Observe that, 1 = 1, 4 = 2 x 2, 9 = 3 x 3, 100 = 10 x 10.

If a is an integer and b = a x a, we say b is a perfect square.

Hence 1, 4, 9, 16, 25 are all perfect squares. Since 0 = 0 x 0, we see that 0 is a perfect square.

If a is an integer, we denote a x a = a^{2}. We read it as square of a or simply a square. Thus 36 = 6^{2} and 81 = 9^{2}. Thus a perfect square is of the m^{2}, where m is an integer.

For example, 4 = 2 x 2 and 4 = (-2) x (-2); in the second representation, we again have equal integers, but negative this time.

**Thus a perfect square is either equal to 0 or must be a positive integer. It can be a negative integer.**

Look at the following table

a | 1 | 2 | 3 | 8 | -7 | -12 | 20 | -15 |

a^{2} |
1 | 4 | 9 | 64 | 49 | 144 | 400 | 225 |

We see that squares of 2, 8, -12, 20 are even numbers and the squares of 1, 3, -7, -15 are odd numbers.

**Statement 1: The Square of an even integer is even and the square of an odd integer is odd.**

Consider first 10 perfect squares,

1^{2} |
2^{2} |
3^{2} |
4^{2} |
5^{2} |
6^{2} |
7^{2} |
8^{2} |
9^{2} |
10^{2} |

1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 |

If we observe that the units place in these squares are 1, 4, 9, 6, 5, 6, 9, 4, 1 and 0 in that order. Thus only the digits which can occupy digit’s place in perfect squares are 1, 4, 5, 6 and 9.

**Statement 2: A perfect square always ends in one of the digits 0, 1, 4, 5, 6 and 9. If the last digit of a number is 2, 3, 7 or 8, it cannot be a perfect square.**

** **

**Exercise 1.2.2**

- Express the following statements mathematically:

i) square of 4 is 16

**Solution:**

4^{2} = 16

ii) square of 8 is 64

**Solution:**

8^{2} = 64

iii) square of 15 is 225

**Solution:**

15^{2} = 225

- Identify the perfect squares among the following numbers:

1, 2, 3, 8, 36, 49, 65, 67, 71, 81, 169, 625, 125, 900, 100, 1000, 100000

**Solution:**

Perfect squares among the above are:

1, 36, 49, 81, 169, 625, 125, 900, 100

- Make a list of all perfect squares from 1 to 500

**Solution:**

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484

- Write the 3 digit numbers ending with 0, 1, 4, 5, 6 , 9 one for each digit but one of them is a perfect square.

**Solution:**

We can take 200, 201, 204, 205, 206 and 209. None of these is perfect square lies between 14^{2} = 196 and 15^{2} = 225.

We can also take 300, 301, 304, 305, 306 and 309. None of these is a perfect square lies between 17^{2} = 289 and 18^{2} = 324.

- Find the numbers from 100 to 400 that end with 0, 1, 4, 5, 6 or 9, which are perfect squares.

**Solution:**

10^{2} = 100, 11^{2} = 121, 12^{2} = 144, 13^{2} = 169, 14^{2} = 196, 15^{2 }= 225, 16^{2} = 256, 17^{2} = 289, 18^{2} = 324, 19^{2} = 361, 20^{2} = 400

**1.2.3 Some facts related to perfect squares**

There are some nice properties about perfect squares. Let us study them here:

- Look at following table:

a | 4 | 10 | 20 | 25 | 100 | 300 | 1000 |

a^{2} |
16 | 100 | 400 | 625 | 10000 | 90000 | 1000000 |

The number of zeros at the end of a^{2} |
0 | 2 | 2 | 0 | 4 | 4 | 6 |

We observe that, the number of zeros at end of a square is always an even number.

**STATEMENT 3: /if a number has k zeros at the end, then its square ends in 2k zeros**

Thus, if a number has odd number of zeros, it cannot be perfect square.

