Key Concepts & Formulae at a Glance
- A number obtained by multiplying a number by itself is a square number. Squares of natural numbers are perfect squares: 1, 4, 9, 16, 25, …
- A number is a perfect square exactly when it has an odd number of factors — every factor pairs with a partner, except the one factor (the square root) that pairs with itself.
- All perfect squares end in 0, 1, 4, 5, 6 or 9 (never 2, 3, 7 or 8), and can only have an even number of zeros at the end.
- The square of an even number is even; the square of an odd number is odd.
- Every perfect square is a sum of consecutive odd numbers starting from 1: \(1+3+5+\dots+(2n-1) = n^2\).
- Square root is the inverse of squaring. Every perfect square has two integer square roots, \(+n\) and \(-n\); the positive one is denoted \(\sqrt{\phantom{x}}\).
- A number is a perfect square if its prime factors can be split into two identical groups.
- A number obtained by multiplying a number by itself three times is a cube: 1, 8, 27, 64, …, and perfect cubes are ghana in Sanskrit.
- A number is a perfect cube if its prime factors can be split into three identical groups (each prime factor appears in multiples of 3 in the factorisation).
- A cube can only end with a multiple of 3 zeros (never exactly 1 or 2 zeros).
- Cube root is denoted \(\sqrt[3]{\phantom{x}}\). For example, \(\sqrt[3]{27}=3\).
- A taxicab number can be written as the sum of two positive cubes in two different ways — the smallest is the Hardy–Ramanujan number, 1729.
The Locker Puzzle
Queen Ratnamanjuri leaves her son Khoisnam and his 99 relatives a puzzle involving 100 lockers, numbered 1 to 100. Person k toggles every k-th locker (opens it if closed, closes it if open). Person 1 opens every locker, Person 2 toggles every 2nd locker, Person 3 toggles every 3rd, and so on until all 100 people have had their turn. Khoisnam realises he can predict which lockers stay open before the process even begins.
A locker is toggled once for every factor of its number — e.g. locker #6 is toggled by persons 1, 2, 3 and 6 (the factors of 6), so it is toggled 4 times and ends up closed. A locker toggled an odd number of times ends up open; one toggled an even number of times ends up closed. Since factors normally pair up (1↔6, 2↔3 for the number 6), most numbers have an even number of factors — except perfect squares, where one factor pairs with itself (e.g. 36 = 6 × 6), leaving an odd count.
MTDoes every number have an even number of factors?
No. Most numbers do have an even number of factors, because their factors split into partner pairs (like 1↔6 and 2↔3 for 6). But numbers like 1, 4, 9, 16, 25, … have one factor that pairs with itself (e.g. \(6 \times 6 = 36\)), which leaves an odd total count of factors.
MTCan you use this insight to find more numbers with an odd number of factors?
Yes — every number that can be written as a factor pair with both factors equal has an odd number of factors: \(1\times1,\ 2\times2,\ 3\times3,\ 4\times4,\ \dots\) In other words, every perfect square has an odd number of factors, because one of its factors (its square root) pairs with itself instead of with a different partner.
MTWrite the locker numbers that remain open.
Only lockers whose numbers are perfect squares are toggled an odd number of times, so only they remain open at the end.
Khoisnam then finds ten lockers with word clues and reads: "The passcode consists of the first five locker numbers that were touched exactly twice." A locker is touched exactly twice only if it has exactly two factors — that is, only 1 and itself — which is precisely the definition of a prime number. So the code is the first five primes: 2-3-5-7-11.
1.1 Square Numbers
The area of a square of side n units is \(n \times n\), written \(n^2\) ("n squared"). Squares of natural numbers (1, 4, 9, 16, 25, …) are called perfect squares. A square can also have a fractional or decimal side, e.g. \(\left(\frac{3}{5}\right)^2=\frac{9}{25}\) and \((2.5)^2=6.25\).
