Challenge Problem

[jinydu]

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?

[Fusion_power]

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

Fusion

[jinydu]

What magically disappearing negative signs?

[Orgasmic Troll]

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

[Mystwalker]

Originally Posted by 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

This also holds true for all even exponents.

[jinydu]

No, the problem stated that:

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

are allowed to be complex numbers.

[mfgoode]

Jinydu:
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

[mfgoode]

Originally Posted by jinydu

No, the problem stated that:

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

are allowed to be complex numbers.

AnY number can be written in complex form
Take 2 =2+i*0
Mally

[Wacky]

Originally Posted by mfgoode

Jinydu:
In your proof you assume that a^2 + b^2 = 0 when that is what you have to prove.

I think that you have perhaps misread the statement of the problem:

Originally Posted by 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

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.

[jinydu]

Originally Posted by 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.

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.

[wpolly]

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).