MathsCBSE Class 8 · A Square and A Cube

Perfect Squares and Square Roots for Class 8

Updated 17 October 2026 · Fact-checked

A perfect square is the square of a natural number, so 1, 4, 9, 16, ... Its square root is found by pairing equal prime factors, by repeated subtraction of odd numbers, or by estimating between known squares. School exams ask you to test squares by the unit digit, find the smallest multiplier or divisor that makes a perfect square, and use the sum-of-odd-numbers pattern.

Each new L-shaped layer is the next odd number, so the sum of the first n odd numbers is n²; 16 = 1+3+5+7.

Illustration drawn for QuestionBuddy.

Understand Perfect Squares and Square Roots

A square number is a number multiplied by itself: $n \times n = n^{2}$ . The squares of natural numbers are called perfect squares. The name comes from area: a square of side $n$ units has area $n^{2}$ square units. Squares are not limited to whole numbers, since a side of $\frac{3}{5}$ or $2.5$ units gives areas $\frac{9}{25}$ and $6.25$ , but "perfect square" in this chapter means the square of a natural number.

Why only squares have an odd number of factors. Factors of a number come in partner pairs whose product is the number, such as $1 \times 36$ , $2 \times 18$ , $3 \times 12$ , $4 \times 9$ and $6 \times 6$ . A pair breaks only when both partners are the same, which happens exactly when the number is a square. So perfect squares are the only numbers with an odd count of factors. This explains the 100-locker puzzle: a locker stays open exactly when its number is a square.

Patterns you must know.

A perfect square ends in 0, 1, 4, 5, 6 or 9. A number ending in 2, 3, 7 or 8 is never a square. The converse is false: 26 ends in 6 but is not a square.
A number ending in $k$ zeros has a square ending in exactly $2 k$ zeros, so a square always ends in an even number of zeros.
The square of an even number is even; the square of an odd number is odd.
The difference between consecutive squares is an odd number: $(n + 1)^{2} - n^{2} = 2 n + 1$ . Hence the sum of the first $n$ odd numbers is $n^{2}$ , and a number is a perfect square if subtracting 1, 3, 5, ... in turn lands exactly on 0.
Between $n^{2}$ and $(n + 1)^{2}$ there are $2 n$ natural numbers.
Two consecutive triangular numbers add up to a square, for example $3 + 6 = 9$ and $6 + 10 = 16$ .

Finding a square root. The square root of $y$ is the number $x$ with $x^{2} = y$ , written $y = x$ . Every perfect square has two integer square roots, $+ x$ and $- x$ , and this chapter uses the positive one.
1. *Prime factorisation.* If the prime factors can be split into two identical groups, the product of one group is the root. Pair equal primes: $324 = (2 \times 2) (3 \times 3) (3 \times 3)$ , so $324 = 2 \times 3 \times 3 = 18$ . A prime left without a partner means the number is not a perfect square.
2. *Repeated subtraction.* Subtract 1, 3, 5, ... and count the steps until 0.
3. *Estimation.* Trap the number between two known squares, use the unit digit to narrow the choices, then test the midpoint. To estimate a root that is not exact, choose the nearer square.

This page treats squares and square roots; cubes and cube roots are on their own topic page.

Perfect Squares and Square Roots: formulas to remember

Square of a number: $n^{2} = n \times n$; Area of a square of side n.
Gap between consecutive squares: $(n + 1)^{2} - n^{2} = 2 n + 1$; The next square is the previous square plus the next odd number.
Sum of first n odd numbers: $1 + 3 + 5 + \dots + (2 n - 1) = n^{2}$; The nth odd number is 2n - 1.
Numbers between consecutive squares: $count between n^{2} and (n + 1)^{2} = 2 n$; Both squares themselves are excluded.
Zeros at the end: $n ends in k zeros \Rightarrow n^{2} ends in 2 k zeros$; So a square never ends in an odd number of zeros.

How to solve Perfect Squares and Square Roots questions

Use this order for any "is it a perfect square" or "find the root" question.

1Check the unit digit first. If it is 2, 3, 7 or 8, stop: not a square. Check that the number of end zeros is even.
2For a medium number, factorise into primes and write the factors in pairs.
3If a prime has no partner, the number is not a square. To make it one, multiply by the unpaired primes (smallest multiplier) or divide by them (smallest divisor).
4For a root, take one prime from each pair and multiply. Check by squaring the answer.
5For a number that is not a perfect square, bracket it between two consecutive squares and choose the nearer root.
6For sums of odd numbers, count the terms (n) and write n squared; for a range, subtract two such sums.

