Call admissible a set of integers that has the following property:
If (possibly ) then for every integer .
Determine all pairs of nonzero integers such that the only admissible set containing both and is the set of all integers.
[i]
Call admissible a set of integers that has the following property:
If (possibly ) then for every integer .
Determine all pairs of nonzero integers such that the only admissible set containing both and is the set of all integers.
[i]
To solve the problem, we aim to determine all pairs of nonzero integers such that the only admissible set containing both and is the set of all integers. According to the problem statement, a set of integers is admissible if whenever and are in , is also in for every integer .
### Step-by-Step Analysis
1. Definition of Admissible Set
Given the definition, for any integers , the expression must also be in for any integer . Notably, choosing specific values for yields several important cases:
- When , this yields .
- When , we obtain .
2. Exploring Consequences
We compute some values to understand the closure of under these conditions:
- Starting with elements and in :
- Using the condition , both and must be in .
- Utilizing , we derive:
- If we choose such that the expression includes forms like Euclidean algorithms, this could result in generating 1 if and are coprime:
- Particularly, repeated applications will eventually include elements such as the greatest common divisor of and .
3. Condition for Admissibility
The minimal condition for a set containing and to be closed under these operations is . This means:
- With , elliptic stepping continually reduces combinations of down to .
- Hence, this process can eventually generate any integer, showing must be the set of all integers.
4. Conclusion
The problem therefore reduces to determining when any elements and can generate the full set of integers. This happens precisely when:
Thus, the set of pairs such that the only admissible set containing both and is the set of all integers is exactly those pairs for which . Consequently, the answer is: