A **square number** or **perfect square** is an integer that is the square of an integer; in other words, it is the product of some integer with itself.** For example:** 9 is a square number, since it can be written as 3 × 3.

Square numbers are non-negative.

**Example:**

0^{2} = 0

1^{2} = 1

2^{2} = 4

3^{2} = 9

4^{2} = 16

5^{2} = 25

6^{2} = 36

7^{2} = 49

8^{2} = 64

9^{2} = 81

Squares of small numbers like 3, 4, 5, 6, 7, ... etc. are easy to find.

But is it possible to find square of 23 so quickly? We need to multiply 23 by 23.

WEE can find the answer without having to multiply 23 × 23.

We know 23 = 20 + 3

Therefore 23² = (20 + 3)² = 20(20 + 3) + 3(20 + 3)

= 20² + (20 X 3) + (3 X 20 ) + 3²

= 400 + 60 + 60 + 9 = 529

**Example 1: **Find the square of the following numbers without actual multiplication.

(i) 39 (ii) 42

**Solution: **(i) 39² = (30 + 9)² = 30(30 + 9) + 9(30 + 9)

= 30² + 30 × 9 + 9 × 30 + 9²

= 900 + 270 + 270 + 81 = 1521

(ii) 42² = (40 + 2)² = 40(40 + 2) + 2(40 + 2)

= 40² + 40 × 2 + 2 × 40 + 2²

= 1600 + 80 + 80 + 4 = 1764

**Pythagorean triplets**

Consider the following:

32 + 42 = 9 + 16 = 25 = 52

The collection of numbers 3, 4 and 5 is known as **Pythagorean triplet**. 6, 8, 10 is also a Pythagorean triplet, since

62 + 82 = 36 + 64 = 100 = 102

Again, observe that

52 + 122 = 25 + 144 = 169 = 132. The numbers 5, 12, 13 form another such triplet.

**Example: **Write a Pythagorean triplet whose smallest member is 8.

**Solution: **We can get Pythagorean triplets by using general form 2m, m² – 1, m² + 1.

Let us first take m² – 1 = 8

So, m² = 8 + 1 = 9

which gives m = 3

Therefore, 2m = 6 and m² + 1 = 10

The triplet is thus 6, 8, 10. But 8 is not the smallest member of this.

So, let us try 2m = 8

then m = 4

We get m² – 1 = 16 – 1 = 15

and m² + 1 = 16 + 1 = 17

The triplet is 8, 15, 17 with 8 as the smallest member.

The inverse operation of addition is subtraction and the inverse operation of multiplication is division. Similarly, finding the square root is the inverse operation of squaring.

1² = 1, therefore square root of 1 is 1

2² = 4, therefore square root of 4 is 2

3² = 9, therefore square root of 9 is 3

**Finding square root through repeated subtraction**

Consider **√**81. Then,

**Step 1:** 81 – 1 = 80

**Step 2: **80 – 3 = 77

**Step 3:** 77 – 5 = 72

**Step 4:** 72 – 7 = 65

**Step 5:** 65 – 9 = 56

**Step 6:** 56 – 11 = 45

**Step 7:** 45 – 13 = 32

**Step 8:** 32 – 15 = 17

**Step 9:** 17 – 17 = 0

From 81 we have subtracted successive odd numbers starting from 1 and obtained 0 at 9th step.

Therefore **√**81 = 9.

**Finding square root through prime factorisation**

Find the square root of 6400.

**Solution: **Write 6400 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

Therefore **√**6400 = 2 × 2 × 2 × 2 × 5 = 80

**Finding square root by division method**

**Step I:** Group the digits in pairs, starting with the digit in the units place. Each pair and the remaining digit (if any) is called a period.

**Step II:** Think of the largest number whose square is equal to or just less than the first period. Take this number as the divisor and also as the quotient.

**Step III:** Subtract the product of the divisor and the quotient from the first period and bring down the next period to the right of the remainder. This becomes the new dividend.

**Step IV:** Now, the new divisor is obtained by taking two times the quotient and annexing with it a suitable digit which is also taken as the next digit of the quotient, chosen in such a way that the product of the new divisor and this digit is equal to or just less than the new dividend.

**Step V:** Repeat steps (2), (3) and (4) till all the periods have been taken up. Now, the quotient so obtained is the required square root of the given number.

**Example:**

**1. Find the square root of 784 by the long-division method.
Solution: **

Marking periods and using the long-division method,

Therefore, √784 = 28