- Look at the adjoining table:

a | a^{2} |
The remainder of a^{2}when divided by 3 |
The remainder of a^{2}when divided by 4 |

1 | 1 | 1 | 1 |

2 | 4 | 1 | 0 |

3 | 9 | 0 | 1 |

5 | 25 | 1 | 1 |

8 | 64 | 1 | 0 |

11 | 121 | 1 | 1 |

-6 | 36 | 0 | 0 |

We can see that, the remainder of a perfect square is either 1 or 0. we can also note that remainder of a perfect square when divided by 4 are either 0 or 1.

Since, when a number is divided by 3, the reminder used to be 0, 1 or 2. Similarly, when a number is divided by 4 the remainder used to be 0, 1, 2 or 3.

** **

**STATEMENT 4: The reminder of a perfect square when divided by 3 is either 0 or 1, but never be 2. The reminder of a perfect square when divided by 4 is either 0 or 1, but never be 2 and 3.**

**STATEMENT 5: When the product of 4 consecutive integers is added to 1, the resulting number is perfect square.**

Ex:

(1 x 2 x 3 x 4) + 1 = 24 + 1 = 25 = 5^{2}

(5 x 6 x 7 x 8) + 1 = 1680 + 1 = 41^{2}

**STATEMENT 6: The sum of the first n odd natural numbers is equal to n ^{2}, for every natural number n.**

Ex:

1 = 1 = 1^{2}

1 + 3 = 4 = 2^{2}

1 + 3 + 5 = 9 = 3^{2}

1 + 3 + 5 + 7 = 16 = 4^{2}

1 + 3 + 5 + 7 + 9 = 25 = 5^{2}

STATEMENT 7: consider the number N = 1000….01, where zeros appear k times. (For example, for k = 6, you get N = 10000001; there are 6 zeros in the middle.) Then N^{2} = 1000…02000…01, where the number of zeros on both the sides of 2 is k.

Ex:

11^{2} = 121

101^{2} = 10201

1001^{2} = 1002001

10001^{2} = 100020001 and so on.

**STATEMENT 8: The sum of n ^{th} and (n+1)^{th} triangular number is (n+1)^{2}.**

Ex:

The dots are now arranged in shapes. Now count the number of dots in each triangle. (Single dot is considered as the generate triangle.) They are 1, 3, 6, 10, 15, 21, 28, 36 and so on. These are called triangular numbers.

For n^{th }triangular number, we form a triangle of dots with n- rows and each row contains as many points as index of that row. If you want find the 8^{th} triangular number, the number of points in the 8^{th} triangle is

1 + 2 + 3 + 4+ 5 + 6 + 7 + 8 = 36

The first few triangular numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91.

Take any two consecutive triangular number and find their sum. For example : 10 + 15 = 25 = 5^{2}; 21 + 28 = 49 = 7^{2}. Which holds the statement 8.

**Exercise 1.2.3**

- Find the sum 1 + 3 + 5 + ……. +51 (the sum of all odd numbers from 1 to 51) without actually adding them.

**Solution:**

[We have 5 odd numbers from 1 to 10, therefore there are 5×5 = 25 odd numbers are there from 1 to 50, we know 51 is also an odd number…. So, there are 25+1 = 26 ]

There are 26 odd numbers are there from 1 to 51.

Thus, 1 + 3 + 5 + …….. + 51 = 26^{2 }= 676.

- Express 144 as a sum of 12 odd numbers.

**Solution:**

We know that, the sum of the first n odd natural numbers is equal to n^{2}, for every natural number n.

Therefore, the sum of first 12 odd natural numbers is equal to 12^{2} = 144.

Thus,

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 22 = 12^{2} = 144.

- Find the 14
^{th}and 15^{th}triangular numbers, and find their sum. Verify the statement 8 for this sum.

**Solution:**

We know, statement 8: The sum of n^{th} and (n+1)^{th} triangular number is (n+1)^{2}.

