Challenge Problem

[asdf]

Oops. I was wrong...

[mfgoode]

09/01/05

Jinydu: You will realise the many false assumptions you have made by following this simple demonstration.

To Prove: a^4 + b^4 +c^4 +d^4 = 0 when a +b +c +d = 0
Let X = a + b and Y = c + d. Given that X + Y =0. Then expanding
by B.T.

( X + Y ) ^ 4 = X ^ 4 + 4*X ^3*Y + 6 * X ^ 2* Y ^ 2 +4 * X* Y ^ 3 + Y ^ 4
= X^4 + 4( X ^3) Y + 6 (X^2) ( Y ^ 2) + 4 X Y^ 3 + Y^4 [dispensing with the multiplication sign for easy reading]

Thus X^4 + Y^4 = (X + Y) ^ 4 - 4 (X^3) Y - 6 (X^2) (Y^2) - 4 X (Y^3)

Simplifying R.H.S. = 0 - 2 X Y [ 2 (X^2) + 3 (X Y) +2 (Y^2)
If X^4 + Y ^4 is to be = 0 then R.H.S. =0

We then conclude 1) X = 0 or
2) Y = 0 or
3) both X , Y = 0 or
4) [2X^2 + 3 XY +2Y^2] = 0

By analysing the above options we come to the conclusion that X = Y =0
Similarly for n= 5 group a, b, c, d, e, as (X + Y) and work out X^5 + Y ^ 5 = 0 the same way without much trouble.

Then By the Method of Induction Let this be true for X ^n + Y ^n.
Then prove it is also true for exponent (n + 1 ). And you have your proof knowing that it is true for X + Y = 0 ; X^2 + Y^2 = 0 ; X ^3 + Y ^3 and if its true for n it will be true for n + 1. and so on up 1,2,3,4,5 etc. Q. E .D.
Mally.

[Wacky]

Originally Posted by mfgoode

To Prove: a^4 + b^4 +c^4 +d^4 = 0
when a +b +c +d = 0
Let X = a + b and Y = c + d.

<snip>

If X^4 + Y ^4 is to be = 0

Mally,
I am unable to follow your reasoning.

Where do you get "X^4 + Y^4" ?

"X^4 + Y^4" is certainly not the same as "a^4 + b^4 +c^4 +d^4"

[mfgoode]

wink:
Thanks Wacky! I anticipated the objection but you beat me to it before I could put down an amendment.
We have derived X^4 + Y ^4 = 0 which is strictly equivalent to (A + B )^4
+ (c + d) ^4 = 0.
This expression involves (- 2 ab ) and (- 2cd ) as factors on simplifying similar to (-2 X Y ) resuting in the same conclusion . Hence a ^4 + b ^4 + c ^4 + d^4 = 0.
Good observation Wacky! I hope this explanation is satisfactory to you.
I am keen to know of any more flaws in this proof. After all it is converting jinydu's conjecture into a theorem.
Mally

[jinydu]

Originally Posted by mfgoode

wink:
Thanks Wacky! I anticipated the objection but you beat me to it before I could put down an amendment.
We have derived X^4 + Y ^4 = 0 which is strictly equivalent to (A + B )^4
+ (c + d) ^4 = 0.
This expression involves (- 2 ab ) and (- 2cd ) as factors on simplifying similar to (-2 X Y ) resuting in the same conclusion . Hence a ^4 + b ^4 + c ^4 + d^4 = 0.
Good observation Wacky! I hope this explanation is satisfactory to you.
I am keen to know of any more flaws in this proof. After all it is converting jinydu's conjecture into a theorem.
Mally

Thanks for the reply, mfgoode

Are you trying to show that:

A^4 + B^4 + C^4 + D^4 = 0 (the problem's original assumption) implies X^4 + Y^4 = 0

?

Expanding X^4 + Y^4 and applying that assumption, we get the expression

4(A^3)B + 6(A^2)(B^2) + 4A(B^3) + 4(C^3)D + 6(C^2)(D^2) + 4C(D^3)

Factoring out, as you said, we get

2AB(2A^2 + 3AB + 2B^2) + 2CD(2C^2 + 3CD + 2D^2)

But just from the original assumptions of the problem, why can we conclude that the above expression
(= X^4 + Y^4) must equal 0?

Thanks for your efforts, everyone

Jinydu

[mfgoode]

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.

Jinydu: It appears I have put the cart before the horse. However this should be simpler to prove as in a way it goes backwards of the proof I have given.

I will not be with this thread or the others for quite a while as Im leaving for a one horse town for at least two weeks where there are few computers and only one cyber cafe. Anyway I will have plenty of time to think the matter thru. Kindly go thru my personal message and reply as your conjecture has vast possibilities.

