# Square of Sum

## Theorem

- $\forall x, y \in \R: \paren {x + y}^2 = x^2 + 2 x y + y^2$

## Algebraic Proof 1

Follows from the distribution of multiplication over addition:

\(\ds \left({x + y}\right)^2\) | \(=\) | \(\ds \left({x + y}\right) \cdot \left({x + y}\right)\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds x \cdot \left({x + y}\right) + y \cdot \left({x + y}\right)\) | Real Multiplication Distributes over Addition | |||||||||||

\(\ds \) | \(=\) | \(\ds x \cdot x + x \cdot y + y \cdot x + y \cdot y\) | Real Multiplication Distributes over Addition | |||||||||||

\(\ds \) | \(=\) | \(\ds x^2 + 2xy + y^2\) |

$\blacksquare$

## Algebraic Proof 2

Follows directly from the Binomial Theorem:

- $\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$

putting $n = 2$.

$\blacksquare$

## Geometric Proof

In the words of Euclid:

*If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.*

(*The Elements*: Book $\text{II}$: Proposition $4$)

Let the straight line $AB$ be cut at random at $C$.

Construct the square $ADEB$ on $AB$ and join $DB$.

Construct $CF$ parallel to $AD$ through $C$.

Construct $HK$ parallel to $AB$ through $G$.

From Parallelism implies Equal Alternate Angles, $\angle CGB = \angle ADB$.

We have that $BA = AD$.

So from Isosceles Triangle has Two Equal Angles:

- $\angle ADB = \angle ABD$

So $\angle CGB = \angle CBG$, and so from Triangle with Two Equal Angles is Isosceles, $BC = CG$.

From Opposite Sides and Angles of Parallelogram are Equal, we have $CB = GK$ and $CG = KB$ and so $CGKB$ is equilateral.

Now, since $CG \parallel BK$, from Parallelism implies Supplementary Interior Angles we have that $\angle KBC$ and $\angle GCB$ are supplementary.

But as $\angle KBC$ is a right angle, $\angle BCG$ is also a right angle.

So from Opposite Sides and Angles of Parallelogram are Equal, $\angle CGK$ and $\angle GKB$ are also right angles.

So $CGKB$, described on $CB$. is right-angled, and as it is equilateral, by definition it is square.

For the same reason, $HGDF$ is also square, and described on $HG$, which equals $AB$.

So $HGDF$ and $CGKB$ are the squares on $AC$ and $CB$.

Now we have from Complements of Parallelograms are Equal that $\Box ACGH = \Box GKEF$.

But $\Box ACGH$ is the rectangle contained by $AC$ and $CB$, as $CB = GC$.

So $\Box GKEF$ is also equal to the rectangle contained by $AC$ and $CB$.

But the squares $HGFD$ and $CBKG$ are equal to the squares on $AC$ and $CB$.

So the four areas $HGFD$, $CBKG$, $ACGH$ and $GKEF$ are equal to:

- the squares on $AC$ and $CB$

and:

- twice the rectangle contained by $AC$ and $CB$.

But $HGFD$, $CBKG$, $ACGH$ and $GKEF$ are also equal to the square on $AB$.

Hence the result.

$\blacksquare$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 2$: Special Products and Factors: $2.1$ - 1980: David M. Burton:
*Elementary Number Theory*(revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem