Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
sci.math.*
»
sci.math
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
Quadruple with (a, b, c, d)
Replies:
2
Last Post:
Feb 2, 2013 11:33 PM




Quadruple with (a, b, c, d)
Posted:
Feb 2, 2013 5:26 AM


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 (ab, bc, cd, da).
Then at least one number of the quadruple will eventually become arbitrarily large.
 Solution) 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 4space 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.
.... omission For reference, text copy with jpg. http://board2.blueweb.co.kr/user/math565/data/math/olilim.jpg ...
for n >=2, (a_n)^2 + (b_n)^2 + (c_n)^2 + (d_n)^2 >= {2^(n1)}*{(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 and 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 large". I need your logical explanation or proof.