Patterns and Properties of Perfect Squares
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| n² | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 |
| n | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| n² | 121 | 144 | 169 | 196 | 225 | 256 | 289 | 324 | 361 | 400 |
MTWhich of the following numbers have the digit 6 in the units place? (i) 38² (ii) 34² (iii) 46² (iv) 56² (v) 74² (vi) 82²
A square ends in 6 only when the original number ends in 4 or 6 (since \(4^2=16\) and \(6^2=36\) both end in 6). Checking each: 38 ends in 8 (38² ends in 4); 34 ends in 4 ✓; 46 ends in 6 ✓; 56 ends in 6 ✓; 74 ends in 4 ✓; 82 ends in 2 (82² ends in 4).
MTIf a number contains 3 zeros at the end, how many zeros will its square have at the end?
A number ending in 3 zeros can be written as \(m \times 1000 = m \times 10^3\), so its square is \(m^2 \times 10^6\) — six zeros.
MTWhat do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?
The number of zeros at the end of the square is always exactly double the number of zeros at the end of the original number. Since doubling any whole number always gives an even number, a perfect square's trailing zero count is always even — it can never end in exactly 1, 3, 5, … zeros.
MTWhat can you say about the parity of a number and its square?
An even number is \(2k\), so its square \(4k^2\) is also even. An odd number is \(2k+1\), so its square \(4k^2+4k+1\) is also odd (one more than an even number). So squaring never changes parity.
Perfect Squares and Odd Numbers
The differences between consecutive squares are consecutive odd numbers: \(4-1=3\), \(9-4=5\), \(16-9=7\), \(25-16=9\), and so on. Equivalently, every square is a sum of consecutive odd numbers starting from 1:
Each coloured "L-shell" adds the next odd number of dots: 1, then 3 more (total 4), then 5 more (total 9), then 7 more (total 16), then 9 more (total 25), then 11 more (total 36).
MTUsing the pattern above, find 36², given that 35² = 1225. How do we find the 36th odd number? What is the nth odd number?
The nth odd number is \(2n-1\) (1st is 1, 2nd is 3, 3rd is 5, …, 6th is 11, and so on). So the 36th odd number is \(2(36)-1=71\).
Since \(35^2=1225\) is the sum of the first 35 odd numbers, adding the 36th odd number gives the sum of the first 36 odd numbers, which is \(36^2\): \(1225+71=1296\).
A number can also be tested for being a perfect square by successively subtracting consecutive odd numbers starting from 1. If you reach exactly 0, the number is a perfect square (and the count of odd numbers subtracted is its square root); if you cross below 0 without hitting it exactly, the number is not a perfect square.
MTFind how many numbers lie between two consecutive perfect squares. Do you notice a pattern?
If \(p\) and \(q\) are two consecutive perfect square numbers (say \(p=n^2\) and \(q=(n+1)^2\)), the count of numbers strictly between them is \(q-p-1\). Since \((n+1)^2-n^2=2n+1\), this simplifies to \(2n+1-1=2n\) — exactly twice the smaller number's square root. For example, between 16 (\(4^2\)) and 25 (\(5^2\)) lie \(2 \times 4 = 8\) numbers (17 to 24).
MTHow many square numbers are there between 1 and 100? Between 101 and 200? Tabulate the number of squares in each block of 100. What is the largest square less than 1000?
Counting perfect squares landing in each block of 100 (using the squares table above and beyond):
| 1–100 | 101–200 | 201–300 | 301–400 | 401–500 |
|---|---|---|---|---|
| 10 | 4 | 3 | 3 | 2 |
| 501–600 | 601–700 | 701–800 | 801–900 | 901–1000 |
| 2 | 2 | 2 | 2 | 1 |
As expected from the previous pattern, the perfect squares get more spread out (fewer per block of 100) as the numbers grow larger, since the gap between consecutive squares \(n^2\) and \((n+1)^2\) keeps growing (it equals \(2n+1\)).
Perfect Squares and Triangular Numbers
Triangular numbers (1, 3, 6, 10, 15, …) are the counts of dots arranged in a triangle, where the nth triangular number \(T_n = 1+2+3+\dots+n = \frac{n(n+1)}{2}\). Adding two consecutive triangular numbers always gives a perfect square: \(1+3=4=2^2\), \(3+6=9=3^2\), \(6+10=16=4^2\), and continuing the pattern, \(10+15=25=5^2\). In general, \(T_{n-1}+T_n = n^2\).
