Squares and Square Roots | FIO

Question 18

(i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq. units, and (d) 5 sq. units?

(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

Solution

Understand the Question

If a square is formed on a grid with vertices at grid points, let the horizontal and vertical distances between adjacent vertices be whole numbers $a$ and $b$ .
By the Pythagoras theorem, the side length $s$ satisfies:

s^2 = a^2 + b^2

Since the area of a square is $\text{Area} = s^2$ , any valid grid square must have an area that can be written as the sum of two squares of whole numbers ( $a^2 + b^2$ ).

(i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq. units, and (d) 5 sq. units?

Step 1 · Check Area of 2 sq. units

We need whole numbers $a$ and $b$ such that $a^2 + b^2 = 2$ . Diagram 1

For $a = 1$ and $b = 1$ :

\begin{aligned} 1^2 + 1^2 &= 1 + 1 \\[0.6em] &= 2 \end{aligned}

Therefore, a square of area 2 sq. units is possible.

Step 2 · Check Area of 3 sq. units

We need whole numbers $a$ and $b$ such that $a^2 + b^2 = 3$ .

Testing possible whole numbers:

If $a = 0 \implies 0^2 + b^2 = 3 \implies b^2 = 3$ (no whole number solution)
If $a = 1 \implies 1^2 + b^2 = 3 \implies b^2 = 2$ (no whole number solution)
If $a \ge 2 \implies a^2 \ge 4 > 3$

Since no whole numbers satisfy $a^2 + b^2 = 3$ , a square of area 3 sq. units is not possible.

Step 3 · Check Area of 4 sq. units

We need whole numbers $a$ and $b$ such that $a^2 + b^2 = 4$ . Diagram 2

For $a = 2$ and $b = 0$ :

\begin{aligned} 2^2 + 0^2 &= 4 + 0 \\[0.6em] &= 4 \end{aligned}

Therefore, a square of area 4 sq. units is possible.

Step 4 · Check Area of 5 sq. units

We need whole numbers $a$ and $b$ such that $a^2 + b^2 = 5$ . Diagram 3

For $a = 1$ and $b = 2$ :

\begin{aligned} 1^2 + 2^2 &= 1 + 4 \\[0.6em] &= 5 \end{aligned}

Therefore, a square of area 5 sq. units is possible.

Answer

(i) (a) Possible (2 sq. units)

(b) Not possible (3 sq. units)

(d) Possible (5 sq. units)

(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

Step 1 · Determine all possible integer-valued areas

The area must be of the form $a^2 + b^2$ , where $a$ and $b$ are whole numbers (not both zero):

\begin{aligned} 1^2 + 0^2 &= 1 \\ 1^2 + 1^2 &= 2 \\ 2^2 + 0^2 &= 4 \\ 2^2 + 1^2 &= 5 \\ 2^2 + 2^2 &= 8 \\ 3^2 + 0^2 &= 9 \\ 3^2 + 1^2 &= 10 \\ 3^2 + 2^2 &= 13 \\ 4^2 + 0^2 &= 16 \\ 4^2 + 1^2 &= 17 \\ 3^2 + 3^2 &= 18 \\ 4^2 + 2^2 &= 20 \end{aligned}

Thus, any positive integer that can be expressed as the sum of two squares of whole numbers is a possible area.

Answer

(ii) All positive integers expressible as $a^2 + b^2$ for whole numbers $a, b$ : $1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, \dots$

Common Mistakes

Assuming Grid Squares Must Be Upright: Squares on a grid do not have to be parallel to grid lines; tilted (slanted) squares with vertices on grid points are also valid and have areas like $2, 5, 8, 10,$ etc.
Assuming All Integers Are Possible: Thinking any whole number area can be drawn, overlooking that the area must satisfy $s^2 = a^2 + b^2$ for whole numbers $a$ and $b$ .

Question 18

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