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ICSE • Class 8 • Mathematics • Ch 4
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Squares and Square Roots

In ICSE Class 8 Mathematics, "Squares and Square Roots" provides an authoritative, mathematically rigorous master study guide investigating perfect squares, properties of square numbers, methods of extracting square roots, and Pythagorean triplets. This comprehensive chapter explores Properties of Perfect Squares (Definition: $n = m^2$ for $m \in \mathbb{Z}$; End digit rules: numbers ending in $2, 3, 7,$ or $8$ can NEVER be perfect squares; Numbers ending in an odd number of zeros are never perfect squares; Squares of even numbers are always even; Squares of odd numbers are always odd; Sum of first $n$ odd natural numbers: $1 + 3 + 5 + \dots + (2n - 1) = n^2$; Number of non-perfect square numbers between $n^2$ and $(n + 1)^2$ is strictly $2n$), Pythagorean Triplets (Form: $2m, m^2 - 1, m^2 + 1$ satisfying $(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2$), Methods of Finding Square Roots: 1. Repeated Subtraction of consecutive odd numbers, 2. Prime Factorisation Method (pairing identical prime factors: $\sqrt{a^2 b^2} = ab$), 3. The Long Division Method (pairing digits from right to left, quotient doubling algorithm; Indispensable for large integers and decimals), Square Roots of Decimals and Fractions (Pairing integral part leftward and fractional part rightward), Finding the Least Number to be Added or Subtracted to make a number a perfect square, and Word Problems on Square Roots aligned with the 2026–27 CISCE ICSE curriculum.

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How Did an Indian Math Genius in 628 CE Calculate the Exact Square Root of a 20-Digit Giant Number in Seconds Using Just Two Sticks in the Sand?

In 628 CE in Rajasthan, the great Indian astronomer Brahmagupta sat in the red sandstone courtyard of Bhinmal. While scholars in Europe were struggling to multiply double-digit Roman numerals, Brahmagupta developed the Long Division Algorithm for Square Roots! He realized that every digit of a square root controls two digits of the square ($10^2 = 100, 100^2 = 10,000$), meaning digits must be grouped in pairs from right to left! If you look at any number ending in the digits 2, 3, 7, or 8, can it EVER be a perfect square? Mathematics answers with an absolute, unbreakable NO! Why does adding the first ten odd numbers ($1 + 3 + 5 + \dots + 19$) instantly equal $10^2 = 100$? How do you find the least number that must be added to 5,607 to make it a perfect square? Let's master squares and square roots.

Why This Chapter Matters

Square roots are essential across geometry (Pythagorean theorem, diagonals, circle areas), physics (kinetic energy velocity $v = \sqrt{2K/m}$, pendulum periods $T = 2\pi\sqrt{l/g}$), civil architecture foundation surveying, and statistical standard deviation. Mastering prime factorisation and long division square roots is an essential ICSE mathematics milestone.

Before You Begin (Prerequisites)

  • Multiplication tables up to 20.
  • Prime factorisation using the division tree.
  • Basic decimal operations.

What You Will Learn (Core Objectives)

  • Identify perfect squares by inspecting unit digits and zero counts.
  • Apply the properties of square numbers (sum of odd numbers, numbers between consecutive squares).
  • Generate Pythagorean triplets using the algebraic template $(2m, m^2 - 1, m^2 + 1)$.
  • Calculate square roots using Prime Factorisation and Repeated Subtraction.
  • Extract square roots of large multi-digit integers and decimals using the Long Division Algorithm.
  • Solve word problems involving least numbers to be added or subtracted to achieve perfect squares.

Chapter Roadmap & Progression

1 1. Properties of Perfect Squares
2 2. Pythagorean Triplets
3 3. Methods of Finding Square Roots
4 4. Least Number Added or Subtracted...

