Algebraic Manipulation Skill II

Another algebraic manipulation skill that I want to share with students today:

Rewrite $a^2c^2+b^2c^2+a^2d^2+b^2d^2$ as the sum of two squares.

The vast majority students would say, hey! Isn't it obvious that we can factor out $a^2$ in $a^2c^2+a^2d^2$ and $b^2$ in $b^2c^2+b^2d^2$ such that we get:

$\begin{align*}a^2c^2+b^2c^2+a^2d^2+b^2d^2&=(a^2c^2+a^2d^2)+(b^2c^2+b^2d^2)\\&=a^2(c^2+d^2)+b^2(c^2+d^2)\\&=(a^2+b^2)(c^2+d^2)\end{align*}$

They were not thinking well and their pattern was to jump right into action to begin doing what they could "see" in their head. Little did they know we're asked to rewrite the problem as the sum of two squares, i.e.

 $a^2c^2+b^2c^2+a^2d^2+b^2d^2=\text{something}^2+\text{some other thing}^2$

So their previous attempt at a solution has completely failed.

Now, we are back to the drawing board, let's look at what was given, $a^2c^2+b^2c^2+a^2d^2+b^2d^2$, and see if we have detected something useful, a pattern, a relationship, anything that could give us perspective and idea to start working at a solution:

If you're observant enough, you would try to regroup the terms so we have the following two groups:


$\displaystyle \color{yellow}\bbox[5px,purple]{a^2c^2+b^2d^2}$

$\displaystyle \color{yellow}\bbox[5px,green]{b^2c^2+a^2d^2}$

Note that their product is the same, $\displaystyle \color{yellow}\bbox[5px,purple]{a^2c^2\times b^2d^2=a^2b^2c^2d^2}$ and $\displaystyle \color{yellow}\bbox[5px,green]{b^2c^2\times a^2d^2=a^2b^2c^2d^2}$.

This is a real meaningful discovery, since we could safely introduce the new and helpful term into the original equation and this gives:

$a^2c^2+b^2c^2+a^2d^2+b^2d^2$

$\displaystyle \color{black}=\color{yellow}\bbox[5px,purple]{a^2c^2+b^2d^2}\color{black}+\color{yellow}\bbox[5px,green]{b^2c^2+a^2d^2}$

$\displaystyle \color{black}=\color{yellow}\bbox[5px,purple]{a^2c^2+2abcd+b^2d^2}\color{black}+\color{yellow}\bbox[5px,green]{b^2c^2-2abcd+a^2d^2}$

$\displaystyle \color{black}=\color{yellow}\bbox[5px,purple]{(ac+bd)^2}\color{black}+\color{yellow}\bbox[5px,green]{(ac-bd)^2}$

We did it! We've successfully transformed the given expression to become another expression that is a sum of two squares, we're hence done!

But, what if we put the $-2bbcd$ to the group $\displaystyle \color{yellow}\bbox[5px,purple]{a^2c^2+b^2d^2}$?

Don't worry about it, you'll still get the sum of two squares, written differently with the entire different terms now:

$a^2c^2+b^2c^2+a^2d^2+b^2d^2$

$\displaystyle \color{black}=\color{yellow}\bbox[5px,purple]{a^2c^2+b^2d^2}\color{black}+\color{yellow}\bbox[5px,green]{b^2c^2+a^2d^2}$

$\displaystyle \color{black}=\color{yellow}\bbox[5px,purple]{a^2c^2-2abcd+b^2d^2}\color{black}+\color{yellow}\bbox[5px,green]{b^2c^2+2abcd+a^2d^2}$

$\displaystyle \color{black}=\color{yellow}\bbox[5px,purple]{(ac-bd)^2}\color{black}+\color{yellow}\bbox[5px,green]{(ac+bd)^2}$

That means there are two possible solutions for this problem. We could still draw some really useful conclusion, not premature, but a true conclusion for all cases:

$\begin{align*}a^2c^2+b^2c^2+a^2d^2+b^2d^2&=(a^2+b^2)(c^2+d^2)\\&=(ac+bd)^2+(ac-bd)^2\\&\stackrel{\text{or}}{=}(ac-bd)^2+(ac+bd)^2\end{align*}$

It literally means the product of two factors, which each factor is the sum of two squares could be interpreted as the sum of two squares.