<s>[INST] Please help on this Invariant Problem on Authur Engel's Problem Solving Strategies I'm reading the solution to this problem but got stuck, I shall post the problem along with the solution below and will point out where I do not understand.
Problem: 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  $(a - b, b - c, c - d, d - a)$. Prove that at least one number of the quadruple will become arbitrary large.
Solution:
Let $P_{n} = (a_{n}, b_{n}, c_{n}, d_{n})$ be the quadruple after $n$ iterations, then we have $a_{n} + b_{n} + c_{n} + d_{n} = 0$ for $n≥1$. A very important function for point $P_{n}$ in 4-dimensional space is the square of its distance from the origin $(0,0,0,0)$, which is $a^2_{n} + b^2_{n} + c^2_{n} + d^2_{n}$. If we can prove that it has no upper bound, we would be finished.
We try to find a relation between $P_{n+1}$ and $P_{n}$. Hence,
$a^2_{n+1} + b^2_{n+1} + c_{n+1}^2 + d_{n+1}^2 = (a_{n} - b_{n})^2 + (b_{n} - c_{n})^2 + (c_{n} - d_{n})^2 + (d_n - a_n)^2 
= 2(a^2_n + b^2_n + c^2_n + d^2_n) - 2a_{n}b_{n}  - 2b_{n}c_{n} - 2c_{n}d_{n}  - 2d_{n}a_{n}.\tag{1}$
Now, since $a_{n} + b_{n} + c_{n} + d_{n} = 0$, then
$0 = (a_{n} + b_{n} + c_{n} + d_{n})^2 = (a_n + c_n)^2  + (b_n + d_n)^2 + 2a_{n}b_n + 2a_{n}d_n + 2b_{n}c_n + 2c_{n}d_n.\tag{2}$
Okay, I under the steps above, what what I don't understand is this:
Adding (1) and (2) for $a^2_{n+1} + b^2_{n+1} + c_{n+1}^2 + d^2_{n+1}$ we get
$2(a^2_n + b^2_n + c^2_n + d^2_n) + (a_n + c_n)^2 + (b_n + d_n)^2 ≥ 2(a^2_n + b^2_n + c^2_n + d^2_n).\tag{3}$
From this invariant relationship, we conclude for $n≥2$
$a^2_n + b^2_n + c^2_n + d^2_n ≥ 2^{n-1}(a^2_1 + b^2_1 + c^2_1 + d^2_1) . \tag{4}$
Hence, the distance of the point $P_n$ from origin increases without bound, which means that at least one of the component will become arbitrary large.
Okay, I understand (1) and (2) but I don't understand how he obtained (3) and (4). Please someone should provide detailed explanation on the author obtained (3) and (4). Thanks in advance.
 [/INST] As pointed out in the comments, there is an error in (2); you should have something like
$$0 = ((a+c)+(b+d))^2 = (a+c)^2+(b+d)^2 + 2(a+c)(b+d)$$
giving
$$(a+c)^2+(b+d)^2 = -2ab-2ad-2cb-2cd$$
But the RHS here is exactly what appeared in (1).
So substituting (2) into (1) gives
$$a^2_{n+1} + b^2_{n+1} + c^2_{n+1} + d^2_{n+1} = 2 (a^2_{n} + b^2_{n} + c^2_{n} + d^2_{n}) + (a_n+c_n)^2+(b_n+d_n)^2$$
But squares are always non-negative so
$$a^2_{n+1} + b^2_{n+1} + c^2_{n+1} + d^2_{n+1} \ge 2 (a^2_{n} + b^2_{n} + c^2_{n} + d^2_{n})$$
which is (3).
Then by induction or similar, it follows that 
$$\begin{align}
a^2_{n+1} + b^2_{n+1} + c^2_{n+1} + d^2_{n+1}
&\ge 2 (a^2_{n} + b^2_{n} + c^2_{n} + d^2_{n}) \\
&\ge 2 \cdot 2 (a^2_{n-1} + b^2_{n-1} + c^2_{n-1} + d^2_{n-1}) \\
& \ge \cdots \\
& \ge 2^{n} (a^2_{1} + b^2_{1} + c^2_{1} + d^2_{1}) \\
\end{align}$$
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