Complete Concept Guide (100% Curriculum Coverage)

1. Properties of Perfect Squares

Understand
A. The Unit Digit Rule:

A natural number is a Perfect Square if it is the square of some natural number ($n = m^2$):

  • The square of any number can end ONLY in digits $0, 1, 4, 5, 6,$ or $9$.
  • $$\mathbf{\text{A number ending in } 2, 3, 7, \text{ or } 8 \text{ can NEVER be a perfect square!}}$$
  • A number ending in an odd number of zeros (e.g., $1000, 500000$) is never a perfect square. (A square must have an even number of terminal zeros: $100, 10000$).
B. Key Mathematical Patterns:
  1. Sum of Consecutive Odd Numbers: The sum of the first $n$ odd natural numbers is strictly equal to $n^2$: $$1 + 3 + 5 + \dots + (2n - 1) = \mathbf{n^2}$$ Example: $1 + 3 + 5 + 7 + 9 = 5^2 = \mathbf{25}$.
  2. Non-Square Numbers Between Consecutive Squares: Between $n^2$ and $(n + 1)^2$, there lie exactly $2n$ non-perfect square numbers: Example: Between $12^2 = 144$ and $13^2 = 169$, there are $2(12) = \mathbf{24\text{ non-square numbers}}$.

2. Pythagorean Triplets

Pythagorean Triplets

Three natural numbers $(a, b, c)$ form a Pythagorean Triplet if they satisfy Pythagoras' theorem: $a^2 + b^2 = c^2$.

General Algebraic Form (for any integer $m > 1$):
$$\mathbf{2m, \quad m^2 - 1, \quad m^2 + 1}$$ $$\text{Verification: } (2m)^2 + (m^2 - 1)^2 = 4m^2 + m^4 - 2m^2 + 1 = m^4 + 2m^2 + 1 = (m^2 + 1)^2$$

Example: Find a Pythagorean triplet whose smallest member is 6:

  • Set $2m = 6 \implies m = 3$.
  • $m^2 - 1 = 3^2 - 1 = 9 - 1 = \mathbf{8}$.
  • $m^2 + 1 = 3^2 + 1 = 9 + 1 = \mathbf{10}$.
  • The triplet is $(6, 8, 10)$. (Check: $6^2 + 8^2 = 36 + 64 = 100 = 10^2$).

3. Methods of Finding Square Roots

Square Root Extraction
Method 1: Prime Factorisation:

Resolve the number into prime factors, pair identical factors, and take one factor from each pair:
$\sqrt{1764} = \sqrt{(2 \times 2) \times (3 \times 3) \times (7 \times 7)} = 2 \times 3 \times 7 = \mathbf{42}$.

Method 2: The Long Division Method (For Large Numbers & Decimals):
  1. Place bars over pairs of digits starting from the unit's digit (right to left). For decimals, pair leftward from decimal point for the integer part, and rightward for the decimal part.
  2. Find the largest number whose square is less than or equal to the leftmost pair.
  3. Subtract and bring down the next pair of digits.
  4. Double the quotient to form the new trial divisor, find the next matching digit, and repeat!

4. Least Number Added or Subtracted Problems

Optimization
A. Finding the Least Number to be SUBTRACTED:
  1. Apply the long division method to find the square root of the given number.
  2. The Remainder obtained at the end of the long division is the exact least number that must be subtracted!
B. Finding the Least Number to be ADDED:
  1. Apply long division to find the integer root $k$. The number lies between $k^2$ and $(k + 1)^2$.
  2. Square the next integer: $(k + 1)^2$.
  3. The required least number to be added is: $$\mathbf{\text{Least Number Added} = (k + 1)^2 - \text{Given Number}}$$

Key Formulas, Identities & Theorems

Pythagorean Triplet Form
$$2m, \quad m^2 - 1, \quad m^2 + 1 \quad (m > 1)$$
Satisfies (2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2.
Sum of First n Odd Numbers
$$\sum_{i=1}^{n} (2i - 1) = n^2$$
Direct identity for consecutive odd integer sums.