Sum of 14^{th} triangular number is,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 = 105

Sum of 15^{th} triangular number is,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 120

Now, let us verify, statement 8: The sum of n^{th} and (n+1)^{th} triangular number is (n+1)^{2}.

i.e., Sum of 14^{th} triangular number(n) + sum of 15^{th} triangular number(n+1) = 105 + 120

= 225 = 15^{2} = (n+1)^{2}, which holds the statement8.

- What are the remainders of a perfect square when divided by 5?

**Solution:**

a | a^{2} |
The remainders of a^{2} when divided by 5 |

1 | 1 | 1 |

2 | 4 | 4 |

5 | 25 | 0 |

6 | 36 | 1 |

-7 | 49 | 4 |

11 | 121 | 1 |

The remainders of perfect squares when divided by 5 are 0, 1 and 4.

**1.2.4 Methods of squaring a number**

Many times it is easy to find the square of a number without actually multiplying the number to itself.

Consider 42 = 40 + 2

Thus,

42^{2} = (40+2)^{2}

= 40^{2} + 2(40)(2) + 2^{2}

Now it is easy to recognise 40^{2} = 1600; 2x40x2 = 160; 2^{2} = 4.

Therefore, 42^{2} = 1600 + 160 + 4 = 1764.

Example: Find 89^{2}

**Solution:** 89^{2} = (80 + 9)^{2}

= 80^{2} + 2 x 80 x 9 + 9^{2}

= 6400 + 1440 + 81

= 7921

** Exercise 1.2.4**

- Find the squares of:

i) 31

**Solution:**

31 = 30 + 1

31^{2} = (30 + 1)^{2}

= 30^{2 }+ 2 x 30 x 1 +1

= 900 + 60 + 1

= 961

ii) 72

**Solution:**

72 = 70 + 2

72^{2} = (70 + 2)^{2}

= 70^{2} + 2 x 70 x 2 + 2^{2}

= 4900 + 280 + 4

= 5184

iii) 37

**Solution:**

37 = 30 + 7

37^{2} = (30 + 7)^{2}

= 30^{2} + 2 x 30 x 7 + 7^{2}

= 900 + 420 + 49

= 1369

iii) 166

**Solution:**

166 = 160 + 6

166^{2} = (160 + 6)^{2}

= 160^{2} + 2 x 160 x 6 + 6^{2}

= 25600 + 1920 + 36

= 27556

2. Find the squares of:

i) 85

**Solution:**

85^{2} = (80 + 5)^{2}

= 80^{2} + 2 x 80 x 5 + 5^{2}

= 1600 + 800 + 25

=2425

ii) 115

**Solution:**

115^{2} = (100 + 15)^{2}

= 100^{2} + 2 x 100 x 15 + 15^{2}

= 10000 + 3000 + 225

= 13225

iii) 165

165^{2} = (160 + 5)^{2}

= 160^{2} + 2 x 160 x 5 + 5^{2}

= 25600 + 1600 + 25

= 27225

- Find the squares of 1468 by writing this as 1465+3

1468^{2} = (1465 + 3)^{2}

= (1465)^{2} + 2 x 1465 x 3 + 3^{2}

= 2146225 + 8790 + 9

= 2155024

**1.2.5 Square roots**

Consider the following perfect squares:

1 = 1^{2}, 4 = 2^{2}, 9 = 3^{2}, 16 = 4^{2}, 49 = 7^{2}, 196 = 14^{2.}

In each case the number is obtained by the product of two equal numbers. Here we say 1 is the square root of 1; 2 is the square root of 4; 3 is the square root of 9 and so on.

** **

**Suppose N is a natural number such that N= M ^{2}. The number M is called a square root of N.**

** **

We have seen earlier m^{2} = m x m = (-m) x (-m). Thus m^{2 }has 2 roots m and –m. For example, 16 = 4^{2} = (-4)^{2}, thus both 4 and -4 are the roots of 16. Mathematically both 4 and -4 are accepted as the square root of 16. Thus,

Whenever the word square root is used, it is always meant to be the positive square root. The square root on n is denoted by √N.

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