Prove a = b = c

Date: 01/27/2002 at 15:05:55
From: Jim Chang
Subject: Proof

When a^2 + b^2 + c^2 = ab + bc + ca and abc does not equal 0, prove 
that a = b = c.

I tried aa + bb + cc = ab + bc + ca and then substituting all a's for 
b's: bb + bb + cc = bb + bc + cb, giving me bb + cc = 2bc, which is 
correct, but I need algebraic proof. 

Thanks!

Date: 01/28/2002 at 03:07:42
From: Doctor Floor
Subject: Re: Proof

Hi, Jim,

Thanks for your question.

From abc unequal to zero, we see that none of a,b,c is equal to zero.

Now let's suppose that a <= b <= c. Then we can find nonnegative 
integers t and u satisfying b = a + t and c = a + u. Substitution 
yields

 a^2 + b^2 + c^2 = 3a^2 + 2(t+u)a + t^2 + u^2

 ab + bc + ba = 3a^2 + 2(t+u)a + tu

So we derive from a^2+b^2+c^2 = ab+bc+ca that

 t^2 + u^2 = tu   [*]

We now must show that t = u = 0. 

If t = 0, and we substitute in [*], we find that u^2 = 0, and thus 
u = 0. So it is enough to show that one of the two must be 0.

Suppose that both t and u are positive. Then it is not difficult to 
see:

 * if t > u then t^2 > ut and thus also t^2 + u^2 > ut, 
 * if t < u then t^2 + u^2 > ut in the same way.
 * if t = u, clearly t^2 + u^2 = 2ut > ut.

So that leads to a contradiction. And thus t = u = 0, and also 
a = b = c.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   

Date: 02/04/2002 at 18:42:57
From: Doctor Allan
Subject: Re: Proof

Hello Jim,

I would like to offer another solution to your problem besides the 
good solution that Dr. Floor already provided to you.

If we try to multiply the equation by 2 we get

   2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca

Rewrite this as

   2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0
   a^2 + a^2 + b^2 + b^2 + c^2 + c^2 - 2ab - 2bc - 2ca = 0

Rearranging the terms yields

   a^2 + b^2 - 2ab + b^2 + c^2 - 2bc + c^2 + a^2 - 2ca = 0
   (a-b)^2 + (b-c)^2 + (c-a)^2 = 0

When is it the case that this is zero?

- Doctor Allan, The Math Forum
  http://mathforum.org/dr.math/