1.2 Square Numbers

Why “square”?

The word comes straight from geometry. The area of a square is the product of its two equal sides, so a square of side n units covers n \times n square units of space.

Side length (units) Area (sq. units)
1 1\times 1 = 1
2 2\times 2 = 4
3 3\times 3 = 9
6 6\times 6 = 36
9 9\times 9 = 81
12 12\times 12 = 144

We record this product with the squaring notation:

n \times n = n^2, \qquad \text{read as } ``n \text{ squared}".

Squaring is not reserved for whole numbers. A square can have a fractional or decimal side too:

\left(\tfrac{4}{7}\right)^2 = \tfrac{4}{7}\times\tfrac{4}{7} = \tfrac{16}{49}, \qquad (1.5)^2 = 1.5\times 1.5 = 2.25.

The squares of natural numbers, 1, 4, 9, 16, 25, \dots, are called perfect squares.

Patterns in the last digit

Try This Write out the squares of the first 25 natural numbers in a table. Then ignore everything except the units digit (the last digit) of each square. Which digits show up, and, just as interesting, which ones never do?

Collect the units digits of 1^2, 2^2, 3^2, \dots and you always land on one of:

0,\ 1,\ 4,\ 5,\ 6,\ 9.

A perfect square never ends in 2, 3, 7, or 8.

This gives a useful one-way test. The last digit can never guarantee that a number is a square, 36 and 196 both end in 6 and are squares, yet 46 also ends in 6 and is not. But the last digit can certainly rule a number out: anything ending in 2, 3, 7, or 8 cannot be a perfect square.

Math Talk The squares 7^2 = 49,\ 13^2 = 169,\ 17^2 = 289,\ 23^2 = 529 all end in 9. Look at the numbers being squared, what do their last digits have in common? Using that, can you predict the last digit of 33^2 and of 57^2 without doing the full multiplication?

A one-second trick: squaring numbers that end in 5

Here is a shortcut worth memorising. To square a number ending in 5, take the digits before the 5 (call that part a), compute a\times(a+1), and simply write 25 after it.

35^2:\quad 3\times 4 = 12 \ \Rightarrow\ 35^2 = 1225, \qquad 65^2:\quad 6\times 7 = 42 \ \Rightarrow\ 65^2 = 4225.

Why does it work? Writing the number as 10a + 5,

(10a+5)^2 = 100a^2 + 100a + 25 = 100\,a(a+1) + 25,

which is exactly “a(a+1), then 25.” Try it on 85^2: 8\times 9 = 72, so 85^2 = 7225.

Zeros at the end

Watch what trailing zeros do under squaring:

20^2 = 400,\quad 30^2 = 900,\quad 50^2 = 2500 \qquad(\text{one zero} \to \text{two zeros}) 200^2 = 40000,\quad 600^2 = 360000 \qquad(\text{two zeros} \to \text{four zeros}).

The number of trailing zeros doubles. So a perfect square can only finish with an even number of zeros, a number ending in an odd run of zeros (such as 7000, with three) is never a perfect square. Squaring also preserves parity: an even number stays even when squared, and an odd number stays odd.

Every square is a running total of odd numbers

Look at the gaps between neighbouring squares:

4-1=3,\quad 9-4=5,\quad 16-9=7,\quad 25-16=9,\ \dots

The gaps are exactly the odd numbers 3, 5, 7, 9, \dots Turning that around, each square is built by stacking on the next odd number:

\begin{aligned} 1 &= 1\\ 1+3 &= 4\\ 1+3+5 &= 9\\ 1+3+5+7 &= 16\\ 1+3+5+7+9 &= 25. \end{aligned}

In words: the sum of the first n odd numbers is n^2.

A Proof Without Words Imagine a square built from dots. To grow an n\times n square into an (n+1)\times(n+1) square, you wrap an L-shaped border around the top and right edge. That border holds exactly 2n+1 dots, the next odd number. Adding L after L from the centre outward gives 1,\ 1+3,\ 1+3+5,\dots and the running total is always a perfect square. A clear picture can stand in for a paragraph of proof.

