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# Question on well ordering watch

1. Howdy, would like to know how to prove this question, don't know quite which way to go with it >.<

Prove that the equality x2 + y2 + z2 = 2xyz

can hold for whole numbers x, y, z only when x = y = z = 0

thanks for any help.
2. wow wish i could help buddy what is this fp3 ? edexcel
3. nope, just random number theory that I occassionally do....I think I have worked it out, but I am not convinced with myself, so would still like a solution if anyone would be kind enough to provide one 8 )
4. It's obvious that x=y=z=0 satisfies the equation. So let's prove that no other solutions exist. Actually, let's look at the more general case:
x^2 + y^2 + z^2 = 2^u (xyz), where u is an integer >= 1.

Since the RHS is even, then x^2+y^2+z^2 is even. Thus exactly one of x,y,z is even, or all of them are even. However, for the first case we have:
x^2 + y^2 + z^2 = 2 (mod 4)
while 2^u xyz = 0 (mod 4).

Hence all of them must be even. So set x=2a, y=2b and z=2c. Then:
4a^2 + 4b^2 + 4c^2 = 2^u * 2a * 2b * 2c
=> a^2 + b^2 + c^2 = 2^v abc, where v=u+1.

So we see that a,b,c,v satisfy our general equation. Moreover, we have:
0 < a^2 + b^2 + c^2 = (x^2 + y^2 + z^2)/4 < x^2 + y^2 + z^2

Therefore, by the well-ordering of the naturals, the equation can only have a solution when:
x^2 + y^2 + z^2 = 0
=> x=y=z=0

Since this holds for any u>=1, then it holds for u=1.
5. Out of intrest are you working out of the Number Theory Book ? I used to try few questions out of there back in the days but I couldnt hack it in the end .
6. Yep I am, though I only look at it every now and again, have done the first section in it, some pleasing results in there...well, I find them interesting

dvs - Spread rep I must, haven't repped anyone in donkeys. Thanks though
7. (Original post by Exile)
Out of intrest are you working out of the Number Theory Book ? I used to try few questions out of there back in the days but I couldnt hack it in the end .
No kidding... That book looks pretty hardcore.

Thanks for the link though. That will make an excellent summer read.
8. I think after a little initial work, working mod 4 sorts the problem out pretty swiftly.
9. (Original post by Wrangler)
I think after a little initial work, working mod 4 sorts the problem out pretty swiftly.
Care to elaborate?
10. Let s=hcf(x,y,z) then our problem is:

p2 + q2 + r2 = 2spqr

For appropriate {p,q,r}, where exactly one of them is even. Then we have p2=0mod4, q2=1mod4 and r2=1mod4, while 2spqr=0mod4.

I'm absolultely positive I've seen this before though (or some variant of), so I can't really take too much credit.
11. Wow. That's very nice.
12. I've deliberately left out a couple of points, so as not to spoil the question - but I hope I've made the gist of it obvious enough.

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