Arithmetic: Perfect Square Unit Digits & Long Division Pairing

Squares & Square Roots: Properties, Triplets & Division PROPERTIES OF PERFECT SQUARES • The Impossible Unit Digits: Numbers ending in 2, 3, 7, 8 NEVER square! Squares only end in: 0, 1, 4, 5, 6, 9 • Sum of n Odd Numbers = n2 1 + 3 + 5 + 7 + 9 = 52 = 25 • Non-squares between n2 & (n+1)2 = 2n Between 122 (144) & 132 (169): 2(12) = 24 • Pythagorean Triplet: 2m, m2-1, m2+1 Example: m=3 ⇒ (6, 8, 10) LONG DIVISION METHOD ALGORITHM 1. Pair digits in TWOS from right to left: 17 64 → Pair 64 and Pair 17 2. Long Division Extraction: Largest square ≤ 17 is 42 = 16 (Rem 1) Bring down 64 → 164 Double quotient: 4 × 2 = 8 → Trial divisor 82 × 2 = 164 √1764 = 42 3. Least Number to be Subtracted = Remainder 4. Least Number Added = (k+1)2 - Number 2, 3, 7, 8 CANNOT BE SQUARES • PAIR DIGITS FROM RIGHT • SUBTRACT REMAINDER FOR PERFECT SQUARE

Chapter Summary & 10 Key Takeaways

Takeaway 1
A perfect square is the square of a whole integer (n = m^2).
Takeaway 2
Numbers ending in 2, 3, 7, or 8 are NEVER perfect squares.
Takeaway 3
Numbers ending in an odd number of zeros cannot be perfect squares.
Takeaway 4
The sum of the first n consecutive odd numbers equals n^2.
Takeaway 5
Between n^2 and (n+1)^2 lie exactly 2n non-perfect square numbers.
Takeaway 6
Pythagorean triplets follow the algebraic form: 2m, m^2 - 1, m^2 + 1 for any integer m > 1.
Takeaway 7
Square roots can be extracted by prime factorisation (pairing factors) or long division.
Takeaway 8
The long division method pairs digits from right to left (and left-to-right for decimal fractions).
Takeaway 9
To make a number a perfect square, the least number to be subtracted is the long division remainder.
Takeaway 10
The least number to be added is (k + 1)^2 - Number, where k is the integer root quotient.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Find the square root of $5929$ using the Long Division Method.
Reveal Answer & Explanation
Answer:

Step 1: Pair the digits from right to left: $\overline{59} \; \overline{29}$.
Step 2: The largest square less than or equal to $59$ is $7^2 = 49$. Quotient $= 7$, Remainder $= 59 - 49 = 10$.
Step 3: Bring down the next pair $29$ to form the new dividend: $1029$.
Step 4: Double the quotient $7 \times 2 = 14$ to form the trial divisor $14\_$.
Step 5: Test the unit digit: $147 \times 7 = 1029$.
Step 6: Subtract $1029 - 1029 = 0$.
• Final Answer:

$$\mathbf{\sqrt{5929} = 77}$$

.


Pair as 59 and 29. $7^2 = 49$ leaves 10. Divisor $147 \times 7 = 1029$. Square root is 77.
2
Find the least number that must be subtracted from $7581$ so that the remainder is a perfect square. Also find the square root of the resulting perfect square.
Reveal Answer & Explanation
Answer:

Step 1: Perform long division on $7581$ (paired as $\overline{75} \; \overline{81}$):
• $8^2 = 64$. Remainder $= 75 - 64 = 11$.
• Bring down $81$ to get $1181$.
• Double quotient: $8 \times 2 = 16$. Trial divisor: $167 \times 7 = 1169$.
• Remainder $= 1181 - 1169 = \mathbf{12}$.
Step 2: The least number that must be subtracted is the remainder: $12$.
Step 3: Calculate the perfect square:

$$7581 - 12 = 7569$$


$$\mathbf{\sqrt{7569} = 87}$$

.


