Sum of squares equation problem

[The legend]

Homework Statement

Prove only solution in integers of the equation

x² + y ² + z² = 2xyz

is x = y = z =0


2. The attempt at a solution

Well, using common sense got the idea but don't exactly know how to prove it!
Can anyone please help as how to start ...?

Thanks!

 

[HallsofIvy]

Well, a proof is often in the details of common sense. What "common sense" gave you the idea?

 

[The legend]

Suppose we take in negative integers.
then the sum of squares is positive and the product beside is negative.(eliminated possibility)

Then again taking 0 the answer is satisfied.

Then taking positive ints by a bit of trial none (That I found ) satisfy the eqn
(though this trial is the main problem that i faced)

So the thing is I need the exact proof and not trial.

 

[Dick]

The general idea is to show that if x,y,z is a solution then x/2,y/2,z/2 is also a solution. It's an infinite descent thing. One of x, y, and z must be even, right? Now consider divisibility by 4. You won't get the exact proof here. You only get hints. You'll have to work the proof out for yourself.

 

[The legend]

Ok .. here's what i did after your suggestion( thanks!)

since it must keep on working for x/2 , y/2 , z/2 and then by 4 and so on finally it must come to x/x , y/y , z/z (that is 1 ,1 ,1)

So the equation comes as this

1^2 + 1^2 + 1^2 = 2 * 1 * 1 *1

But since the equality doesn't hold true... well the only answer remaining is x = y = z =0.

Is this right??

 

[Dick]

Ooops. Actually my 'hint' isn't even right. The equation isn't homogeneous in x, y, z. Sorry. Here's what I meant to say. Suppose 2^k is the largest power of two dividing all of the numbers x, y and z. Then x=2^k*x', y=2^k*y', z=2^k*z', where at least one of x', y', z' is odd. Write out the equation in terms of x', y' and z' and show you have a problem mod 4.

 

[The legend]

Got it!

writing in terms of x', y', z'

(2^kx')² + (2^ky')² + (2^kz')² = 2 * (2^kx') * (2^ky') * (2^kz')

x' + y' + z' = 2^3k+1-2k * x'* y'* z'

x' + y' + z' = 2^k+1 * x'* y'* z'

But this contradicts the main problem statement..in terms of x'* y'* z'(for all values except 0 = x = y = z)

Right?

And also I want to know what was that mod 4 cause I don't know how to use it in a sum. Could you please tell me how to do this sum by that method?

 

[Dick]

Got it!

writing in terms of x', y', z'

(2^kx')² + (2^ky')² + (2^kz')² = 2 * (2^kx') * (2^ky') * (2^kz')

x' + y' + z' = 2^3k+1-2k * x'* y'* z'

x' + y' + z' = 2^k+1 * x'* y'* z'

But this contradicts the main problem statement..in terms of x'* y'* z'(for all values except 0 = x = y = z)

Right?

And also I want to know what was that mod 4 cause I don't know how to use it in a sum. Could you please tell me how to do this sum by that method?

What happened the squares on x' etc? And how is that a contradiction? If you square an odd number what's its value mod 4? How about squaring an even number?

 

[The legend]

Since I don't know much about this mod 4(nothing at all actually!) I will try this out when I learn about it. Can you suggest from which website can i learn about this mod?? (Its not in my syllabus so no school teacher going to teach me nor do I have the books havin that )

 

[Dick]

A value of x 'mod 4' is just the remainder you get if you divide x by 4. If x is even what's the remainder if you divide x^2 by 4? Answer the same question in the case x is odd. You can also try looking at http://en.wikipedia.org/wiki/Modular_arithmetic or there must be a lot of other references.

Examples:

2 mod 4=2
3 mod 4=3
4 mod 4=0
5 mod 4=1
6 mod 4=2
7 mod 4=3
8 mod 4=0
etc etc