Squares and Square Roots: Properties, Prime Factorization, and Long Division

In geometry, a square is a four-sided polygon with all sides of equal length. If a square has a side length of $a$ units, its area is given by the product of its length and breadth: $a \times a = a^2$. In arithmetic and algebra, this concept gives rise to Square Numbers (or Perfect Squares) and their inverse operation, Square Roots.

Understanding squares and square roots is fundamental to school mathematics. It forms the bedrock for advanced topics such as the Pythagorean Theorem, quadratic equations, coordinate geometry, mensuration, and algebraic expressions. Mastering these concepts equips you with essential numerical mental estimation techniques, pattern recognition skills, and algorithmic precision required for CBSE Class 8 board assessments and competitive examinations.

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1. In-Depth Conceptual Breakdown

1.1 What is a Square Number?

When a natural number is multiplied by itself, the product obtained is called the square of that number.

If $n$ is a natural number, then the square of $n$ is denoted by $n^2$: $n^2 = n \times n$

For instance:

• $1^2 = 1 \times 1 = 1$
• $2^2 = 2 \times 2 = 4$
• $5^2 = 5 \times 5 = 25$
• $12^2 = 12 \times 12 = 144$

A natural number $m$ is called a perfect square or a square number if there exists a natural number $n$ such that $m = n^2$.

Table of Squares from 1 to 20

It is strongly recommended for Class 8 students to memorize the squares of natural numbers up to 20:

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1.2 Important Properties of Square Numbers

Square numbers exhibit unique properties that allow us to test whether a number is a perfect square without complete calculation.

Property 1: Ending Digits (Units Digit Rule)

Look at the units digit of the square numbers above: $1, 4, 9, 6, 5, 6, 9, 4, 1, 0$.

• All perfect squares end in $0, 1, 4, 5, 6,$ or $9$.
• Rule: A natural number ending in $2, 3, 7,$ or $8$ is NEVER a perfect square.
• Example: $1057$ ends in $7$, so it cannot be a perfect square. $23453$ ends in $3$, so it cannot be a perfect square.

Property 2: Relationship Between the Units Digit of a Number and its Square

• Example: The square of $83$ ends in $9$ (since $3^2 = 9$). The square of $277$ ends in $9$ (since $7^2 = 49$).

Property 3: Number of Trailing Zeros

If a number ends with $k$ zeros, its square will end with $2k$ zeros (always an even number of zeros).

• $10$ (1 zero) $\rightarrow 10^2 = 100$ (2 zeros)
• $400$ (2 zeros) $\rightarrow 400^2 = 160000$ (4 zeros)
• Rule: A number ending in an odd number of zeros is never a perfect square (e.g., $5000$ is not a perfect square).

Property 4: Parity of Squares

• The square of an even number is always even. ($6^2 = 36$, $14^2 = 196$)
• The square of an odd number is always odd. ($7^2 = 49$, $13^2 = 169$)

Property 5: Sum of Consecutive Odd Numbers

The square of any natural number $n$ is equal to the sum of the first $n$ consecutive odd natural numbers: $n^2 = 1 + 3 + 5 + 7 + \dots + (2n - 1)$

• $1^2 = 1$ (sum of 1st odd number)
• $2^2 = 1 + 3 = 4$ (sum of first 2 odd numbers)
• $3^2 = 1 + 3 + 5 = 9$ (sum of first 3 odd numbers)
• $4^2 = 1 + 3 + 5 + 7 = 16$ (sum of first 4 odd numbers)

If a number cannot be expressed as a sum of consecutive odd natural numbers starting from $1$, it is not a perfect square.

Property 6: Non-Square Numbers Between Consecutive Squares

Between two consecutive square numbers $n^2$ and $(n+1)^2$, there are $2n$ non-square natural numbers.

• Total numbers between $n^2$ and $(n+1)^2 = (n+1)^2 - n^2 - 1 = n^2 + 2n + 1 - n^2 - 1 = 2n$.
• Example: Between $3^2 = 9$ and $4^2 = 16$, the non-square numbers are $10, 11, 12, 13, 14, 15$.
• Total count $= 2 \times 3 = 6$ numbers.

Property 7: Pythagorean Triplets

A set of three natural numbers $a, b, c$ is called a Pythagorean triplet if: $a^2 + b^2 = c^2$

For any natural number $m > 1$, the triple $(2m, m^2 - 1, m^2 + 1)$ forms a Pythagorean triplet.

• Take $m = 3$:
  • $2m = 2(3) = 6$
  • $m^2 - 1 = 3^2 - 1 = 8$
  • $m^2 + 1 = 3^2 + 1 = 10$
• Check: $6^2 + 8^2 = 36 + 64 = 100 = 10^2$. Hence, $(6, 8, 10)$ is a Pythagorean triplet.