Perfect Squares and Square Roots in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Square roots from area

Suppose a school wants to tile a square courtyard with 1296 identical square tiles, laid in full rows and columns with none cut and none left over.

How many tiles go along each side? If the school later wants the next larger square courtyard using the same tiles, how many extra tiles are needed?

Try it first, then show the solutionSolution

Tiles along a side $= 1296$ . Since $3 0^{2} = 900$ and $4 0^{2} = 1600$ , and the unit digit is 6, try 34 and 36.
$3 6^{2} = 1296$ , so the side has 36 tiles.
Next square is $3 7^{2}$ . Extra tiles $= 3 7^{2} - 3 6^{2} = 2 \times 36 + 1 = 73$ .

Answer: 36 tiles along each side; 73 extra tiles for the next square layout.

Memory hook: Growing a square by one row and one column adds an L-shaped border: that border is the next odd number.

Odd-number pattern

Imagine a stadium section whose seats form a perfect square: each new row added at the back, with a matching column at the side, makes an L-shaped strip of seats.

The section is 45 seats by 45 seats. How many seats does the L-shaped strip contain if it grows to 46 by 46, and what is the new total?

Try it first, then show the solutionSolution

Strip $= 2 \times 45 + 1 = 91$ seats (the 46th odd number is $2 \times 46 - 1 = 91$ ).
New total $= 2025 + 91 = 2116 = 4 6^{2}$ .

Answer: 91 seats in the strip; 2116 seats in all.

Memory hook: Every step from n squared to the next square adds one L of 2n + 1 blocks.

Tricks to remember Perfect Squares and Square Roots

Short memory aids. Say each one aloud once, then use it on a question the same day.

Squares never end in 2, 3, 7, 8
Remember the "unlucky four": 2, 3, 7, 8. Any number ending there is out. Everything else needs more checking.
Use it when: First test in any "which is not a perfect square" question.
Zeros double
Zeros at the end double when you square: 70 has 1 zero, 4900 has 2; 600 has 2 zeros, 360000 has 4.
Use it when: To reject numbers like 8000 or 90000000 that end in an odd number of zeros.
L-shape for odd numbers
Picture an L around a square. The L around an n by n square has 2n + 1 tiles, which is the next odd number.
Use it when: To jump from one known square to its neighbour: 46 squared is 45 squared plus 91.
Pair or fix
In the prime factorisation, any prime left without a partner is the "single sock". Multiply by it to complete the pair, or divide it away.
Use it when: Smallest multiplier or divisor questions.

Perfect Squares and Square Roots: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Spotting a non-square · 1 mark · MCQBoard level

Which one of the following is not a perfect square? (a) 1089 (b) 2048 (c) 3136 (d) 4225

Try it first, then show the solutionSolution

2048 ends in 8, and no square ends in 8.
Check the others: $3 3^{2} = 1089$ , $5 6^{2} = 3136$ , $6 5^{2} = 4225$ .

Answer: (b) 2048

Example 2: Smallest divisor to get a square · 2 marks · Short answerBoard level

A natural number divides 7350 and the quotient turns out to be a perfect square. Name the least such divisor greater than 1, and give the square root of the quotient.

Try it first, then show the solutionSolution

$7350 = 2 \times 3 \times 5^{2} \times 7^{2}$ . (1 mark)
The primes 2 and 3 have no partners, so divide by $2 \times 3 = 6$ . Quotient $= 1225 = 5^{2} \times 7^{2}$ , so $1225 = 5 \times 7 = 35$ . (1 mark)

Answer: Divide by 6. The quotient is 1225 and its square root is 35.

Example 3: Sum of odd numbers over a range · 3 marks · Short answerHarder board level

Find the sum of all odd numbers from 51 to 99 without adding term by term.

Try it first, then show the solutionSolution

The sum of the first $n$ odd numbers is $n^{2}$ . The odd numbers up to 99 are the first 50 (the 50th is $2 \times 50 - 1 = 99$ ), so their sum is $5 0^{2} = 2500$ . (1 mark)
The odd numbers up to 49 are the first 25, with sum $2 5^{2} = 625$ . (1 mark)
Required sum $= 2500 - 625 = 1875$ . Check: 25 terms with average 75 give $25 \times 75 = 1875$ . (1 mark)

Answer: 1875

Example 4: Neighbouring squares from a given square · 3 marks · Short answerHarder board level

Given that $4 5^{2} = 2025$ , find $4 6^{2}$ and $4 4^{2}$ using odd numbers only, and state how many natural numbers lie strictly between $4 5^{2}$ and $4 6^{2}$ .

