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Topic: Quadruple with (a, b, c, d)
Replies: 2   Last Post: Feb 2, 2013 11:33 PM

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Posts: 2,144
Registered: 12/13/04
Quadruple with (a, b, c, d)
Posted: Feb 2, 2013 5:26 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

Hello teacher~

Suppose not all four integers a, b, c, d are equal.
Start with (a, b, c, d) and repeatedly replace (a, b, c, d)
by (a-b, b-c, c-d, d-a).

Then at least one number of the quadruple will eventually
become arbitrarily large.

Let P_n = (a_n, b_n, c_n, d_n) be the quadruple after n iterations.
Then a_n + b_n + c_n + d_n = 0 for n >= 1.

A very importand function for the point P_n in 4-space
is the square of its distance from the origin (0,0,0,0),
which is (a_n)^2 + (b_n)^2 + (c_n)^2 + (d_n)^2.

For reference, text copy with jpg.

for n >=2,
(a_n)^2 + (b_n)^2 + (c_n)^2 + (d_n)^2
>= {2^(n-1)}*{(a_1)^2 + (b_1)^2 + (c_1)^2 + (d_1)^2}.

The distance of the points P_n from the origin increases without bound,
which means that at least one component must become arbitrarily large

Hm, my question is...

I know that a_n + b_n + c_n + d_n = 0
I know that (a_n)^2 + (b_n)^2 + (c_n)^2 + (d_n)^2 increases without bound.

I can't understand that "at least one component must become arbitrarily
I need your logical explanation or proof.

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