Maths Olympiad Prep

Track / Stage 6 / 224 of 400 #1224 of 1964

Problem 1224

National Olympiad, first round
Number theory Difficulty 6.3 Prove it

Suggestion: Find all possible neighbors of the number 16. (a) Show that the numbers from 1 to 16 can be written in a line, such that the sum of any two adjacent numbers is a perfect square.

(b) Show that the numbers from 1 to 16 cannot be written around a circle, such that the sum of any two adjacent numbers is a perfect square.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

The key observation that helps solve (a) and resolves (b) is to look for the possible neighbors for the number 16.

A neighbor of 16 is a number that, when added to 16, results in a perfect square. One candidate is the number 9, since 16+9=5216+9=5^{2}.

There are no others, because the next perfect square after 25 is 36, and the largest sum we can obtain from two numbers between 1 and 16 is 15+16=3115+16=31.

(a) Since 16 has only one possible neighbor, it must be at an end. Starting with 16, we obtain the solution below.

1697214115412133610151816-9-7-2-14-11-5-4-12-13-3-6-10-15-1-8

(b) For it to be possible to place all the numbers from 1 to 16 around a circle, every number would need to have two neighbors. But the only possible neighbor for 16 is 9, making it impossible to construct a circular arrangement.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.