Use the widget below to grow the square one L-shaped layer at a time and watch the odd numbers pile up to n^2.

The same idea gives a test for perfect squares: keep subtracting 1, 3, 5, 7, \dots If you hit exactly 0, the number is a perfect square and the number of subtractions is its square root. For example:

36 \xrightarrow{-1} 35 \xrightarrow{-3} 32 \xrightarrow{-5} 27 \xrightarrow{-7} 20 \xrightarrow{-9} 11 \xrightarrow{-11} 0.

Six subtractions reached 0, so 36 = 6^2. Try a non-square like 44 and you will sail straight past 0, a sure sign it is not a perfect square.

Worked Example 1

Find 29^2, given that 28^2 = 784.

Since 28^2 is the sum of the first 28 odd numbers, 29^2 just needs the 29th odd number added on. The nth odd number is 2n-1, so the 29th is 2(29)-1 = 57. Hence

29^2 = 784 + 57 = 841.

Worked Example 2

Find 50^2, given that 49^2 = 2401.

Add the 50th odd number, 2(50)-1 = 99:

50^2 = 2401 + 99 = 2500. \qquad\text{(Check: } 50^2 = 2500.\text{)}

How many numbers sit between two squares?

Between n^2 and (n+1)^2 there are exactly 2n whole numbers. For instance, between 24^2 = 576 and 25^2 = 625 there are 2\times 24 = 48 of them. More generally, if p and q are consecutive perfect squares, then q - p - 1 numbers lie strictly between them.

Triangular numbers, hiding inside squares

The triangular numbers 1, 3, 6, 10, 15, \dots count dots stacked into ever-larger triangles. A pretty fact connects them to squares: the sum of two neighbouring triangular numbers is always a perfect square.

3+6 = 9 = 3^2,\qquad 6+10 = 16 = 4^2,\qquad 10+15 = 25 = 5^2.

Square roots

If a square has area 81 sq. cm, what is its side? We need a number that, multiplied by itself, gives 81. Since 9\times 9 = 81, the side is 9 cm, and we call 9 the square root of 81.

\text{If } y = x^2, \text{ then } x \text{ is a square root of } y, \text{ written } x = \sqrt{y}.

There is a subtlety, though. Because 9^2 = 81 and (-9)^2 = 81, the number 81 has two square roots, +9 and -9. Every perfect square has two integer square roots, one positive and one negative:

\sqrt{81} = \pm 9,\qquad \sqrt{144} = \pm 12,\qquad \text{in general } \sqrt{n^2} = \pm n.

Throughout this chapter we will normally take the positive square root.

Three ways to find a square root

Given a number such as 729, how do we tell whether it is a perfect square, and if so, find its root?

  1. List the squares. Since 25^2 = 625, keep squaring upward: 26^2 = 676,\ 27^2 = 729. So \sqrt{729} = 27. (Fine for small numbers, slow for big ones.)

  2. Subtract odd numbers (as above), dependable, but tedious for large numbers.

  3. Prime factorisation, the most powerful method. A perfect square is an integer times itself, so its prime factors always split into two identical groups.

Worked Example 3

Is 1764 a perfect square?

1764 = 2\times 2\times 3\times 3\times 7\times 7 = (2\times 3\times 7)\times(2\times 3\times 7).

The factors fall into two identical groups, so

1764 = (2\times 3\times 7)^2 = 42^2 \quad\Rightarrow\quad \sqrt{1764} = 42.

Is 198 a perfect square? Its prime factorisation is 198 = 2\times 3\times 3\times 11. The 2 and the 11 have no partners, so the factors will not split into two identical groups, and 198 is not a perfect square.

Estimating square roots

When a number is not a perfect square, we can still estimate its root by trapping it between squares we know.

Worked Example 4

Estimate \sqrt{2916}.

