Context: Look at the following numbers: . They are arranged such that each pair of adjacent numbers adds up to a square: , , , . Q. Can you do the same with numbers from to (again, without repetition), but this time arranging all the numbers in a circle?

Rational Numbers | A

Question 2

Context: Look at the following numbers: $3, 6, 10, 15, 1$ . They are arranged such that each pair of adjacent numbers adds up to a square: $3 + 6 = 9$ , $6 + 10 = 16$ , $10 + 15 = 25$ , $15 + 1 = 16$ .

Q. Can you do the same with numbers from $1$ to $32$ (again, without repetition), but this time arranging all the numbers in a circle?

Solution

Understand the Question

We need to arrange all numbers from $1$ to $32$ (each used exactly once) in a circle.
Every pair of adjacent numbers in the circle must add up to a perfect square.
Since the arrangement is circular, the first and last numbers in the sequence must also add up to a perfect square.

Step 1 · Identify Possible Perfect Square Sums

The minimum sum of two distinct numbers is $1 + 2 = 3$ . The maximum sum of two distinct numbers is $31 + 32 = 63$ .

The perfect squares between $3$ and $63$ are:

\begin{aligned} 2^2 &= 4 \\ 3^2 &= 9 \\ 4^2 &= 16 \\ 5^2 &= 25 \\ 6^2 &= 36 \\ 7^2 &= 49 \end{aligned}

Thus, each adjacent pair must sum to one of: $4, 9, 16, 25, 36, \text{ or } 49$ .

Step 2 · Construct the Circular Arrangement

By analyzing the possible square-sum connections for each number from $1$ to $32$ , we can form a continuous loop. Question diagram

A valid circular sequence of the $32$ numbers is:

1 - 8 - 28 - 21 - 4 - 32 - 17 - 19 - 30 - 6 - 3 - 13 - 12 - 24 - 25 - 11 - 5 - 31 - 18 - 7 - 29 - 20 - 16 - 9 - 27 - 22 - 14 - 2 - 23 - 26 - 10 - 15 - (1)

Verifying adjacent pair sums:

\begin{aligned} 1 + 8 &= 9 = 3^2, & 8 + 28 &= 36 = 6^2, & 28 + 21 &= 49 = 7^2, & 21 + 4 &= 25 = 5^2 \\ 4 + 32 &= 36 = 6^2, & 32 + 17 &= 49 = 7^2, & 17 + 19 &= 36 = 6^2, & 19 + 30 &= 49 = 7^2 \\ 30 + 6 &= 36 = 6^2, & 6 + 3 &= 9 = 3^2, & 3 + 13 &= 16 = 4^2, & 13 + 12 &= 25 = 5^2 \\ 12 + 24 &= 36 = 6^2, & 24 + 25 &= 49 = 7^2, & 25 + 11 &= 36 = 6^2, & 11 + 5 &= 16 = 4^2 \\ 5 + 31 &= 36 = 6^2, & 31 + 18 &= 49 = 7^2, & 18 + 7 &= 25 = 5^2, & 7 + 29 &= 36 = 6^2 \\ 29 + 20 &= 49 = 7^2, & 20 + 16 &= 36 = 6^2, & 16 + 9 &= 25 = 5^2, & 9 + 27 &= 36 = 6^2 \\ 27 + 22 &= 49 = 7^2, & 22 + 14 &= 36 = 6^2, & 14 + 2 &= 16 = 4^2, & 2 + 23 &= 25 = 5^2 \\ 23 + 26 &= 49 = 7^2, & 26 + 10 &= 36 = 6^2, & 10 + 15 &= 25 = 5^2, & 15 + 1 &= 16 = 4^2 \end{aligned}

Answer

Yes, they can be arranged in a circle as:

$1, 8, 28, 21, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 5, 31, 18, 7, 29, 20, 16, 9, 27, 22, 14, 2, 23, 26, 10, 15$

Common Mistakes

Forgetting the Circular Condition: Ensure the last number ( $15$ ) and the first number ( $1$ ) also add up to a square ( $15 + 1 = 16$ ).
Duplicate or Missing Numbers: Each integer from $1$ to $32$ must appear exactly once.
Ignoring Constrained Numbers: Numbers with only two valid square partners (such as $16, 18, 25, 26, 27, 28, 29, 30, 31, 32$ ) must be placed adjacent to their exact paired neighbors.

Question 2

More questions in A