Showing S = ℤ Using 13 & 1000 in Subring

[Bachelier]

Let S be a subring of the ring of integers ℤ. Show that if 13 ∈ S and 1000 ∈ S, then S equals ℤ.

I'm thinking we should construct a bijection using both numbers. Any ideas? thanks

 

[Bachelier]

I see that ∃ no takers. I know the problem looks weird but it is a problem I read in an old paper.

The only thing I can think of is that we need to show that 1 ∈ S. To do this I can use 12 ∈ S because 12 = 1000 - 13*76, therefore 13 - 12 ∈ S and hence S has an identity and therefore we can extend to Z.

 

[lavinia]

Let S be a subring of the ring of integers ℤ. Show that if 13 ∈ S and 1000 ∈ S, then S equals ℤ.

I'm thinking we should construct a bijection using both numbers. Any ideas? thanks

13 and 1000 are relatively prime so 1 is in the subring that they generate.

 

[Bachelier]

Thank you Lavinia. You're becoming my favorite Science adviser. :)