Challenge Problem
 

Try to prove (or disprove) this conjecture:

If A(1), A(2), A(3), A(4), ... A(n) are complex numbers, and:

A(1) + A(2) + A(3) + ... + A(n) = 0
A(1)^2 + A(2)^2 + A(3)^2 + ... + A(n)^2 = 0
A(1)^3 + A(2)^3 + A(3)^3 + ... + A(n)^3 = 0
...
A(1)^n + A(2)^n + A(3)^n + ... + A(n)^n = 0

then the only possible solution is A(1) = A(2) = A(3) = ... = A(n) = 0.

The case of n = 1 is trivial, because the entire system of equations is just:

A = 0

Of course, the only possible solution to that is A = 0.

The case of n = 2 is easy:

A + B = 0
A^2 + B^2 = 0

From the first equation, we have B = -A. Substituting that into the second equation, we have:

A^2 + (-A)^2 = 0
2A^2 = 0
A^2 = 0
A = 0

And from the equation B = -A, B must also equal 0.

The case of n = 3:

A + B + C = 0
A^2 + B^2 + C^2 = 0
A^3 + B^3 + C^3 = 0

does take more effort; but I still managed to solve it and show that A = B = C = 0 (feel free to try it yourself).

Can you solve the problem for all positive integers, n?

I've always been leery of magically disappearing negative signs.

Fusion

What magically disappearing negative signs?

all A(k)^2 >= 0

if any A(k)^2 > 0, then Sum[1 to n]A(k)^2 > 0

therefore A(k) must equal zero for all k

[QUOTE=TravisT]all A(k)^2 >= 0

if any A(k)^2 > 0, then Sum[1 to n]A(k)^2 > 0

therefore A(k) must equal zero for all k[/QUOTE]
This also holds true for all even exponents.

No, the problem stated that:

A(1), A(2), A(3), A(4) ... A(n)

are allowed to be [B]complex numbers[/B].

Challenge Problem
 

Jinydu: :rolleyes:
In your proof you assume that a^2 + b^2 = 0 when that is what you have to prove.
Lets try it again.
Given a + b = 0
Hence (a+b)^2 = a^2 + B^2 +2ab
heance a^2 + b^2 =-2ab and not 0
Test case a =2 ; b=-2
a^2 +b^2 = 4 +4 = 8
-2ab = -2*2*-2 =8
LHS =RHS and eqn is balanced
Similarly for a + b +c= 0
a^3 +b^3 +c^3 = 3a*b*c. [This can be derived ]
Subs for a = 3 ; b=-2 ; c =-1. You will find both sides balance.
Hope thius is clear.
Mally :coffee:

Challenge Problem
 

[QUOTE=jinydu]No, the problem stated that:

A(1), A(2), A(3), A(4) ... A(n)

are allowed to be [B]complex numbers[/B].[/QUOTE]
:smile: AnY number can be written in complex form :wink:
Take 2 =2+i*0
Mally :coffee:

[QUOTE=mfgoode]Jinydu: :rolleyes:
In your proof you assume that a^2 + b^2 = 0 when that is what you have to prove.[/QUOTE]

I think that you have perhaps misread the statement of the problem:
[QUOTE=jinydu]
If A(1), A(2), A(3), A(4), ... A(n) are complex numbers, and:

A(1) + A(2) + A(3) + ... + A(n) = 0
A(1)^2 + A(2)^2 + A(3)^2 + ... + A(n)^2 = 0
A(1)^3 + A(2)^3 + A(3)^3 + ... + A(n)^3 = 0
...
A(1)^n + A(2)^n + A(3)^n + ... + A(n)^n = 0

then [/QUOTE]

As I read it, we are to assume that a+b =0 and a^2 + b^2 = 0 and prove that a=0, b=0 is the solution in the field of complex numbers.

[QUOTE=Wacky]I think that you have perhaps misread the statement of the problem:

As I read it, we are to assume that a+b =0 and a^2 + b^2 = 0 and prove that a=0, b=0 is the solution in the field of complex numbers.[/QUOTE]

Yes, that's right.

Now, the challenge is to show that this is also true for the analogous case of n = 3 (which I have already done), and the general case of any positive integer n (which I haven't done).

As for my statement that the variables are allowed to be complex numbers, my point was that you can't automatically rule out the case where n is even. For instance, if A = i, A^2 = -1 and hence, it is possible for some of the other variables to be positive.

By Newton's Formula, we have that all the elementary symmetric polynomials of A1, A2, ... An is equal to zero.
Thus these A's are roots to the equation x^n=0 and thus are zero(By Vieta's Theorem).