If possible, exhibit and integral basis for whose elements are conjugate for .
Let be a squarefree integer, and let if mod 4 and if mod 4. Recall that the integers in have the form where . Let if mod 4 and if mod 4. Note that if mod 4 and if mod 4 (so that ), that if mod 4 and if mod 4, and that if mod 4 and if mod 4.
Note that . In particular, if then is the minimal polynomial of , so that the conjugates of are itself and .
Recall that the discriminant of is if mod 4 and if mod 4, and that any integral basis will have this discriminant.
Suppose now that is an integral basis of , with and . (Certainly if then cannot be a basis, much less an integral basis; if , then (for example) we cannot generate 1.) Note that . Then we must have
Note that if mod 4 and if mod 4. Evidently, this determinant is , and so we have .
If mod 4, then , and we have . This equation has no solutions in , and thus does not have an integral basis whose elements are conjugate.
If mod 4, then , and we have , so that . This gives a necessary condition on and if we wish to be an integral basis.
Recall that is an integral basis of . If is another integral basis, then there is a (unique) invertible matrix such that . Say , , and ; comparing coefficients, and recalling that (since mod 4), we have . Computing the determinant, we have . Combined with our necessary condition, . If , then , and if , then . Thus our only candidates for are and . Or, in terms of , and . It is easy to see that both and are integral bases.
To summarize, if mod 4, then is an integral basis whose elements are conjugate, where . If mod 4, then no such integral basis exists.