Square Roots
If \(y=x^2\), then x is the square root of y. Every positive perfect square has two integer square roots, one positive and one negative (since \((-n)^2=n^2\) as well); the symbol \(\sqrt{\phantom{x}}\) conventionally denotes the positive root.
MTWhat is the square root of 64?
\(8 \times 8 = 64\) and \((-8) \times (-8) = 64\) as well, so both +8 and −8 square to 64.
To check whether a number is a perfect square and find its square root, three methods work: (1) listing squares in sequence until you reach or pass the number; (2) successively subtracting consecutive odd numbers starting from 1 until you hit exactly 0; and (3) prime factorisation — if the prime factors can be split into two identical groups, the number is a perfect square, and the product of one group is the square root.
MTIs 324 a perfect square? Is 156 a perfect square?
324: \(324 = 2 \times 2 \times 3 \times 3 \times 3 \times 3\), which groups into two identical sets \((2\times3\times3)\times(2\times3\times3) = 18 \times 18\). So 324 is a perfect square, and \(\sqrt{324}=18\).
156: \(156 = 2 \times 2 \times 3 \times 13\). These factors cannot be split into two identical groups (the 3 and 13 have no partners), so 156 is not a perfect square.
MTFind whether 1156 and 2800 are perfect squares using prime factorisation.
1156: \(1156 = 2\times2\times17\times17 = (2\times17)^2\). The factors split evenly into \((2\times17)\times(2\times17)\), so 1156 is a perfect square: \(\sqrt{1156}=34\).
2800: \(2800 = 2\times2\times2\times2\times5\times5\times7 = 2^4 \times 5^2 \times 7\). The prime 7 appears only once, with no partner, so the factors cannot be split into two identical groups — 2800 is not a perfect square.
For larger numbers, we can estimate the square root by bracketing between known squares and narrowing the interval — e.g. for \(\sqrt{1936}\): since \(40^2=1600\) and \(50^2=2500\), we know \(40 \lt \sqrt{1936} \lt 50\). The last digit 6 tells us the root ends in 4 or 6. Testing \(45^2 = (40+5)^2 = 1600+400+25=2025 \gt 1936\), so \(40 \lt \sqrt{1936} \lt 45\), which points to 44. Checking, \(44^2=1936\) — confirmed.
Figure it Out — Squares
Nine questions from page 10–11 of the textbook, covering perfect square identification, patterns, and factorisation.
1Which of the following numbers are not perfect squares? (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
Perfect squares can only end in 0, 1, 4, 5, 6 or 9. 2032 ends in 2 (not possible), 2048 ends in 8 (not possible), 1027 ends in 7 (not possible), and 1089 ends in 9 (possible — and indeed \(1089=33^2\)).
2Which one among 64², 108², 292², 36² has last digit 4?
A square ends in 4 only when the original number ends in 2 or 8. 64 ends in 4 (64² ends in 6); 108 ends in 8 ✓; 292 ends in 2 ✓; 36 ends in 6 (36² ends in 6).
3Given 125² = 15625, what is the value of 126²?
\(126^2 - 125^2 = (126-125)(126+125) = 1 \times 251 = 251\). So \(126^2 = 15625 + 251\).
4Find the length of the side of a square whose area is 441 m².
\(441 = 3\times3\times7\times7 = (3\times7)^2 = 21^2\), so \(\sqrt{441}=21\).
5Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
LCM of 4, 9, 10: \(4=2^2\), \(9=3^2\), \(10=2\times5\). LCM \(=2^2\times3^2\times5=180\). Prime factorisation of 180 is \(2^2\times3^2\times5^1\) — the 5 appears to an odd power, so 180 itself is not a perfect square. Multiplying by 5 makes every prime's power even: \(180\times5=900=2^2\times3^2\times5^2=(2\times3\times5)^2=30^2\).
6Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
\(9408 = 2^6 \times 3 \times 7^2\) (since \(9408 = 2\times2\times2\times2\times2\times2\times3\times7\times7\)). The powers of 2 and 7 are already even, but 3 appears to an odd power (power 1). Multiplying by 3 makes every power even: \(9408\times3=28224=2^6\times3^2\times7^2=(2^3\times3\times7)^2=168^2\).
