Constructing Sets of Positive Integers Where GCD Equals Difference

It is indeed possible to find a set of n positive integers such that for any two elements in the set, their greatest common divisor (GCD) is equal to their difference. This article will explore the construction of such sets and the underlying mathematical principles involved.

Introduction

The problem can be approached by constructing a set with a common difference d. Let's explore the approach and its verification:

Construction of the Set

To construct a set ( S ) of n positive integers, we choose a positive integer ( d ). This ( d ) will serve as the common difference for the elements in the set. The set ( S ) is defined as:

S {d, 2d, 3d, ..., nd}

where ( n ) is the number of elements in the set.

Verification of the Construction

Let's verify that this construction meets the requirement for any two elements ( a kd ) and ( b md ) in the set, where ( k ) and ( m ) are distinct integers from 1 to ( n ):

The difference between ( a ) and ( b ) is:

a - b kd - md d(k - m)

The GCD of ( a ) and ( b ) is:

gcd(a, b) gcd(kd, md) d cdot gcd(k, m)

To satisfy the condition that the GCD equals the difference, we need:

d cdot gcd(k, m) d(k - m)

Dividing both sides by ( d ) (since ( d ) is positive and non-zero), we get:

gcd(k, m) k - m

This equality holds if ( k ) and ( m ) are consecutive integers such that ( gcd(k, m) 1 ) or ( k - m 1 ).

However, if we specifically choose ( d 1 ), then the set ( S {1, 2, 3, ..., n} ) ensures that for any two integers in the set, their GCD will equal their difference. Thus, it is indeed always possible to find such a set of integers.

Conclusion

A specific example for ( n 3 ) would be the set ( S {1, 2, 3} ).

General Case and Mathematical Induction

For each positive integer ( n ), it is possible to construct a set of ( n ) positive integers satisfying the given property. This can be demonstrated through mathematical induction.

For n 1, any set consisting of a single positive integer would vacuously satisfy the property since no pair of distinct elements could be obtained from that set.

For n 2, any set of the form ( S {a, 2a} ) would satisfy.

Assume S {a_1, a_2, ..., a_k} is a set satisfying the property for n k.

We now proceed to construct from S a set of k 1 distinct positive integers that would also hold true for the property.

Let t be the greatest element from S. Define b_1 a_1 - t! , b_2 a_2 - t! , ..., b_k a_k - t!, where t! denotes the product of all consecutive integers from 1 to t.

My claim is that the k-integer set {b_1, b_2, ..., b_k} satisfies the given property.

Choose an arbitrary pair of elements b_i, b_j from the set. Since a_i divides t!, it must divide b_i. Similarly, a_j divides b_j. Since the difference between a_i and a_j (denoted as d) is their greatest common divisor, d divides both b_i and b_j. Consequently, d must be the greatest common divisor of these two numbers since d equals their difference.

Let b_{k 1} t!. It is evident that b_i - b_{k 1} gives the greatest common divisor of b_i and b_{k 1} for any element b_i.

Therefore, the set {b_1, b_2, ..., b_k, b_{k 1}} satisfies the provided condition.

Mathematical induction then allows us to claim that such a construction is possible for any positive integer value of n.