Try it first, then show the solutionSolution

$4 6^{2} = 4 5^{2} + (the 46th odd number) = 2025 + 91 = 2116$ . (1 mark)
$4 4^{2} = 4 5^{2} - (the 45th odd number) = 2025 - 89 = 1936$ . (1 mark)
Numbers strictly between $4 5^{2}$ and $4 6^{2}$ : $2 \times 45 = 90$ . Check: $2116 - 2025 - 1 = 90$ . (1 mark)

Answer: 46 squared is 2116, 44 squared is 1936, and 90 numbers lie between 2025 and 2116.

Example 5: How many four-digit squares? · 4 marks · Long answerHarder board level

Find the smallest and the greatest four-digit perfect squares, and hence the number of four-digit perfect squares.

Try it first, then show the solutionSolution

Smallest four-digit number is 1000. $3 1^{2} = 961$ has three digits and $3 2^{2} = 1024$ has four, so the smallest is 1024. (1 mark)
Greatest four-digit number is 9999. $9 9^{2} = 9801$ and $10 0^{2} = 10000$ has five digits, so the greatest is 9801. (1 mark)
The squares of 32, 33, ..., 99 are four-digit squares. (1 mark)
Count $= 99 - 32 + 1 = 68$ . (1 mark)

Answer: Smallest 1024, greatest 9801; there are 68 four-digit perfect squares.

Where students lose marks in Perfect Squares and Square Roots

The slips examiners see most often, and what to write instead.

Mistake: Declaring a number a square because of its unit digit
The unit digit can only rule a number out. Numbers ending in 0, 1, 4, 5, 6, 9 may or may not be squares.
Wrong working
"1026 ends in 6, so it is a perfect square."
Right working
$3 2^{2} = 1024$ and $3 3^{2} = 1089$ , so 1026 lies between consecutive squares and is not a square.
Do this: Use the unit digit only to eliminate; confirm with factorisation or estimation.
Usually costs: The full mark of an MCQ or the reasoning marks.
Mistake: Multiplying by every prime instead of only the unpaired ones
Only primes with an odd power need fixing.
Wrong working
"7350 = 2 x 3 x 5^2 x 7^2, so multiply by 2 x 3 x 5 x 7."
Right working
Only 2 and 3 are unpaired. Multiply (or divide) by $2 \times 3 = 6$ .
Do this: Write the exponents and circle only the odd ones.
Usually costs: 1 to 2 marks.
Mistake: Off-by-one in the odd-number pattern
The nth odd number is 2n - 1, not 2n + 1.
Wrong working
"46 squared = 45 squared + 93."
Right working
The 46th odd number is $2 \times 46 - 1 = 91$ , so $4 6^{2} = 2025 + 91 = 2116$ .
Do this: Test with n = 1: first odd number is 1 = 2(1) - 1.
Usually costs: The answer mark and the following steps.
Mistake: Counting the squares themselves as lying between them
Between n squared and (n+1) squared there are 2n numbers, excluding both ends.
Wrong working
"Numbers between 45^2 and 46^2: 2 x 45 + 1 = 91."
Right working
$2116 - 2025 - 1 = 90$ .
Do this: Subtract the two squares and then subtract 1.
Usually costs: 1 mark.

Exam tips

Open every "is it a perfect square" answer with the unit digit and the end zeros: two quick checks that can finish the question in one line.
Write prime factorisations in power form and circle the odd exponents; that picture gives the smallest multiplier or divisor at once.
For sums of consecutive odd numbers, find how many terms there are before writing n squared.
In estimation answers, show the two bracketing squares and the unit-digit argument; those lines carry the step marks.
Always verify by squaring your root. It takes five seconds and catches slips.