7How many numbers lie between the squares of the following numbers? (i) 16 and 17 (ii) 99 and 100
Using the earlier result, between \(n^2\) and \((n+1)^2\) there are \(2n\) numbers.
(i) Between 16² and 17²: \(2\times16=32\) numbers.
(ii) Between 99² and 100²: \(2\times99=198\) numbers.
8In the following pattern, fill in the missing numbers: 1²+2²+2²=3², 2²+3²+6²=7², 3²+4²+12²=13², 4²+5²+20²=(__)², 9²+10²+(__)²=(__)²
The pattern follows \(k^2+(k+1)^2+[k(k+1)]^2 = [k(k+1)+1]^2\). Checking: for \(k=4\), \(k(k+1)=20\) and \(k(k+1)+1=21\), so \(4^2+5^2+20^2=21^2\) — matching the given row. For \(k=9\), \(k(k+1)=90\) and \(k(k+1)+1=91\).
9How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
The figure is a 9 × 9 grid of alternating icons (81 cells total), alternating between a diamond pattern and a grid-of-tiny-squares pattern, starting and ending each odd-numbered row with a diamond. Counting the two types across all 81 cells gives 41 diamond cells and 40 grid cells. Each grid cell shown is itself made up of a 4 × 4 arrangement of tiny squares, i.e. 16 tiny squares per grid cell.
Total tiny squares \(= 40 \times 16 = 640\).
Prime factorisation: \(640 = 2^7 \times 5\).
1.2 Cubic Numbers
A cube of edge length n units is built from \(n \times n \times n = n^3\) unit cubes (each square layer has \(n\times n\) unit cubes, and there are n such layers). Numbers of the form \(n^3\) — 1, 8, 27, 64, 125, … — are called perfect cubes. In Sanskrit, this idea is called ghana.
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| n³ | 1 | 8 | 27 | 64 | 125 | 216 | 343 | 512 | 729 | 1000 |
Is 9 a cube? No — \(2^3=8\) and \(3^3=27\), so 9 (and every number from 10 to 26) falls strictly between two consecutive cubes and is not itself a perfect cube.
MTCan a cube end with exactly two zeroes (00)? Explain.
If a number ends in k zeros, it is divisible by \(10^k\) but not \(10^{k+1}\), so its cube is divisible by \(10^{3k}\) but not \(10^{3k+1}\) — meaning a cube's trailing zero count is always a multiple of 3 (0, 3, 6, 9, …). Two zeros is not a multiple of 3, so it's impossible.
Cubes of fractions and decimals follow the same rule: \(\left(\frac{4}{6}\right)^3=\frac{64}{216}\), \((13.08)^3=2237.810112\), and \((-6)^3=-216\) (a negative number cubed stays negative).
Try This — Taxicab Numbers
When Ramanujan was ill in hospital, Hardy mentioned he'd arrived in taxicab number 1729, calling it "rather a dull number." Ramanujan immediately replied that it was actually very interesting — the smallest number expressible as the sum of two cubes in two different ways: \(1729 = 1^3+12^3 = 9^3+10^3\). Such numbers are called taxicab numbers, and 1729 is the Hardy–Ramanujan number.
Find the two ways to write the next two taxicab numbers, 4104 and 13832, as sums of two positive cubes.
\(4104 = 2^3+16^3 = 8+4096 = 4104\), and also \(4104=9^3+15^3=729+3375=4104\).
\(13832 = 2^3+24^3=8+13824=13832\), and also \(13832=18^3+20^3=5832+8000=13832\).
Perfect Cubes and Consecutive Odd Numbers
Just as squares are sums of consecutive odd numbers, cubes are sums of consecutive odd numbers too — but a different count of them each time:
MTLater in this series we get: 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109. Can you tell what this sum is without doing the calculation?
Following the pattern, the sum of k consecutive odd numbers in this series equals \(k^3\). This row has 10 consecutive odd numbers, so it must equal \(10^3\).
Cube Roots
If \(y=x^3\), then x is the cube root of y, written \(x=\sqrt[3]{y}\). Unlike square roots, every real number has exactly one real cube root (no ± ambiguity), because a negative number cubed stays negative. A number is a perfect cube exactly when its prime factors split into three identical groups (equivalently, each prime factor's power is a multiple of 3).
