TITLE: Can we decide a conjecture is decidable without knowing a conjecture is correct or false?
QUESTION [6 upvotes]: Can we decide a conjecture is decidable without knowing a conjecture is correct or false?
I asked this question because I assume that the millenium prize problem is already to be decidable, otherwise the mathematician would need to consider infinity numbers of cases in order to get the money or the question's logic makes it a fake question.

REPLY [2 votes]: Here's another perspective on a related question: a problem where we can prove that an 'answer' in some sense exists but we can also prove that we can't find the answer.
A graph is a set of vertices and a set of edges between those vertices - for instance, the vertices might be 'Hollywood actors' with an edge between any two actors who have appeared in a movie together, or the vertices might be 'US cities with more than 100,000 people', with an edge between any two cities within 500 miles of each other.
One graph is a minor of another if, essentially, you can tag some of the vertices in the larger graph matching the vertices of the smaller graph and 'draw edges' between them matching the edges of the smaller graph - in essence, it's like a subgraph (though there are obviously some minor technical details).
And finally, we say a collection of graphs is minor-closed if every graph that's in the collection also brings along all of its minors.  For instance, the set of planar graphs (the ones that can be drawn in the plane without any of their edges crossing) is minor-closed this way; any subgraph, or any contraction of a subgraph, is still planar.
Now, the Robinson-Seymour theorem says that any minor-closed set of graphs can be defined by a small set of graphs that aren't in it, the so-called 'forbidden minors'; any graph that has one of these forbidden minors as a minor isn't in the collection, and all graphs that don't have any of the forbidden minors are in the collection.  For instance, for the planar graphs there are two forbidden minors: $K_5$ (a graph with five vertices, and edges between every pair of vertices) and $K_{3,3}$ (a graph with three 'red' nodes and three 'blue' nodes, and every possible edge from one color to the other).
So, we know that for any collection of graphs we give, there's some set of forbidden minors defining those graphs; in this sense we have a 'yes' answer to the problem.  But that doesn't mean that we can actually find the set of forbidden minors; in fact, even though that set is known to exist, it's also known that there can't be any procedure for finding them; any algorithm for taking (a description of) a set of graphs and finding its forbidden minors could also solve the halting problem.