The single question is “What would the other say if I asked him if he is a liar?”

This is easy because it is only a matter of logic, but I suspect that Daniel intent is more mischievous, to highlight the ambiguity of natural language use for stating logic/math problems.
Because everything depends on how you define “always disagree” and this cannot be elucidated from the problem statement alone.

I go with Anthony. Although some may say the question “are you a Falsian” is another problem, it fits this problem well: both should answer “no” and we know here that they don’t always disagree. Hence Byp #2 must be the liar.

When asked : “Are you a Falsian ?”, they should both answer “No”. Therefore, the two Bypolarians don’t always disagree.
Therefore, the second Bypolarians is a Falsian, and you should not worry.

Now you’re warmed up, you can try the Hardest Logic Puzzle Ever. (Watch out for the spoilers further down on the same page).

Case 1, Both are Truans or both are Falsians: They can not meet with different greetings. Conflict!

Case 2, FB is Truan and SB is Falsian: SB is telling the truth at the end of the conversation but he should’t. So this can not be the case.

Case 3, FB is Falsian and SB is Truan: All statements seem convenient, so we should be worried.

I do not remember but there was a similar puzzle where the person should find a single question that will allow him to identify which is Truan and which is Falsian.

Stmt 3 False True
Stmt 4 True ****

It cannot be that Stmt 4 by Second Byp be true if Falsians ALWAYS lie, therefore the First Byp is the Falsian => worried.

What’s most interesting is that such a simple problem appear puzzling, there must be some step in our deductions which goes against our natural tendencies, r.e. Johnson Laird’s Mental Models.