Prove that for $a ,b, c, d : a_{n+1} = |a_n-b_n| , b_{n+1} = |b_n-c_n|, c_{n+1} = |c_n-d_n|, d_{n+1} = |d_n-a_n|$ recursively results in $0,0,0,0$. Please prove that:
For $a ,b, c, d : a_{n+1} = |a_n-b_n| , b_{n+1} = |b_n-c_n|, c_{n+1} = |c_n-d_n|, d_{n+1} = |d_n-a_n|$ repeated recursively results in $0,0,0,0$ .
The recursion happens all at once, so in effect $a_{n+1} = |a_n-b_n| , b_{n+1} = |b_n-c_n|, c_{n+1} = |c_n-d_n|, d_{n+1} = |d_n-a_n|$
For example,
$1 2 3 5, 1 1 2 4, 0 1 2 3, 1 1 1 3, 0 0 2 2, 0 2 0 2, 2 2 2 2, 0 0 0 0$
.
It would seem to work for all real numbers.
 A: This is not true in general. The key observation is that if the sequence stops for some initial condition $(a, b, c, d)$, then it stops for $(\lambda a+\mu, \lambda b+\mu, \lambda c+\mu, \lambda d + \mu)$.
Let
$$ q=\frac{1}{3} \left( 1+\sqrt[3]{19+3\sqrt{33}}+\sqrt[3]{19-3\sqrt{33}} \right)\approx 1.8393 $$
be the unique real root of $q^3-q^2-q-1=0$, and consider
$$ a_1=0, \quad b_1=1, \quad c_1=q(q-1)\approx 1.54369, \quad d_1=q \approx 1.8393. $$
We can easily check that
$$ a_2=1, \quad b_2=q^2-q-1, \quad c_2=2q-q^2, \quad d_2=q. $$
Now, subtract $q$ and then divide everything by $1-q$. Using that
$$ q^2-2q-1=-q^3+2q^2-q=-q(1-q)^2  $$
thanks to $q^3-q^2-q-1=0$, we get
$$ (1, q(q-1), q, 0), $$
which is the initial condition up to a rotation. This way, the original sequence does not stop.
See The Convergence of Difference Boxes by A. Behn, C. Kribs-Zaleta and V. Ponomarenko for more discussion in this problem; in particular, they show that this is essentially the "only" counterexample.
A: The following arguments works if $a_n,\dots$ are integers (so the conclusion also holds for rational by considering integers $ra_n,rb_n,rc_n,rd_n$ where $r\in\mathbb N$ is some number).
Let $a_n'=a_n-b_n,\,b_n'=b_n-c_n$ and so on.
Then $a_{n+1}'=a_{n+1}-b_{n+1}=|a_n'|-|b_n'|$.
Then
$$\sum_{\rm sym}(a_{n+1}'+b_{n+1}')^2=\sum_{\rm sym}(|a_n'|-|c_n'|)^2\le\sum_{\rm sym}(a_n'+c_n')^2=\sum_{\rm sym}(a_n'+b_n')^2,$$
where the symetric sum means that each of $a,b$ runs through all (possibly the same) $a,b,c,d$.
The equality holds only if $a_n'=b_n'=c_n'=d_n'=0$.
So, $\sum_{\rm sym}(a_n'+b_n')^2$ is decreasing unless $a_n'=b_n'=c_n'=d_n'=0$, in which case $a_{n+1}=b_{n+1}=c_{n+1}=d_{n+1}=0$.
With our assumption, $\sum_{\rm sym}(a_n'+b_n')^2$ only takes non-negative integer values;
hence it cannot decreases forever, and we are done.
