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

Consider a grid of dots with distance $1\text{ unit}$ between adjacent horizontal and vertical dots.
If the horizontal distance between two vertices is $a$ and the vertical distance is $b$ (where $a$ and $b$ are whole numbers), by Pythagoras theorem the side length $s$ of the square satisfies:

s^2 = a^2 + b^2

The area of the square is $\text{Area} = s^2 = a^2 + b^2$ . Therefore, the area of any square formed on a grid must be expressible as the sum of two whole number squares.

(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 · Area of 2 sq. units

Check if $a^2 + b^2 = 2$ for whole numbers $a$ and $b$ . Diagram 1

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

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

Since $a = 1, b = 1$ are whole numbers, a square with area 2 sq. units is possible.

Step 2 · Area of 3 sq. units

Check if $a^2 + b^2 = 3$ for whole numbers $a$ and $b$ .

Testing whole numbers:

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

Since $3$ cannot be written as the sum of two squares of whole numbers, a square with area 3 sq. units is not possible.

Step 3 · Area of 4 sq. units

Check if $a^2 + b^2 = 4$ for whole numbers $a$ and $b$ . Diagram 2

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

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

Since $a = 2, b = 0$ are whole numbers, a square with area 4 sq. units is possible.

Step 4 · Area of 5 sq. units

Check if $a^2 + b^2 = 5$ for whole numbers $a$ and $b$ . Diagram 3

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

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

Since $a = 1, b = 2$ are whole numbers, a square with area 5 sq. units is possible.

Answer

(i) (a) Possible, (b) Not possible, (c) Possible, (d) Possible

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

Step 1 · Find Possible Integer-Valued Areas

The possible integer-valued areas are numbers that can be written as the sum of two whole number squares ( $a^2 + b^2$ , where $a, b$ are whole numbers and 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}

Answer

(ii) Numbers expressible as $a^2 + b^2$ for whole numbers $a$ and $b$ (e.g., $1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, \dots$ )

Common Mistakes

Assuming squares must be grid-aligned: Assuming that squares can only have sides parallel to the grid lines (which would only give areas $1, 4, 9, 16, \dots$ ). Tilted squares can form non-perfect square areas such as $2, 5, 8, 10, 13, \dots$
Non-integer vertex coordinates: Overlooking that all grid vertices must have whole number coordinates, so $a$ and $b$ must strictly be whole numbers.

Question 18

More questions in FIO