MTFind the cube roots of these numbers: (i) ∛64 (ii) ∛512 (iii) ∛729
(i) \(64=4^3\), so \(\sqrt[3]{64}=4\). (ii) \(512=8^3\), so \(\sqrt[3]{512}=8\). (iii) \(729=9^3\), so \(\sqrt[3]{729}=9\).
MTSimilar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?
1-digit cubes: \(1^3=1\), \(2^3=8\) — 2 cubes. 2-digit cubes: \(3^3=27\), \(4^3=64\) — 2 cubes. 3-digit cubes: \(5^3=125\) through \(9^3=729\) — 5 cubes.
Prime factorisation confirms this: each prime factor of a number appears exactly three times in the prime factorisation of its cube. For instance, \(15=3\times5\), so \(15^3=3375=3^3\times5^3\); and \(12=2^2\times3\), so \(12^3=1728=2^3\times2^3\times3^3\) (i.e. \(2^6\times3^3\)).
MTCompute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?
Cubes: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.
Level 1 (1st differences): 7, 19, 37, 61, 91, 127, 169, 217, 271.
Level 2 (2nd differences): 12, 18, 24, 30, 36, 42, 48, 54.
Level 3 (3rd differences): 6, 6, 6, 6, 6, 6, 6.
Perfect squares needed two levels of successive differences to become constant; perfect cubes need three levels — the differences become constant one level later than for squares, matching the fact that cubes are a "degree higher" than squares.
Figure it Out — Cubes
Five questions from page 16–17 of the textbook, covering cube roots, cube identification, and comparisons.
1Find the cube roots of 27000 and 10648.
\(27000 = 27 \times 1000 = 3^3 \times 10^3 = (3\times10)^3 = 30^3\), so \(\sqrt[3]{27000}=30\).
\(10648 = 2^3 \times 11^3 = (2\times11)^3=22^3\) (check via prime factorisation: \(10648 = 2\times2\times2\times11\times11\times11\)), so \(\sqrt[3]{10648}=22\).
2What number will you multiply by 1323 to make it a cube number?
\(1323 = 3^3 \times 7^2\) (since \(1323 = 3\times3\times3\times7\times7\)). The power of 3 is already a multiple of 3, but 7 has power 2, needing one more factor of 7 to reach power 3.
3State true or false. Explain your reasoning. (i) The cube of any odd number is even. (ii) There is no perfect cube that ends with 8. (iii) The cube of a 2-digit number may be a 3-digit number. (iv) The cube of a 2-digit number may have seven or more digits. (v) Cube numbers have an odd number of factors.
(i) False. An odd number times itself three times (odd × odd × odd) is always odd — e.g. \(3^3=27\) is odd.
(ii) False. \(2^3=8\) and \(12^3=1728\) both end in 8, so perfect cubes ending in 8 certainly exist.
(iii) False. The smallest 2-digit number is 10, and \(10^3=1000\), which already has 4 digits. So every 2-digit number's cube has at least 4 digits, never just 3.
(iv) False. The largest 2-digit number is 99, and \(99^3=970299\), which has only 6 digits — well short of seven.
(v) False. Having an odd number of factors is a special property of perfect squares, not cubes. For instance, \(8=2^3\) has factors 1, 2, 4, 8 — that's 4 factors, an even count.
4You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.
Shortcut: split the number into its last 3 digits and the rest. The last digit of the "last 3 digits" block tells you the units digit of the cube root (via the pairing 1↔1, 8↔2, 7↔3, 4↔4, 5↔5, 6↔6, 3↔7, 2↔8, 9↔9, 0↔0). The remaining leading digits, bracketed between two consecutive cubes, give the tens digit.
1331: last 3 digits "331" end in 1 → units digit 1. Remaining "1": \(1^3=1\), so tens digit 1. Root = 11.
4913: "913" ends in 3 → units digit 7. Remaining "4": \(1^3=1 \le 4 \lt 2^3=8\), so tens digit 1. Root = 17.
12167: "167" ends in 7 → units digit 3. Remaining "12": \(2^3=8 \le 12 \lt 3^3=27\), so tens digit 2. Root = 23.
32768: "768" ends in 8 → units digit 2. Remaining "32": \(3^3=27 \le 32 \lt 4^3=64\), so tens digit 3. Root = 32.
