, I'm not sure I fully get how this works something about it seems like magic. I see how you're piecing the parts of the \(P_i\) together to get a unique cubic polynomial, but I don't know why the existence of a unique polynomial necessarily connects to the only solution.

Answer 23

I think what I have shown is there exists a bijection between "first n powers of n numbers" and "nth degree monic polynomials"

Given a polynomial, its roots are unique. Since a bijection exists, given a set of "first n powers of n numbers", the corresponding monic polynomial is unique.