  • 2916 lies between 2500 = 50^2 and 3600 = 60^2, so 50 < \sqrt{2916} < 60.
  • It ends in 6, so its root ends in 4 or 6 → either 54 or 56.
  • Test the midpoint: 55^2 = (50+5)^2 = 2500 + 500 + 25 = 3025 > 2916, so the root is below 55 → it must be 54.

Indeed 54^2 = 2916, so \sqrt{2916} = 54.

Estimate \sqrt{170}. Since 13^2 = 169 and 14^2 = 196, we get 13 < \sqrt{170} < 14. As 169 is extremely close to 170, the root is only a whisker above 13, so \sqrt{170} \approx 13.

Try This A square floor tile has an area of 200\ \text{cm}^2. Karthik wants to cut from it the largest possible square coaster with a whole-number side. Since 200 is not a perfect square and the nearest squares are 14^2 = 196 and 15^2 = 225, the largest whole-number side that fits is 14 cm. What is the side if the tile’s area is 250\ \text{cm}^2 instead?

Exam Tip In an exam, always read the sign carefully. The symbol \sqrt{\ } asks only for the positive root, so \sqrt{49} = 7. But the equation x^2 = 49 has two answers, x = +7 and x = -7. Many marks are lost by writing just one root when both are required, or by adding a stray “\pm” when only \sqrt{\ } is asked for.

Figure it Out: Square Numbers

Practice

  1. Which of the following are not perfect squares?   (i) 1058   (ii) 4225   (iii) 3072   (iv) 2023
  2. Which one among 73^2,\ 88^2,\ 117^2,\ 45^2 ends in the digit 9?
  3. Given 60^2 = 3600, what is 61^2?   (i) 3600+60   (ii) 3600+120   (iii) 3600+121   (iv) 3600+61
  4. Find the side of a square whose area is 729\ \text{m}^2.
  5. Find the smallest square number divisible by each of 6, 8 and 15.
  6. Find the smallest number by which 2700 must be multiplied to make a perfect square, and find the square root of the product.
  7. How many numbers lie between the squares of (i) 24 and 25,   (ii) 50 and 51?
  8. Fill in the missing numbers: \;5^2 + 6^2 + 30^2 = (\underline{\ \ })^2,\qquad 6^2 + 7^2 + (\underline{\ \ })^2 = (\underline{\ \ })^2.
  9. Use the “ends-in-5” shortcut to compute 95^2 and 115^2 mentally.
  1. Not perfect squares: (i) 1058 (ends in 8), (iii) 3072 (ends in 2), (iv) 2023 (ends in 3). Only 4225 = 65^2 is a perfect square.
  2. 73^2 and 117^2, both base numbers end in 3 or 7, so their squares end in 9. (88^2 ends in 4, 45^2 ends in 5.)
  3. (iii) 3600 + 121, because 61^2 = 60^2 + (60 + 61) = 3600 + 121 = 3721.
  4. 27 m, since 27^2 = 729.
  5. LCM(6,8,15) = 120 = 2^3 \times 3 \times 5. To make every exponent even, multiply by 2\times 3\times 5 = 30: 120\times 30 = \mathbf{3600} = 60^2.
  6. 2700 = 2^2 \times 3^3 \times 5^2. The lone extra 3 is unpaired, so multiply by 3: 2700\times 3 = 8100 and \sqrt{8100} = \mathbf{90}.
    1. 2\times 24 = \mathbf{48}.   (ii) 2\times 50 = \mathbf{100}.
  8. 5^2+6^2+30^2 = (\mathbf{31})^2 and 6^2+7^2+(\mathbf{42})^2 = (\mathbf{43})^2. (Pattern: a^2 + (a+1)^2 + [a(a+1)]^2 = [a(a+1)+1]^2.)
  9. 95^2: 9\times 10 = 90 \Rightarrow \mathbf{9025}.   115^2: 11\times 12 = 132 \Rightarrow \mathbf{13225}.