Show that there is a large value 
Suppose not all 4 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 show that at least one number of the quadruple will become arbitrarily large.

Let $x_k$ be the pair with:
$x_k = (a_k, b_k, c_k, d_k)$ and $x_{k+1} = (a_k - b_k, b_k - c_k, c_k - d_k, d_k - a_k)$ then it follows that,
$a_k - b_k + b_k - c_k + c_k - d_k + d_k - a_k = 0 = L$
But I cant show any further. The idea is to find an invariant. I did find one but I can't do anything with it.
Actually using
$x_k = (a_k, b_k, c_k, d_k)$
It is seen that,
$a_k = a_{k-1} - b_{k-1}$, then
$a_k + b_k + c_k + d_k = a_{k-1} - b_{k-1} + b_{k-1} - c_{k-1} + ... = 0$
So I have found the invariant.
$(a_k + b_k + c_k + d_k)^2 = a_k^2 + 2a_kb_k + 2a_kc_k + 2a_kd_k + b_k^2 + 2b_kc_k + 2b_kd_k + c_k^2 + 2c_kd_k + d_k^2 $
 A: This is problem $E5$ under The invariance principle in Arthur Engel's book titled "Problem Solving Strtegies"

We have
$$(a_{n+1},b_{n+1},c_{n+1},d_{n+1}) = (a_n-b_n,b_n-c_n,c_n-d_n,d_n-a_n)$$
As you have rightly observed, the invariant is $a_n+b_n+c_n+d_n = 0$ for all $n \geq 1$, where $a_0 = a$, $b_0=b$, $c_0 = c$ and $d_0 = d$. Let us now look at $a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2$. We have
\begin{align}
a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 & = (a_n-b_n)^2+(b_n-c_n)^2+(c_n-d_n)^2 + (d_n-a_n)^2\\
& = 2(a_n^2+b_n^2+c_n^2+d_n^2) - 2(a_nb_n+b_nc_n+c_nd_n+d_na_n) & (\spadesuit)
\end{align}
We shall now show that $ - 2(a_nb_n+b_nc_n+c_nd_n+d_na_n)$ is non-negative for $n \geq 1$. Since $a_n+b_n+c_n+d_n=0$, we have
\begin{align}
0 & = (a_n+b_n+c_n+d_n)^2 = (a_n+c_n)^2 + (b_n+c_n)^2 + 2(a_n+c_n)(b_n+d_n)\\
& = (a_n+c_n)^2 + (b_n+c_n)^2 + 2a_nb_n + 2b_nc_n + 2c_nd_n + 2d_na_n
\end{align}
This gives us
$$-2(a_nb_n+b_nc_n+c_nd_n+d_na_n) = (a_n+c_n)^2 + (b_n+c_n)^2 \geq 0 \,\,\, (\clubsuit)$$
Making use of $(\clubsuit)$ in $(\spadesuit)$, we obtain that
$$a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \geq 2(a_n^2+b_n^2+c_n^2+d_n^2)$$
This gives us
$$a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \geq 2^n(a_1^2+b_1^2+c_1^2+d_1^2)$$
Since $a, b,c,d$ are distinct, one of the terms $a_1,b_1,c_1$ or $d_1$ is non-zero. Hence, $a_{n+1}^2+b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2$ grows unbounded, which means the largest of the terms must keep growing unbounded.
A: It's about the transformation $T:\>{\bf a}\mapsto{\bf a}'$ where $a_j':=a_j-a_{j+1}$ for $j\in{\mathbb Z}_4$. Since $T$ commutes with cyclic shifts $j\to j+1$ we are led to consider  the discrete Fourier transform ${\bf c}:=\hat{\bf a}$ of the ${\bf a}$'s. Since $\omega:=e^{2\pi i/4}=i$ one has
$$c_k'=\sum_j a_j'i^{-jk}=\sum_j(a_j-a_{j+1})i^{-jk}=(1-i^k)c_k\ .$$
It follows that the application of $T$  has the effect
$$\tilde T:\quad (c_0,c_1,c_2,c_3)\mapsto\bigl(0, \ (1-i)c_1,\ 2c_2, \ (1+i)c_3\bigr)$$
on ${\bf c}$, in other words: $T$ is now diagonalized. If the starting values $(a_j)_{j\in{\mathbb Z}_4}$ are not all equal then at least one starting $c_k$,  $k\ne0$, is different from zero. As $|1\pm i|>1$ such $c_k$ will be blown up by successive applications of $\tilde T$. Using Parseval's theorem it then follows that $\|T^n{\bf a}\|^2$ converges to $\infty$ when $n\to\infty$.
