Have a look at the Properties of Perfect Squares and know how to solve problems on Perfect Square concepts. You can easily learn to solve problems on your own once you get a complete grip on the Perfect Square Concept. We have clearly mentioned every property of a perfect square along with examples. Check out the examples for a better understanding. All concepts available on Square are given on our website for free of cost and you can prepare anytime and anywhere.
Different Properties of Perfect Squares
See different properties of perfect squares given below.
Property 1:
Numbers those end with 2, 3, 7, or 8 will never a perfect square. Also, all the numbers ending in 1, 4, 5, 6, 9, 0 are not square numbers.
Examples:
The numbers 20, 32, 73, 167, 298 end in 0, 2, 3, 7, 8 respectively.
So, none of them is a perfect square.
Property 2:
A number that ends with an odd number of zeros is never a perfect square.
Examples:
The numbers 150, 3000, 700000 end in one zero, three zeros, and five zeros respectively.
So, none of them is a perfect square.
Property 3:
The square of an even number is always even.
Examples:
12² = 144, 14² = 196, 16² = 256, 18² = 324, etc.
Property 4:
The square of an odd number is always odd.
Examples:
11² = 121, 13² = 169, 15² = 225, 17² = 289, 19² = 361, etc. All the numbers are odd.
Property 5:
The square of a proper fraction is smaller than the fraction.
Examples:
(3/4)² = (3/4 × 3/4) = 9/16 and 9/16 < 3/4, since (16 × 3) < (9 × 4).
Property 6:
For every natural number n, we have
(n + 1)² – n² = (n + 1 + n)(n + 1 – n) = {(n + 1) + n}.
Therefore, {(n + 1)² – n²} = {(n + 1) + n}.
Examples:
(i) {1 + 3 + 5 + 7 + 9} = sum of first 5 odd numbers = 5²
(ii) {1 + 3 + 5 + 7 + 9 + 11 + 13 + 15} = sum of first 8 odd numbers = 8²
Property 7:
For every natural number n, we can write as the sum of the first n odd numbers = n²
Examples:
(i) {1 + 3 + 5 + 7 + 9} = sum of first 5 odd numbers = 5²
(ii) {1 + 3 + 5 + 7 + 9 + 11 + 13 + 15} = sum of first 8 odd numbers = 8²
Property 8 (Pythagorean Triplets):
Three natural numbers m, n, p is said to form a Pythagorean triplet (m, n, p) if (m² + n²) = p².
Note:
For every natural number m > 1, we have (2m, m² – 1, m² + 1) as a Pythagorean triplet.
Examples:
(i) Putting m = 6 in (2m, m² – 1, m² + 1) we get (12, 35, 37) as a Pythagorean triplet.
(ii) Putting m = 7 in (2m, m² – 1, m² + 1) we get (14, 48, 50) as a Pythagorean triplet.
Solved Examples on Properties of Perfect Squares
1. Without adding, find the sum (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19).
Solution:
(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19) = sum of first 10 odd numbers
Find the square of the 10 to get the answer.
= (10)² = 100
100 is the sum of the given numbers 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19.
2. Express 81 as the sum of nine odd numbers.
Solution:
81 = 9² = sum of first nine odd numbers
Let us write the first nine odd numbers and add them naturally.
= (1 + 3 + 5 + 7 + 9 + 11 + 13) = 81.
3. Find the Pythagorean triplet whose smallest member is 4.
Solution:
For every natural number m > 1. (2m, m² – 1, m² + 1) is a Pythagorean triplet.
Putting 2m = 8, i.e., m = 4, we get the triplet (8, 15, 17).
The final answer is (8, 15, 17).