Finding 5 Positive Integers with GCD Difference

[Mr Davis 97]

Homework Statement

Do five positive integers exist such that the positive difference between any two is the greatest common divisor of those two numbers?

Homework Equations

The Attempt at a Solution

I found four such numbers, $\{6,8,9,12\}$. I did this in an ad hoc way though without any real method. What ways can I attack this problem, given that I've tried to find 5 numbers without much progress?

 

[fresh_42]

I don't have a proof, but my guess is, that it is not possible.

Just an idea:
Let $x,y$ be such two numbers. The requirement is then $x\cdot \mathbb{Z}+y \cdot \mathbb{Z} = (x-y)\cdot \mathbb{Z}$ or that there are integers $\alpha,\beta$ such that $x-y=\alpha x +\beta y$, i.e. $y=\dfrac{1-\alpha}{1+\beta}x$. Hence we have found a pair $(x,y)$ if we would have found a pair $(\alpha,\beta)$ of integers, such that the quotient $\dfrac{1-\alpha}{1+\beta} \in \mathbb{Z}$. My idea is to translate the problem into a geometric one. We need five numbers $\alpha_i$ such that $q_{ij} := \dfrac{1-\alpha_i}{1+\alpha_j} \in \mathbb{Z}$. If we can identify each number with a point on the lattice $\mathbb{Z}^2$ and draw an edge, if $q_{ij} \in \mathbb{Z}$, then we get a pentagram. Now my idea is, that there can't be such a pentagram in $\mathbb{Z}^2.$ Or we set the pairs $(\alpha_i,\alpha_j)$ as points and show that there cannot be ten of them. Maybe it's also possible to show that given four such numbers, i.e. six pairs, that there cannot be added a fifth integer one by algebraic methods (calculations).

 

[haruspex]

I believe I found a solution using five numbers in the range [2³.3.5.7.11-24, 2³.3.5.7.11].

To arrive at this, I considered how many of the numbers can be odd, how many 2 mod 4, how many 1 mod 3, etc.

 

[haruspex]

Given this is not my thread or homework assignment, could you please explain the notation of your solution?
I'm having difficulty understanding the list within the square brackets. Thanks.

I mean the range [x-24,x], i.e. from x-24 to x inclusive, where x=2³.3.5.7.11.

 

[The Bill]

I mean the range [x-24,x], i.e. from x-24 to x inclusive, where x=2³.3.5.7.11.

Are you using periods to denote multiplication? If so, I think that might be the notation Klystron was asking about.

 

[haruspex]

Are you using periods to denote multiplication? If so, I think that might be the notation Klystron was asking about.

Yes. Isn't that standard in number theory?

 

[haruspex]

Found a much simpler solution by basing it around 2³ and 3².
Using the approach I indicated in post #3, it looks good to start with three consecutive numbers, the middle one being an odd multiple of 3. The other numbers must all be divisible by 4, and at least one divisible by 8.
So we can make a guess like 12, 14, 15, 16, 24.
Clearly 24-14 isn't working, and neither is 24-15. We can add any multiple of 24 to all the numbers without disturbing any of the other relationships. To fix 24-14 we need to make those two multiples of 10, so add 96:
108, 110, 111, 112, 120.
120-111 is still not right. We need to make those divisible by 9 by adding a suitable multiple of 120:
348, 350, 351, 352, 360.