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.

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 
$$\begin{align}
a^2_{n+1} + b^2_{n+1} + c^2_{n+1} + d^2_{n+1}
&\ge 2 (a^2_{n} + b^2_{n} + c^2_{n} + d^2_{n}) \\
&\ge 2 \cdot 2 (a^2_{n-1} + b^2_{n-1} + c^2_{n-1} + d^2_{n-1}) \\
& \ge \cdots \\
& \ge 2^{n} (a^2_{1} + b^2_{1} + c^2_{1} + d^2_{1}) \\
\end{align}$$
