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NCERT Solutions Ganita Prakash (Part 1) Chapter 1 Square Pairs! — Puzzle Time — Square Pairs!

Book page 18 Updated on2026-09-05

Q1.
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. Try arranging the numbers 1 to 17 (without repetition) in a row in a similar way — the sum of every adjacent pair of numbers should be a square.
Answer

Here is an arrangement that works:

16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17

Every adjacent pair sums to 9, 16 or 25:

16+9 = 259+7 = 167+2 = 92+14 = 1614+11 = 2511+5 = 16
5+4 = 94+12 = 1612+13 = 2513+3 = 163+6 = 96+10 = 16
10+15 = 2515+1 = 161+8 = 98+17 = 25
How to find it rather than stumble on it: Two numbers from 1 to 17 add to at least 3 and at most 33, so the only possible square sums are 9, 16 and 25. For each number, list its allowed neighbours — 1 goes with 8 and 15; 2 goes with 7 and 14; 16 goes with 9 only; 17 goes with 8 only. Then join the numbers up like a chain, starting from the most restricted ones.
Q2.
Can you arrange them in more than one way? If not, can you explain why?
Answer

No — apart from reading the same row backwards, the arrangement is the only one.

16 + p is a square only for p = 9 (16 + 9 = 25)
17 + p is a square only for p = 8 (17 + 8 = 25)
Why that forces everything: A number with only one allowed neighbour cannot sit in the middle of the row, since a middle position needs two neighbours. So 16 and 17 must be the two ends. That fixes the start as 16, 9, … and the finish as …, 8, 17. From there each step is forced too: after 9 the untried partners are 7 (16) and 16 (25), but 16 is already used, so 7 must follow — and the same reasoning at every step leaves exactly one chain. Reversing it gives the same row read the other way, not a genuinely new arrangement.
Tip: Check the possible partners of a few numbers: 4 pairs with 5, 12; 6 pairs with 3, 10; 13 pairs with 3, 12. Numbers with two choices sit inside the row; the two with one choice must be the ends.
Q3.
Can you do the same with numbers from 1 to 32 (again, without repetition), but this time arranging all the numbers in a circle?
Answer

Yes. Here is one circular arrangement of 1 to 32 in which every neighbouring pair adds to a square:

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 — and then back to 1

The sums going round are all 9, 16, 25, 36 or 49:

1+8 = 98+28 = 3628+21 = 4921+4 = 254+32 = 3632+17 = 49
17+19 = 3619+30 = 4930+6 = 366+3 = 93+13 = 1613+12 = 25
12+24 = 3624+25 = 4925+11 = 3611+5 = 165+31 = 3631+18 = 49
18+7 = 257+29 = 3629+20 = 4920+16 = 3616+9 = 259+27 = 36
27+22 = 4922+14 = 3614+2 = 162+23 = 2523+26 = 4926+10 = 36
10+15 = 2515+1 = 16
Why a circle is harder than a row: In a row the two ends need only one neighbour each, so a number with a single partner can be parked at an end. In a circle every number needs two neighbours, so no number may have only one allowed partner. That is why 1 to 17 cannot be closed into a circle — 16 and 17 have one partner each — while 32 is large enough for every number to have at least two.
Try This: Check that a circle is impossible for 1 to 31: 31 pairs only with 5 and 18, which is fine, but 30 pairs only with 6 and 19 — trace further and see where the chain is forced to break. Circular arrangements exist for 32 and for every number above it.
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