Question 1

We need to arrange the numbers from 1 to 17 in a row so that every adjacent pair adds up to a perfect square.

Step 1 — Identify possible square sums

First, let us list the perfect squares that can be formed by adding two numbers from 1 to 17. A perfect square is a number obtained by multiplying an integer by itself. The smallest possible sum is $1 + 2 = 3$. The largest possible sum is $16 + 17 = 33$. So, we look for perfect squares between 3 and 33.

$2^2 = 4$ $3^2 = 9$ $4^2 = 16$ $5^2 = 25$

$\boxed{\text{Possible square sums are 4, 9, 16, 25.}}$

Step 2 — List all valid pairs

Now, let us find all pairs of numbers from 1 to 17 that add up to one of these square sums. We will list each number and its possible neighbours.

For sum = 4: (1,3) For sum = 9: (1,8), (2,7), (3,6), (4,5) For sum = 16: (1,15), (2,14), (3,13), (4,12), (5,11), (6,10), (7,9) For sum = 25: (8,17), (9,16), (10,15), (11,14), (12,13)

Step 3 — Map connections for each number

Let us list each number from 1 to 17 and all the numbers it can connect to. We also count how many connections each number has. This is called its degree. Numbers with only one connection must be at the ends of our row.

1: (3, 8, 15) - 3 connections 2: (7, 14) - 2 connections 3: (1, 6, 13) - 3 connections 4: (5, 12) - 2 connections 5: (4, 11) - 2 connections 6: (3, 10) - 2 connections 7: (2, 9) - 2 connections 8: (1, 17) - 2 connections 9: (7, 16) - 2 connections 10: (6, 15) - 2 connections 11: (5, 14) - 2 connections 12: (4, 13) - 2 connections 13: (3, 12) - 2 connections 14: (2, 11) - 2 connections 15: (1, 10) - 2 connections 16: (9) - 1 connection 17: (8) - 1 connection

The numbers 16 and 17 have only one connection each. This means they must be the two ends of our arrangement.

Step 4 — Find the arrangement

We must start our arrangement with either 16 or 17. Let us start with 16.

1. Start with 16. Its only neighbour is 9. Arrangement so far: 16 - 9

2. From 9, its neighbours are 7 and 16. Since 16 is already used, we must go to 7. Arrangement: 16 - 9 - 7

3. From 7, its neighbours are 2 and 9. Since 9 is used, we must go to 2. Arrangement: 16 - 9 - 7 - 2

4. From 2, its neighbours are 7 and 14. Since 7 is used, we must go to 14. Arrangement: 16 - 9 - 7 - 2 - 14

5. From 14, its neighbours are 2 and 11. Since 2 is used, we must go to 11. Arrangement: 16 - 9 - 7 - 2 - 14 - 11

6. From 11, its neighbours are 5 and 14. Since 14 is used, we must go to 5. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5

7. From 5, its neighbours are 4 and 11. Since 11 is used, we must go to 4. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4

8. From 4, its neighbours are 5 and 12. Since 5 is used, we must go to 12. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12

9. From 12, its neighbours are 4 and 13. Since 4 is used, we must go to 13. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13

10. From 13, its neighbours are 3 and 12. Since 12 is used, we must go to 3. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3

11. From 3, its neighbours are 1, 6, and 13. Since 13 is used, we have two choices: 1 or 6. Let us try going to 1 first. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 1

12. From 1, its neighbours are 3, 8, and 15. Since 3 is used, we have two choices: 8 or 15. Let us try going to 8. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 1 - 8

13. From 8, its neighbours are 1 and 17. Since 1 is used, we must go to 17. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 1 - 8 - 17 This arrangement has 14 numbers. We need 17 numbers. This path is too short and does not use all numbers. So, this choice (going to 8 from 1) is a dead end.

    Let us backtrack to step 12 and try going to 15 from 1. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 1 - 15

14. From 15, its neighbours are 1 and 10. Since 1 is used, we must go to 10. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 1 - 15 - 10

15. From 10, its neighbours are 6 and 15. Since 15 is used, we must go to 6. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 1 - 15 - 10 - 6

16. From 6, its neighbours are 3 and 10. Both 3 and 10 are already used in the arrangement. This means we cannot extend the path from 6. This choice (going to 15 from 1) is also a dead end.

    Let us backtrack to step 11 and try going to 6 from 3. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 6

17. From 6, its neighbours are 3 and 10. Since 3 is used, we must go to 10. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 6 - 10

18. From 10, its neighbours are 6 and 15. Since 6 is used, we must go to 15. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 6 - 10 - 15

19. From 15, its neighbours are 1 and 10. Since 10 is used, we must go to 1. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 6 - 10 - 15 - 1

20. From 1, its neighbours are 3, 8, and 15. Since 3 and 15 are used, we must go to 8. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 6 - 10 - 15 - 1 - 8

21. From 8, its neighbours are 1 and 17. Since 1 is used, we must go to 17. Arrangement: 16 - 9 - 7 - 2 - 14 - 11 - 5 - 4 - 12 - 13 - 3 - 6 - 10 - 15 - 1 - 8 - 17 This arrangement has 17 numbers. All numbers from 1 to 17 are used exactly once. This is a valid arrangement.

$\boxed{\text{One arrangement is 16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17.}}$

Step 5 — Check for other arrangements

We systematically explored all possible paths starting from 16. The only point where we had a choice was at number 3. Both choices (going to 1 or 6) were explored. The path through 1 led to dead ends because we could not use all numbers or got stuck. The path through 6 led to a complete arrangement using all 17 numbers and ending at 17. Therefore, there is only one unique sequence (ignoring the reverse order). The reverse order of the arrangement is also valid: 17, 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16. These two sequences are considered the same arrangement.

Answer

There is only one unique way to arrange the numbers 1 to 17 in a row such that the sum of every adjacent pair of numbers is a square. The arrangement is: 16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17. (The reverse of this sequence is considered the same arrangement.) We found this by systematically checking all possible connections, starting from an endpoint (16) and exploring all branches. All other branches led to incomplete sequences or dead ends.