$a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2$ can be expressed as the sum of two squares in at least two different ways The question: Every expression of the form $a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2$ can be expressed as the sum of two squares in at least two different ways. Find any one of the three possible ordered pairs of
positive integers $(x, y)$, with $x > y$, that satisfies $x^2 + y^2 = 44^2 \cdot 10^2 + 10^2 \cdot 33^2 + 33^2 \cdot 5^2 + 5^2 \cdot 44^2$.
What I found: From factorizing, we see that the first expression is just $(a^2 + c^2)(b^2 + d^2)$. $1 = (0 + 1)(0 + 1)$ and $2 = (0 + 1)(1 + 1)$ are of this form, but they cannot be expressed two different ways. Also, if $2$ is the sum of two squares, then it's also implied that $4 = 2 \cdot 2$, $8 = 4 \cdot 2$, and generally all powers of two should be expressible as a sum of two squares two different ways, but $8$ is another counterexample.
Am I misinterpreting the question, or is it wrong? Also, how do I actually tackle finding $x$ and $y$ for the last sentence of the question?

$$\text{First way} $$
$$a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2=\color{red}{a^2b^2  + c^2d^2 + b^2c^2+ d^2a^2+2(ab)(cd) - 2(bc)(da)}=\color{blue}{(ab+cd)^2+(bc-da)^2} $$
$$\text{Second way} $$
$$a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2= \color{red}{a^2b^2+c^2d^2 -2(ab)(cd)+b^2c^2+ d^2a^2+2(bc)(da)}=\color{blue}{(ab-cd)^2+(bc+da)^2}$$ 
$$\text{Our desired minimum two ways}$$
${\text{Now,}}$
$$x^2 + y^2 = 44^2 \cdot 10^2 + 10^2 \cdot 33^2 + 33^2 \cdot 5^2 + 5^2 \cdot 44^2$$ $$\text{Use the reasoning which has been used till now:}$$
$$x^2 + y^2 = 44^2 \cdot 10^2 + 10^2 \cdot 33^2 + 33^2 \cdot 5^2 + 5^2 \cdot 44^2=\color{red}{(44^2 \cdot 10^2) + (    33^2 \cdot 5^2 ) +2\cdot (44 \cdot 10) \cdot (33 \cdot 5)+  (10^2 \cdot 33^2)+(5^2 \cdot 44^2 )-2(  10 \cdot 33)( 5 \cdot 44)}$$ 
$$\color{blue}{(44\cdot10+33\cdot5)^2+(33\cdot10-44\cdot5)^2=?+?}$$
