krit.club logo

Squares and Square Roots - Properties of Square Numbers

Grade 8ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Definition of a Perfect Square: A number nn is called a perfect square if there exists an integer mm such that n=m2n = m^2. Visually, a perfect square can be represented as a square grid of dots where the number of rows equals the number of columns, such as 3×3=93 \times 3 = 9.

Ending Digits (Units Place): Perfect squares always end with the digits 0,1,4,5,6,0, 1, 4, 5, 6, or 99 at the unit's place. A number ending in 2,3,7,2, 3, 7, or 88 is never a perfect square. For example, 121121 (ends in 11) is a square, while 127127 is not.

Property of Zeroes: A perfect square always ends in an even number of zeroes. For instance, 100100 (22 zeroes) and 10,00010,000 (44 zeroes) are perfect squares, but 1,0001,000 (33 zeroes) is not. Visually, this corresponds to pairs of powers of 1010.

Sum of Odd Numbers: The sum of the first nn consecutive odd natural numbers is exactly n2n^2. This can be visualized by adding 'L-shaped' borders of dots (gnomons) to a single dot: 11 (which is 121^2), 1+3=41+3=4 (which is 222^2), 1+3+5=91+3+5=9 (which is 323^2).

Numbers between Successive Squares: Between any two consecutive square numbers n2n^2 and (n+1)2(n+1)^2, there are 2n2n non-perfect square numbers. For example, between 22=42^2=4 and 32=93^2=9, there are 2(2)=42(2) = 4 non-square numbers: 5,6,7,5, 6, 7, and 88.

Parity of Squares: The square of an even number is always even, and the square of an odd number is always odd. For example, 122=14412^2 = 144 (even) and 152=22515^2 = 225 (odd).

Pythagorean Triplets: For any natural number m>1m > 1, the numbers 2m,m21,2m, m^2-1, and m2+1m^2+1 form a Pythagorean Triplet. This means (2m)2+(m21)2=(m2+1)2(2m)^2 + (m^2-1)^2 = (m^2+1)^2, which geometrically represents the sides of a right-angled triangle.

Square of a Number ending in 5: If a number has 55 in its units place, its square ends in 2525. The digits before 2525 can be calculated by multiplying the tens digit aa by (a+1)(a+1). For example, 252=(2×3)25^2 = (2 \times 3) followed by 2525, which is 625625.

📐Formulae

n2=n×nn^2 = n \times n

1+3+5++(2n1)=n21 + 3 + 5 + \dots + (2n-1) = n^2

Non-perfect square numbers between n2 and (n+1)2=2n\text{Non-perfect square numbers between } n^2 \text{ and } (n+1)^2 = 2n

Pythagorean Triplet: (2m,m21,m2+1) where m>1\text{Pythagorean Triplet: } (2m, m^2 - 1, m^2 + 1) \text{ where } m > 1

(a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2

(ab)2=a22ab+b2(a-b)^2 = a^2 - 2ab + b^2

💡Examples

Problem 1:

Find the number of non-perfect square numbers lying between 40240^2 and 41241^2.

Solution:

We know that the number of non-perfect square numbers between n2n^2 and (n+1)2(n+1)^2 is given by the formula 2n2n. Here, n=40n = 40. Therefore, the number of non-square numbers is 2×40=802 \times 40 = 80.

Explanation:

This approach uses the property that between consecutive squares, the count of intermediate integers is twice the base of the smaller square.

Problem 2:

Write a Pythagorean triplet whose smallest member is 1414.

Solution:

We use the general form of a Pythagorean triplet (2m,m21,m2+1)(2m, m^2 - 1, m^2 + 1).

  1. Let 2m=142m = 14, which gives m=142=7m = \frac{14}{2} = 7.
  2. Calculate the second member: m21=721=491=48m^2 - 1 = 7^2 - 1 = 49 - 1 = 48.
  3. Calculate the third member: m2+1=72+1=49+1=50m^2 + 1 = 7^2 + 1 = 49 + 1 = 50. The triplet is (14,48,50)(14, 48, 50). Check: 142+482=196+2304=2500=50214^2 + 48^2 = 196 + 2304 = 2500 = 50^2.

Explanation:

By equating the given even number to 2m2m, we find the value of mm and then derive the other two members using the standard algebraic identities for triplets.