Long division gives quotient 87 with remainder 12. Subtract 12 to get $7569$, whose square root is 87.
3
Find the least number that must be added to $6412$ to make it a perfect square. Also find the square root of the perfect square.
Reveal Answer & Explanation
Answer:

Step 1: Find the square root of $6412$ by long division (paired as $\overline{64} \; \overline{12}$):
• $8^2 = 64$. Remainder $= 0$.
• Bring down $12$. Divisor is $160 \times 0 = 0$. Remainder $= 12$.
• Integer root is $80$, meaning $80^2 < 6412 < 81^2$.
Step 2: Calculate the square of the next integer $(80 + 1) = 81$:

$$81^2 = 81 \times 81 = 6561$$


Step 3: Calculate the least number to be added:

$$\text{Least Number Added} = 6561 - 6412 = \mathbf{149}$$


• The resulting perfect square is $6561$, and $\sqrt{6561} = 81$.


$80^2 = 6400 < 6412$. Next square is $81^2 = 6561$. Subtract: $6561 - 6412 = 149$.
4
Write a Pythagorean Triplet whose smallest member is $14$.
Reveal Answer & Explanation
Answer:

The general form of a Pythagorean Triplet is $(2m, m^2 - 1, m^2 + 1)$.
Step 1: Set the smallest member $2m = 14$:

$$m = \frac{14}{2} = 7$$


Step 2: Find the other two members:

$$m^2 - 1 = 7^2 - 1 = 49 - 1 = \mathbf{48}$$


$$m^2 + 1 = 7^2 + 1 = 49 + 1 = \mathbf{50}$$


• Verification: $14^2 + 48^2 = 196 + 2304 = 2500 = 50^2$. Correct!
• The required Pythagorean Triplet is $(14, 48, 50)$.


Set $2m = 14 \implies m = 7$. Other two terms are $7^2 - 1 = 48$ and $7^2 + 1 = 50$.
5
Evaluate the square root of the decimal: $\sqrt{42.25}$.
Reveal Answer & Explanation
Answer:

Step 1: Pair digits: Integral part $\overline{42}$, decimal part $\overline{25}$.
Step 2: Long division on $42$ gives $6^2 = 36$ (Quotient $= 6$, Remainder $= 42 - 36 = 6$).
Step 3: Place decimal point in quotient ($6.$) and bring down the pair $25$ to form $625$.
Step 4: Double quotient: $6 \times 2 = 12$. Trial divisor: $125 \times 5 = 625$.
Step 5: Remainder $= 0$.
• Final Answer:

$$\mathbf{\sqrt{42.25} = 6.5}$$

.


Pair as 42 and 25. $6^2 = 36$, remainder 6. Divisor $125 \times 5 = 625$. Result is 6.5.
6
How many non-perfect square natural numbers lie between $25^2$ and $26^2$?
Reveal Answer & Explanation
Answer:

• The formula for the number of non-perfect square numbers between $n^2$ and $(n + 1)^2$ is strictly $2n$.
• Here $n = 25$.

$$\text{Number of non-square numbers} = 2 \times 25 = \mathbf{50}$$

.
(Check: $26^2 - 25^2 - 1 = 676 - 625 - 1 = 50$).


Apply $2n$: $2 \times 25 = 50$.
7
Find the side of a square field whose area is $7744\text{ square meters}$.
Reveal Answer & Explanation
Answer: Area of square field $= \text{side}^2 = 7744\text{ m}^2$.
$$\text{Side} = \sqrt{7744}$$
By long division (paired as $\overline{77} \; \overline{44}$):
• $8^2 = 64$. Remainder $= 77 - 64 = 13$.
• Bring down $44 \to 1344$.
• Double quotient: $8 \times 2 = 16$. Trial divisor: $168 \times 8 = 1344$.
$$\text{Side} = \mathbf{88\text{ meters}}$$.
$\sqrt{7744} = 88\text{ meters}$.
8
Without calculating, state with reasons whether $1057$ and $23453$ can be perfect squares.
Reveal Answer & Explanation
Answer:

• $1057$: Cannot be a perfect square because its unit's digit is $7$. The square of any natural number can only end in $0, 1, 4, 5, 6,$ or $9$.
• $23453$: Cannot be a perfect square because its unit's digit is $3$.
• Numbers ending in $2, 3, 7,$ or $8$ can never be perfect squares.


Neither can be a perfect square; numbers ending in 2, 3, 7, or 8 are never perfect squares.
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