# Properties of Perfect Squares, Examples | How to Find a Perfect Square?

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:
12Â² = 144, 14Â² = 196, 16Â² = 256, 18Â² = 324, etc.

Property 4:
The square of an odd number is always odd.
Examples:
11Â² = 121, 13Â² = 169, 15Â² = 225, 17Â² = 289, 19Â² = 361, etc. All the numbers are odd.

Property 5:
The square of a proper fraction is smaller than the fraction.
Examples:
(3/4)Â² = (3/4 Ã— 3/4) = 9/16 and 9/16 < 3/4, since (16 Ã— 3) < (9 Ã— 4).

Property 6:
For every natural number n, we have
(n + 1)Â² – nÂ² = (n + 1 + n)(n + 1 – n) = {(n + 1) + n}.
Therefore, {(n + 1)Â² – nÂ²} = {(n + 1) + n}.
Examples:
(i) {1 + 3 + 5 + 7 + 9} = sum of first 5 odd numbers = 5Â²
(ii) {1 + 3 + 5 + 7 + 9 + 11 + 13 + 15} = sum of first 8 odd numbers = 8Â²

Property 7:
For every natural number n, we can write as the sum of the first n odd numbers = nÂ²
Examples:
(i) {1 + 3 + 5 + 7 + 9} = sum of first 5 odd numbers = 5Â²
(ii) {1 + 3 + 5 + 7 + 9 + 11 + 13 + 15} = sum of first 8 odd numbers = 8Â²

Property 8 (Pythagorean Triplets):
Three natural numbers m, n, p is said to form a Pythagorean triplet (m, n, p) if (mÂ² + nÂ²) = pÂ².
Note:
For every natural number m > 1, we have (2m, mÂ² â€“ 1, mÂ² + 1) as a Pythagorean triplet.
Examples:
(i) Putting m = 6 in (2m, mÂ² â€“ 1, mÂ² + 1) we get (12, 35, 37) as a Pythagorean triplet.
(ii) Putting m = 7 in (2m, mÂ² â€“ 1, mÂ² + 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)Â² = 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 = 9Â² = 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, mÂ² â€“ 1, mÂ² + 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).

Scroll to Top
Scroll to Top