Students will learn what the sum and difference formulas are and see several ways of proving them. Then students will learn how to use the formulas to evaluate certain expressions, solve certain equations, and prove certain identities.

Conclude by giving your students these challenges:

- What a Joke by NRICH
- Tubular Stand by NRICH
- 2020 AMC 8, Problem 4

How Many Eggs? by NRICH

Here's the solution: From the problem, we can deduce the following:

$$\begin{align} & P + M + A + J = 38 \\[0.5em] & J = P - 1 \\[0.5em] & P = M - 5 \\[0.5em] & A = \dfrac{M}{2} \\[0.5em] & P = A + 2 \end{align}$$Next, notice that \(P\) and \(A\) are expressed in terms of \(M.\) Thus, if we could express \(J\) in terms of \(M,\) we would have all quantities in terms of \(M.\) This is easily done:

$$J = P - 1 = (M - 5) - 1 = M - 6$$Next, we rewrite our very first equation in terms of \(M,\) then simplify.

$$\begin{align} & P + M + A + J = 38 \\[0.5em] & M - 5 + M + \dfrac{M}{2} + M - 6 = 38 \\[0.5em] & 2M + \dfrac{M}{2} - 11 = 38 \\[0.5em] & \dfrac{7M}{2} - 11 = 38 \\[0.5em] & \dfrac{7M}{2} = 49 \\[0.5em] & \dfrac{M}{2} = 7 \end{align}$$\(A = M/2,\) and \(M/2 = 7,\) tells us \(A = 7.\) Multiplying both sides of \(M/2 = 7\) by \(2,\) we also find \(M = 14.\) Following the equations for \(J\) and \(P,\) we find \(J = M - 6 = 8,\) and \(P = M - 5 = 9.\) In summary,

$$\begin{align} & A = 7 \\ & M = 14 \\ & J = 8 \\ & P = 9 \end{align}$$Is there any square with integer side lengths, such that its area is equal to its perimeter? If no such square exists, explain why. If at least one such square exists, can you find one? Can you find more than one? Can you find exactly how many exist, and explain why? What if instead of squares, we asked the same questions about rectangles?

Here's a solution using geometry: For any rectangle with integer side lengths, having its width \(\ge 2,\) and its height \(\ge 2,\) let's consider the unit squares whose edges contribute to the perimeter of the rectangle. Notice that each corner square contributes \(2\) to the perimeter. Every non-corner square contributes \(1.\) Because a rectangle has \(4\) corners, there are \(4\) less unit squares than the perimeter of the rectangle. So to compensate, there must be exactly 4 squares inside the rectangle, which do not contribute to its perimeter. There are exactly two ways to arrange 4 squares. Namely, \(1 \times 4\) and \(2 \times 2.\) Thus, when width \(\ge 2,\) and height \(\ge 2,\) there are exactly \(2\) rectangles with integer side lengths, such that their area matches their perimeter. What if width or height is \(\lt 2?\) In this case, each of the two squares at the end contribute \(3\) to the perimeter of the rectangle, and each of the middle squares contribute \(2.\) But there is no way to compensate for this excess by adding squares to the inside of the rectangle, because such rectangles have no inside. Thus, no rectangle with integer side lengths, having width or height \(\lt 2,\) has its area equal to its perimeter.

Here's a solution for squares, using algebra: Let \(s\) be one side of the square. Then putting together our area and perimeter formulas together, gives us \(s^2 = 4s.\) Because \(s\) is a length, we can divide both sides by \(s,\) yielding \(s = 4.\) Thus, this is the only square with its area equal to its perimeter.

Here's a solution for rectangles, using algebra:

$$\begin{align} & wh = 2w + 2h \\[1em] \end{align}$$Solving for \(w,\) then separating the rational expression into its integer and fractional parts:

$$\begin{align} w &= \dfrac{2h}{h - 2} \\[1em] &= \dfrac{2h - 4 + 4}{h - 2} \\[1em] &= \dfrac{2(h - 2) + 4}{h - 2} \\[1em] &= 2 + \dfrac{4}{h - 2} \end{align}$$So \(h - 2\) must divide \(4.\) This will only be the case when \(h - 2\) is \(1,\) \(2,\) or \(4.\) When \(h = 1\) we get the same rectangle as when \(h = 4.\) Hence, this method also gives us two unique rectangles.

TODO: Should we teach this skill of removing the variable from the numerator? Could we generalize the algebraic or geometric solution to cubes or cuboids?