Practice questions on Perfect Squares and Square Roots

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 3 marks · Short answerHarder board level
Show by estimation and the unit digit that $2116$ is a whole number, and find it, without a calculator or full prime factorisation.
Hint
Bracket between 40 and 50, use the unit digit 6, then test 45.
Answer
$4 0^{2} = 1600 < 2116 < 2500 = 5 0^{2}$ , so the root is between 40 and 50. The unit digit 6 means the root ends in 4 or 6. Since $4 5^{2} = 2025 < 2116$ , the root is above 45, so it is 46. Check: $4 6^{2} = 2116$ . So $2116 = 46$ .
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): 1 + 3 + 5 + ... + 39 = 400.
Reason (R): The sum of the first n odd numbers is n squared.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Which odd number is 39? Find its position.
Answer
(a). The nth odd number is $2 n - 1$ ; $39 = 2 \times 20 - 1$ , so 39 is the 20th odd number and the sum is $2 0^{2} = 400$ . R is true and explains A.
Q3 · 4 marks · Long answerHOTS / competency
Show that the sum of two consecutive triangular numbers $T_{n} = \frac{n ( n + 1 )}{2}$ and $T_{n + 1}$ is always a perfect square. Use your result to find which two consecutive triangular numbers add to 1156.
Hint
Add n(n+1)/2 and (n+1)(n+2)/2 and factorise. For 1156, find its square root.
Answer
$T_{n} + T_{n + 1} = \frac{n ( n + 1 )}{2} + \frac{( n + 1 ) ( n + 2 )}{2} = \frac{( n + 1 ) ( n + n + 2 )}{2} = \frac{( n + 1 ) \times 2 ( n + 1 )}{2} = (n + 1)^{2}$ . For 1156: $1156 = 34$ (since $3 4^{2} = 1156$ ), so $n + 1 = 34$ , $n = 33$ . The numbers are $T_{33} = \frac{33 \times 34}{2} = 561$ and $T_{34} = \frac{34 \times 35}{2} = 595$ , and $561 + 595 = 1156$ .
Q4 · 4 marks · Case-basedHOTS / competency
Case study: a school has 200 lockers numbered 1 to 200 and 200 students numbered 1 to 200. Student k toggles (opens if closed, closes if open) every kth locker, starting with all lockers closed.
(i) Which lockers remain open, and how many are they? (1 mark)
(ii) How many times is locker 144 toggled? (1 mark)
(iii) How many lockers are toggled exactly three times? Name them. (2 marks)
Hint
Toggles equal the number of factors. Squares have an odd count; exactly 3 factors means the square of a prime.
Answer
(i) Squares from $1^{2}$ to $1 4^{2} = 196$ (as $1 5^{2} = 225 > 200$ ): 14 lockers. (ii) $144 = 2^{4} \times 3^{2}$ has $(4 + 1) (2 + 1) = 15$ factors, so it is toggled 15 times. (iii) A number with exactly three factors is $p^{2}$ for a prime $p$ . We need $p^{2} \leq 200$ , so $p \leq 14$ : $p = 2, 3, 5, 7, 11, 13$ . Lockers 4, 9, 25, 49, 121 and 169: six lockers.
Q5 · 3 marks · Short answerHarder board level
What is the least square number that 12, 18 and 30 all divide exactly? Justify why no smaller square works.
Hint
Find the LCM, then make every prime exponent even.
Answer
$12 = 2^{2} \times 3$ , $18 = 2 \times 3^{2}$ , $30 = 2 \times 3 \times 5$ , so LCM $= 2^{2} \times 3^{2} \times 5 = 180$ . The prime 5 is unpaired; multiply by 5: $180 \times 5 = 900 = 2^{2} \times 3^{2} \times 5^{2} = 3 0^{2}$ . Any number divisible by all three is a multiple of 180, and the smallest multiple of 180 that is a square is 900. Check: $900 \div 12 = 75$ , $900 \div 18 = 50$ , $900 \div 30 = 30$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Which unit digits can a perfect square have?Show answer
0, 1, 4, 5, 6 or 9. Never 2, 3, 7 or 8.
2.How many natural numbers lie between n squared and (n+1) squared?Show answer
2n.
3.What is the sum of the first n odd numbers?Show answer
n squared.
4.Why do only squares have an odd number of factors?Show answer
Factors pair up with partners, and a pair collapses to one factor only when both partners are equal, which happens for squares.
5.How many zeros can the end of a square have?Show answer
Only an even number, twice the zeros at the end of its root.
6.What does an unpaired prime in a factorisation tell you?Show answer
The number is not a perfect square.

Perfect Squares and Square Roots: frequently asked questions

Is 0 a perfect square?

The chapter defines perfect squares as squares of natural numbers, starting at 1. Zero squared is zero, but questions in this book start from 1.

Why does a square have two square roots?

Because a negative number squared is also positive: 8 squared and (-8) squared are both 64. The radical sign means the positive root, which is what this chapter uses.

Which square root method should I use in an exam?

Prime factorisation for medium numbers, estimation with the unit digit for larger perfect squares, and repeated subtraction only for small numbers or when the question asks for it.

Can a square end with exactly one zero?

No. A number ending in k zeros has a square ending in 2k zeros, so the count of end zeros of a square is always even.