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1.3 What is a Square Root?

The operation of finding the square root is the inverse operation of squaring.

If $n^2 = m$, then the square root of $m$ is $n$, written as: $\sqrt{m} = n$

The symbol $\sqrt{\phantom{x}}$ is called the radical sign.

• Since $7^2 = 49$, the square root of $49$ is $\sqrt{49} = 7$.
• Since $15^2 = 225$, the square root of $225$ is $\sqrt{225} = 15$.

Note on Signs: Every positive real number has two square roots: one positive and one negative. For example, $(+5)^2 = 25$ and $(-5)^2 = 25$. However, at the Class 8 level, $\sqrt{m}$ represents the principal (positive) square root.

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1.4 Finding Square Root by Prime Factorization Method

This method is suitable for finding the square root of perfect squares.

Step-by-Step Procedure:

1. Express the given number as a product of its prime factors.
2. Group identical prime factors into pairs.
3. Take one factor from each pair and multiply them together.
4. The product obtained is the square root of the given number.

Concept of Smallest Multiplier/Divisor:

If a number is not a perfect square, its prime factors will have unpaired terms.

• To make it a perfect square by multiplication, multiply the number by the product of the unpaired factors.
• To make it a perfect square by division, divide the number by the product of the unpaired factors.

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1.5 Finding Square Root by Long Division Method

When numbers are very large, prime factorization becomes time-consuming and tedious. The Long Division Method provides a systematic approach to finding the square root of any number (perfect squares, non-perfect squares, and decimals).

Step-by-Step Algorithm for Whole Numbers:

Step 1: Group digits into pairs (bars) starting from the units digit (right to left).
        Example: 5476 -> 54 76

Step 2: Find the largest single digit whose square is <= the leftmost period.
        Left period = 54. Largest square <= 54 is 49 (7^2).
        Quotient = 7, Divisor = 7. Subtract 49 from 54 -> Remainder = 5.

Step 3: Bring down the next pair of digits to the right of the remainder.
        New Dividend = 576.

Step 4: Double the current quotient to get the starting part of the new divisor.
        New Quotient so far = 7 -> 7 x 2 = 14.
        New Divisor candidate = 14_ (where _ is a digit to be found).

Step 5: Find a digit 'x' such that (14x * x) <= 576.
        Try x = 4: 144 x 4 = 576.
        Quotient becomes 74. Remainder = 0.

Step 6: The final quotient is the square root. -> sqrt(5476) = 74.

Long Division for Decimal Numbers:

• Place bars on the integral part from right to left (starting from the units place).
• Place bars on the decimal part from left to right (starting right after the decimal point).
• Add trailing zeros if a decimal period contains an odd number of digits.
• Place a decimal point in the quotient as soon as you cross the decimal point in the dividend.

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2. Real-World Applications

Application 1: Tiling a Square Room or Field

Suppose a civil engineer needs to construct a square assembly hall with a total area of $1296 \text{ m}^2$. To determine how many meters of boundary wall or skirting material are required, they must find the length of one side: $\text{Side} = \sqrt{\text{Area}} = \sqrt{1296} = 36 \text{ meters}$ The perimeter can then be calculated as $4 \times 36 = 144 \text{ meters}$.

Application 2: Distance Calculations via Pythagoras Theorem

Navigation systems (like GPS) calculate the straight-line distance between two perpendicular vectors using the Pythagorean triplet property ($c = \sqrt{a^2 + b^2}$). If a vehicle travels $9 \text{ km}$ East and then $12 \text{ km}$ North, the direct distance from the starting point is: $\text{Distance} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \text{ km}$

Application 3: Physics - Kinetic Energy and Vehicle Stopping Distance

The kinetic energy of a moving object is given by $E_k = \frac{1}{2}m v^2$. The speed $v$ is proportional to the square root of kinetic energy: $v = \sqrt{\frac{2 E_k}{m}}$. Forensic investigators use this square root relationship to calculate a vehicle's speed before braking, based on the length of skid marks left on the road.

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3. Step-by-Step Solved Textbook Examples

Example 1: Testing and Finding Smallest Multiplier

Problem: Find the smallest natural number by which $252$ must be multiplied so that the product becomes a perfect square. Also, find the square root of the square number so obtained.

Solution:

Step 1: Express $252$ as a product of prime factors using prime factorization.

\begin{array}{l|n} 2 & 252 \\ \hline 2 & 126 \\ \hline 3 & 63 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}

$252 = 2 \times 2 \times 3 \times 3 \times 7 = (2 \times 2) \times (3 \times 3) \times 7$

Step 2: Group the prime factors into pairs.

• $2$ is in a pair $(2 \times 2)$.
• $3$ is in a pair $(3 \times 3)$.
• $7$ has no pair.

Step 3: To make $252$ a perfect square, every prime factor must be in pairs. Thus, we must multiply $252$ by $7$.

