Property 1:-
A number having 2, 3, 7 or 8 at the unit’s place is never a perfect square: i.e., a square number never ends in 2,3,7 or 8.
If you have a look at all the squares, you will observe that all squares of numbers have 0, 1, 4, 5, 6, at units place. None of the square numbers ends in 2, 3, 7 or 8.
Test : Look at the unit’s place of a number. If it ends in 2, 3, 7 and 8, then it is not a perfect square.
For example: The numbers 732, 2093, 82097 and 236918 are not perfect squares.
Note that:
A number having a digit other than, 2, 3, 7 and 8 at the unit’s place is not necessarily a perfect square. It may not be a perfect square. For example, 91, 2305, 2504, 26490 are not perfect squares.
Property 2 :-
The number of zeroes at the end of a perfect square is always even.
In other words, a number ending in an odd number of zeroes is never a perfect square.
Verification : Consider the following tables:
|
Number |
Number of zeros at units place |
Square of the number |
Number of zeros at the unit’s place of the square number |
|
10 20 200 2500 5000 20000 |
1 1 2 2 3 4 |
100 400 40000 6250000 25000000 400000000 |
2 2 4 4 6 8 |
It is evident from the above table that the number of zeros at the end of the square of number is twice the number of zeros at the end of the given number.
Application: By just looking at a number which ends in an odd number of zeros, we can say that it cannot be perfect square. For example, 45000, 16000, 90 etc. are not perfect square.
Remark :It should be noted that numbers ending in an even number of zeros may or may not be a perfect square. For example, 2500 is a perfect square but, 2600, 44700 etc. are not perfect squares.
Property 3 :-
Squares of even numbers are always even numbers and squares of odd numbers are always odd numbers. You may verify this fact from the table of squares. Thus,
\({ 33 }^{ 2 } \) = 1089 , \({ 84 }^{ 2 } \)= 7056, \({ 315 }^{ 2 } \)= 99225, \({ 836 }^{ 2 } \)= 698896,
So, the squares of odd numbers 33 and 315 are odd numbers.Similarly, squares of even numbers 84 and 836 are also even numbers.
Property 4 :-
The square of a natural number other than one is either a multiple of 3 or exceeds a multiple of 3 by 1.
In other words, a perfect square leaves remainder 0 or 1 on division by 3.
Verification : The above property can be easily verified from the following computations:
Square number Remainder when divided by 3
\({ 2 }^{ 2 } \)=4=3×1+1 1
\({ 3}^{ 2 } \)=9=3×3+0 0
\({ 4 }^{ 2 } \)=16=3×5+1 1
\({ 5 }^{ 2 } \)=25=3×8+1 1
\({ 6 }^{ 2 } \)=36=3×12+0 0
\({ 7 }^{ 2 } \)=49=3×16+1 1
\({ 8 }^{ 2 } \)=64=3×21+1 1
Application: If a number when divided by 3 leaves remainder 2, then it is not a perfect square.
Illustration: 635, 98, 122 are not perfect squares as they leave remainder 2 when divided by 3.
Remark If a number leaves remainder 0 or 1 when divided by 3, it is not necessarily a perfect square. However, when it leaves remainder other than 0 and 1, it is not a perfect square.
Property 5:-
The square of a natural number other than one is either a multiple of 4 or exceeds a the multiple of 4 by 1.
In other words, a perfect square leaves remainder 0 or 1 on division by 4.
Verification: The above property can be easily verified from the following computations:
Square number Remainder when divided by 4
\({ 2 }^{ 2 } \)= 4=4×1+0 0
\({ 3 }^{ 2 } \)=9=4×2+1 1
\({ 4 }^{ 2 } \)=16=4×4+0 0
\({ 5 }^{ 2 } \)=25=4×6+1 1
\({ 6 }^{ 2 } \)=36=4×9+0 0
\({ 7 }^{ 2 } \)=49=4×12+1 1
\({ 8 }^{ 2 } \)=64=4×16+0 0
\({ 9 }^{ 2 } \)=81=4×20+1 1
Application: If a number when divided by 4 leaves remainder 2 or 3, it cannot be a perfect square.
Illustration : 3 67, 146, 363, 10003 are not perfect squares as they leave remainder 3, 2, 3 and 3 respectively, when divided by 4.
Remark If a number leaves remainder 0 or 1 when divided by 4, it is not necessarily a perfect square. However, when it leaves remainder other than 0 and 1, it is not a
perfect square.
Property 6:-
If n is a perfect square, then 2n can never be a perfect square.
Property 7:-
A perfect square number is never negative.
Property 8:-
For every natural number n, we have, The sum of first n odd natural numbers = \({ n }^{ 2 } \).
Thus, \({ 1 }^{ 2 } \) = 1 = sum of first 1 odd natural number
= \({ 1 }^{ 2 } \)
\({ 1 }^{ 2 } \) = 4 = 1 +3 = sum of first two odd natural numbers
= 4 = \({ 2}^{ 2 } \)
\({ 1 }^{ 2 } \) = 9 = 1 + 3 + 5 = sum of first three odd natural numbers
=1+3 +5 =9= \({ 3 }^{ 2 } \), and so on.
Pythagorean Triplets :
A triplet (m, n, p) of three natural numbers m, n and p is called a Pythagorean triplet, if \({ m}^{ 2 } \) + \({ n}^{ 2 } \) = \({ p}^{ 2 } \) .
For example: (3, 4, 5), (5, 12, 13), (8, 15, 17) etc., are Pythagorean triplets, because
\({ 3}^{ 2 } \) +\({ 4}^{ 2 } \) = 9+16=25= \({ 5}^{ 2 } \)
\({ 5}^{ 2 } \) + \({ 12}^{ 2 } \) =25+ 144 = 169 = \({ 13}^{ 2 } \),
\({ 8}^{ 2 } \) + \({ 15}^{ 2 } \) = 64+225=289= \({ 17}^{ 2 } \) .
Property 9 :
For any natural number m greater than 1, (2m, \({ m }^{ 2 } -1\) , \({ m }^{ 2 }+1\)) is a Pythagorean triplet.
Proof : In order to prove that (2m, \({ m }^{ 2 } -1\) , \({ m }^{ 2 }+1\)) is a Pythagorean triplet, it is sufficient to prove that
\(2{ m }^{ 2 }\quad +\quad ({ m }^{ 2 }-1)^{ 2 }={ ({ m }^{ 2 }+1) }^{ 2 }\)
We have, \(2{ m }^{ 2 }\quad +\quad ({ m }^{ 2 }-1)^{ 2 }\)=\(4{ m }^{ 2 }+{ m }^{ 4 }-2{ m }^{ 2 }+{ 1 }^{ 2 }\)
= \({ m }^{ 4 }+2{ m }^{ 2 }+{ 1 }^{ 2 }\)
=\({ ({ m }^{ 2 }+1 })^{ 2 }\)
Hence, (2m, \({ m }^{ 2 } -1\) , \({ m }^{ 2 }+1\)) is Pythagorean triplet for any natural number in> 1.
Remark If m and n are relatively prime natural numbers such that m > n and exactly
of them is even and other is odd, then (2mn, m2 — n2, m2 + n2) is a primitive Pythagorean triple.
Here, the word ‘primitive’ means that the three numbers contain no common factor.