Mally

[graeme]

A quick and easy proof:

All cases with n=> 4:

The line A^4 + B^4 + C^4 + D^4 + .... = 0 occurs

any complex number that is raised to the 4th power must be either zero or positive.
So each term must be either zero or a positive value.
No term can be positve therefore they are all individually = 0

A^4=0 has only one solution A=0
therefore all A,B,C,D,... are zero

case with n=3
A+B+C = 0 (1)
A^2+B^2+C^2 =0 (2)
A^3 +B^3 +C^3 = 0 (3)
from (1)
A + B = - C
(A + B)^2 = C^2 (4)
sub (4) into (2)
A^2 + B^2 + (A+B)^2 =0 (5)
2A^2 + 2B^2 + 2AB =0 (6)
A^2 +B^2 +AB = 0 (7)
AB=C^2 (8)

Similarly A^2=BC (9)
and B^2=AC (10)

from (8)
C^4 = A^2B^2 = A^3C (11)
C^3 = A^3 (12)
so A^3 = B^3 = C^3 (13)

sub into (3)
3A^3 = 0 (14)
A=B=C =0 Q.E.D


Case with n=1 and 2 already proven

hope you enjoyed it. (hope it passes muster)
Graeme

[Wacky]

Originally Posted by graeme

any complex number that is raised to the 4th power must be either zero or positive.

(hope it passes muster)
Graeme

Sorry, this assertion is incorrect.

Consider, X = sqrt(2)/2 * (1+i)
Then X^2 = 2/4 * (2*i) = i
And X^4 = i^2 = -1

[graeme]

Bah - Wacky ruins a good proof by pointing out that it's rubbish. Serves me right for using a physicists notion of z^2 (=zz*) rather than those hoity-toity mathematicians version. (zz* is guaranteed to be real; and squaring this is guranteed to be +ve or 0)

OK one final stab and then I'll crawl back under my rock:

consider the genral case:

A + B + C + D + ... = 0 (1)
A^2 + B^2 + C^2 + D^2 + ... = 0 (2)

Multiply (1) by A

A^2 + AB + AC + AD + .... = 0 (3)

Equate (2) and (3)

AA + AB + AC + AD + ... = AA + BB + CC + DD + ...

Can we now compare the co-efficents on each side, and say that

A=A (!)
A=B
A=C
A=D ?

and therefore A,B,C,D... =0


Graeme

[Wacky]

Originally Posted by graeme

then I'll crawl back under my rock:

Down on your hands and knees! The rock is over to your left.

AA + AB + AC + AD + ... = AA + BB + CC + DD + ...

Can we now compare the co-efficents on each side, and say that

This isn't Chemistry either ... No, we cannot.

In order to compare coefficients, each equation would have to hold "For all B" , etc.

Consider:

A = 1, B = i, C = -1, D = -i

A + B + C + D = 1 + i - 1 - i = 0
and
A^2 = 1, B^2 = -1, C^2 = 1, D^2 = -1
therefore
A^2 + B^2 + C^2 + D^2 = 1 - 1 + 1 - 1 = 0
further
A^3 + B^3 + C^3 + D^3 = 1 - i - 1 + i = 0
but
A^4 + B^4 + C^4 + D^4 = 1 + 1 + 1 + 1 = 4

[jinydu]

I still haven't given up!

I've made another attempt on the n = 4 case using a modified version of the Division Method.

http://www.sosmath.com/CBB/viewtopic...12231&start=15

Note that I'm not just looking for any proof with n = 4, I'm looking for a simple proof, in the sense that it doesn't require expanding a giant expression.

Also, I tried to show that one of the variables must be zero without using the second and third equation.

Note that in a proof by contradiction, I can assume that all of the variables are nonzero. If not, the problem would reduce to one with a lower value of n, and we have already solved n = 1, 2 and 3. Since, all the variables are nonzero, there should be no reason why we can't divide by them.

As usual, I started by dividing the first equation by D and the 4th equation by D^4. I then apply the substitutions A/D = A', B/D = B', C/D = C'. The equations then become:

A' + B' + C' = -1
A'^4 + B'^4 + C'^4 = -1

I still haven't been able to arrive at a solution yet, but I have managed to derive the equation:

A'B' * (2A'^2 + 3A'B' + 2B'^2) = C'^4 + 2C'^3 + 3C'^2 + 2C' + 1

Unfortunately, I haven't been able to get any farther than this. But since A'B' can't be 0 (since we've assumed that A and B are nonzero), if I could show that either (2A'^2 + 3A'B' + 2B'^2) or C'^4 + 2C'^3 + 3C'^2 + 2C' + 1 must be 0, then it follows that the other must also be zero. That would be a significant step forward.