$\text{New Perfect Square} = 252 \times 7 = 1764$

Step 4: Find the square root of $1764$. $1764 = (2 \times 2) \times (3 \times 3) \times (7 \times 7)$ $\sqrt{1764} = 2 \times 3 \times 7 = 42$

Final Answer:

• Smallest multiplier $= \mathbf{7}$
• Square root of new number $= \mathbf{42}$

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Example 2: Prime Factorization Method

Problem: Find the square root of $7056$ using the Prime Factorization Method.

Solution:

Step 1: Resolve $7056$ into prime factors.

\begin{array}{l|n} 2 & 7056 \\ \hline 2 & 3528 \\ \hline 2 & 1764 \\ \hline 2 & 882 \\ \hline 3 & 441 \\ \hline 3 & 147 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}

Step 2: Write down the prime factorization: $7056 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7$

Step 3: Group the prime factors into pairs: $7056 = (2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (7 \times 7)$

Step 4: Take one prime factor from each pair to compute the square root: $\sqrt{7056} = 2 \times 2 \times 3 \times 7$ $\sqrt{7056} = 4 \times 21 = 84$

Final Answer: $\mathbf{\sqrt{7056} = 84}$

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Example 3: Long Division Method for Large Whole Numbers

Problem: Find the square root of $15876$ using the Long Division Method.

Solution:

Step 1: Pair the digits from right to left: $\overline{1}\ \overline{58}\ \overline{76}$.

Step 2: Perform long division:

        1   2   6
     -------------
  1  |  1  58  76
     | -1
     -------------
 22  |  0  58
     | -   44
     -------------
246  |     14  76
     | -   14  76
     -------------
     |          0

Step-by-step breakdown of division steps:

1. The first period is $1$. The largest number whose square is $\le 1$ is $1$.
   • Write $1$ in quotient and divisor. Subtract $1 - 1 = 0$.
2. Bring down the next pair, $58$. New dividend $= 58$.
3. Double the quotient ($1 \times 2 = 2$). The new divisor is $2\_$.
4. Test digits for $2\_ \times \_$ to get $\le 58$:
   • $22 \times 2 = 44 \le 58$. (Note: $23 \times 3 = 69 > 58$).
   • Put $2$ in quotient and divisor. Subtract $58 - 44 = 14$.
5. Bring down the next pair, $76$. New dividend $= 1476$.
6. Double the current quotient ($12 \times 2 = 24$). The new divisor is $24\_$.
7. Test digits for $24\_ \times \_$ to get $1476$:
   • $246 \times 6 = 1476$.
   • Put $6$ in quotient and divisor. Remainder $= 0$.

Final Answer: $\mathbf{\sqrt{15876} = 126}$

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Example 4: Long Division Method for Decimals

Problem: Find the square root of $42.25$.

Solution:

Step 1: Pair the integral part from right to left ($\overline{42}$) and decimal part from left to right ($\overline{25}$). The paired representation is $\overline{42}.\overline{25}$.

Step 2: Perform long division:

        6 .  5
     -------------
  6  | 42 . 25
     | -36
     -------------
125  |  6   25
     | -6   25
     -------------
     |       0

Detailed Working:

1. First period $= 42$. Largest square $\le 42$ is $36$ ($6^2$).
   • Quotient $= 6$, Divisor $= 6$. Remainder $= 42 - 36 = 6$.
2. Place decimal point in quotient after $6$.
3. Bring down the decimal period $25$. New Dividend $= 625$.
4. Double the quotient ($6 \times 2 = 12$). New divisor candidate $= 12\_$.
5. Test $x = 5$: $125 \times 5 = 625$.
   • Write $5$ in quotient. Remainder $= 0$.

Final Answer: $\mathbf{\sqrt{42.25} = 6.5}$

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4. Common Student Mistakes to Avoid

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5. Practice Questions for Self-Assessment

Question 1

Find the least number that must be subtracted from $5607$ so as to get a perfect square. Also, find the square root of the resulting perfect square.

<details> <summary><b>Click to view Step-by-Step Solution</b></summary>

Step 1: Apply Long Division on $5607$ to find the remainder.

        7   4
     -------------
  7  | 56  07
     | -49
     -------------
144  |  7  07
     | -5  76
     -------------
     |  1  31

1. Pair digits: $\overline{56}\ \overline{07}$.
2. Largest square $\le 56$ is $49$ ($7^2$). Remainder $= 56 - 49 = 7$.
3. Bring down $07$. Dividend $= 707$.
4. Double quotient: $7 \times 2 = 14$. New divisor candidate $= 14\_$.
5. $144 \times 4 = 576 \le 707$ (since $145 \times 5 = 725 > 707$).
6. Remainder $= 707 - 576 = 131$.