5Which of the following is the greatest? Explain your reasoning. (i) 67³ − 66³ (ii) 43³ − 42³ (iii) 67² − 66² (iv) 43² − 42²
Using \(a^3-b^3=(a-b)(a^2+ab+b^2)\) and \(a^2-b^2=(a-b)(a+b)\), with \(a-b=1\) in every case:
(i) \(67^3-66^3 = 67^2+67\times66+66^2 = 4489+4422+4356 = 13267\)
(ii) \(43^3-42^3 = 1849+1806+1764 = 5419\)
(iii) \(67^2-66^2 = 67+66 = 133\)
(iv) \(43^2-42^2 = 43+42 = 85\)
1.3 A Pinch of History
The earliest known lists of perfect squares and perfect cubes were compiled by the Babylonians as far back as 1700 BCE on clay tablets, used to quickly find square roots and cube roots for land measurement and architectural design.
In ancient Sanskrit texts (from at least the third century BCE), the term varga meant both the square figure/its area and the "square power," while ghana meant both the solid cube and the product of a number with itself three times. The fourth power was called varga-varga. Āryabhaṭa (499 CE) states: "A square figure of four equal sides and the number representing its area are called varga. The product of two equal quantities is also called varga."
The word "root" (as in square root, cube root) traces back to the Sanskrit mūla, meaning root of a plant, basis, cause, or origin — used in India for these operations since at least the first century BCE. Varga-mūla meant square root and ghana-mūla meant cube root; these were later carried into Arabic (jidhr) and Latin (radix), both also meaning "root of a plant." Another Sanskrit term, pada (foot, basis, cause), was used similarly — Brahmagupta (628 CE) explains: "The pada (root) of a kṛti (square) is that of which it is a square."
Summary
- A number obtained by multiplying a number by itself is called a square number. Squares of natural numbers are called perfect squares.
- All perfect squares end with 0, 1, 4, 5, 6 or 9. Squares can only have an even number of zeros at the end.
- Square root is the inverse operation of square. Every perfect square has two integral square roots. The positive square root of a number is denoted by the symbol √. For example, √9 = 3.
- A number obtained by multiplying a number by itself three times is called a cube. For example 1, 8, 27, … etc., are cubes.
- A number is a perfect square if its prime factors can be split into two identical groups.
- A number is a perfect cube if its prime factors can be split into three identical groups.
- The symbol ∛ denotes cube root. For example, ∛27 = 3.
Puzzle Time — Square Pairs!
The Challenge
The numbers 3, 6, 10, 15, 1 are arranged so that every adjacent pair sums to a perfect square: \(3+6=9\), \(6+10=16\), \(10+15=25\), \(15+1=16\).
Part 1: Arrange the numbers 1 to 17 (without repetition) in a row so every adjacent pair sums to a perfect square. Can you find more than one arrangement? If not, explain why.
Part 2: Can you do the same with 1 to 32, arranged in a circle this time (so the first and last numbers must also sum to a square)?
One valid row arrangement for 1–17 (check each adjacent sum against the nearby perfect squares 4, 9, 16, 25, 36):
Check a few links: \(17+8=25=5^2\), \(8+1=9=3^2\), \(1+15=16=4^2\), \(15+10=25=5^2\), \(10+6=16=4^2\), \(6+3=9=3^2\), …, and finally \(2+7=9=3^2\), \(7+9=16=4^2\), \(9+16=25=5^2\) — every consecutive pair here sums to 9, 16, or 25. (Rearrangements of this type are not unique — this is one valid solution among several; try building your own using the graph of which numbers can be neighbours of which.)
The circular version with numbers 1 to 32 is a genuinely harder combinatorial challenge — it asks you to find a full "cycle" through all 32 numbers where every step (including wrapping from the last number back to the first) lands on a perfect square sum. Approach it by first listing, for each number from 1 to 32, every other number in the range it can be paired with to form a square sum, then hunt for a closed loop that visits every number exactly once (a Hamiltonian cycle) — squared paper and a lot of patient trial-and-error (or a systematic graph-search) is the way in.
Frequently Asked Questions
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Move on to Power Play — exponents, laws of powers, and scientific notation.