Step 2: Interpret remainder. The remainder $131$ represents the excess amount preventing $5607$ from being a perfect square.

$\text{Required Number to Subtract} = 131$ $\text{Perfect Square} = 5607 - 131 = 5476$ $\sqrt{5476} = 74$

Final Answer:

• Number to be subtracted $= \mathbf{131}$
• Square root of the new number $= \mathbf{74}$

</details>

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Question 2

Find the smallest square number that is divisible by each of the numbers $8$, $15$, and $20$.

<details> <summary><b>Click to view Step-by-Step Solution</b></summary>

Step 1: Find the Least Common Multiple (LCM) of $8, 15,$ and $20$.

$8 = 2^3 = 2 \times 2 \times 2$ $15 = 3 \times 5$ $20 = 2^2 \times 5 = 2 \times 2 \times 5$

$\text{LCM}(8, 15, 20) = 2 \times 2 \times 2 \times 3 \times 5 = 120$

Step 2: Write the prime factorization of $120$ and group into pairs. $120 = (2 \times 2) \times 2 \times 3 \times 5$

Step 3: Identify unpaired prime factors. The prime factors $2$, $3$, and $5$ are unpaired.

Step 4: Multiply $120$ by the product of unpaired factors to get the smallest perfect square. $\text{Required Square Number} = 120 \times (2 \times 3 \times 5) = 120 \times 30 = 3600$

Final Answer: $\mathbf{3600}$

</details>

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Question 3

Find the length of the side of a square whose area is equal to the area of a rectangle of length $13.6 \text{ m}$ and breadth $3.4 \text{ m}$.

<details> <summary><b>Click to view Step-by-Step Solution</b></summary>

Step 1: Calculate the area of the rectangle. $\text{Area of rectangle} = \text{length} \times \text{breadth}$ $\text{Area} = 13.6 \times 3.4 = 46.24 \text{ m}^2$

Step 2: Set the area of the square equal to the area of the rectangle. $\text{Area of square} = s^2 = 46.24 \text{ m}^2$ $s = \sqrt{46.24}$

Step 3: Find $\sqrt{46.24}$ using long division. Pairs: $\overline{46}.\overline{24}$

        6 .  8
     -------------
  6  | 46 . 24
     | -36
     -------------
128  | 10   24
     | -10  24
     -------------
     |       0

1. $6^2 = 36 \le 46$. Remainder $= 10$.
2. Bring down $24$. New Dividend $= 1024$.
3. Double quotient: $6 \times 2 = 12$.
4. $128 \times 8 = 1024$.

Final Answer: The side of the square is $\mathbf{6.8\text{ meters}}$.

</details>

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6. Exam Revision & FAQs

FAQ 1: How can I quickly determine if a given number is NOT a perfect square just by looking at it?

Look for these primary indicators:

1. Units digit test: If the number ends in $2, 3, 7,$ or $8$, it is never a perfect square.
2. Trailing zeros test: If the number ends in an odd number of zeros (e.g., $1000$ or $400000$), it is never a perfect square.
3. Digital Root (Sum of Digits) Test: Repeatedly add the digits of the number until you get a single digit. A perfect square always has a digital root of $1, 4, 7,$ or $9$. (Note: While necessary, this is not a sufficient condition on its own, but it quickly eliminates non-squares).

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FAQ 2: What is the difference between finding the "smallest multiplier" versus the "least number to be subtracted" to make a number a perfect square?

• Smallest Multiplier/Divisor: Use the Prime Factorization Method. Look for unpaired prime factors and multiply or divide by those unpaired factors.
• Least Number to be Subtracted/Added: Use the Long Division Method.
  • For subtraction: The answer is directly the remainder left at the end of long division.
  • For addition: Find the square of $(\text{Quotient} + 1)$ and subtract the original number from it.

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FAQ 3: How many non-square numbers lie between $15^2$ and $16^2$?

Using the property that there are $2n$ non-square numbers between $n^2$ and $(n+1)^2$: Here, $n = 15$. $\text{Number of non-square numbers} = 2n = 2 \times 15 = 30$ (Verification: $15^2 = 225$, $16^2 = 256$. The numbers are $226, 227, \dots, 255$, total $256 - 225 - 1 = 30$).

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FAQ 4: How do you write a Pythagorean triplet whose smallest member is 12?

Using the general form $(2m, m^2 - 1, m^2 + 1)$:

Case 1: Set $2m = 12 \implies m = 6$.

• First member: $2m = 12$
• Second member: $m^2 - 1 = 6^2 - 1 = 36 - 1 = 35$
• Third member: $m^2 + 1 = 6^2 + 1 = 36 + 1 = 37$

Check if $12$ is the smallest member: $12 < 35 < 37$. Yes!

Therefore, the required Pythagorean triplet is $(12, 35, 37)$.