Let $\sigma(x)$ denote the sum of the divisors of the positive integer $x$. That is, $$\sigma(x) = \sum_{d \hspace{0.01in} \mid \hspace{0.01in} x}{d}.$$
For example, the divisors of $28$ are $$1, 2, 4, 7, 14, \text{and} \hspace{0.05in} 28,$$ and $\sigma(28)=1+2+4+7+14+28=56$.
We now state the:
Odd Perfect Number Problem
There is no odd (positive) integer $N$ satisfying $\sigma(N)=2N$.
Given that no one has found an odd perfect number (OPN), and given that nobody (as of yet) has proved that there is indeed none, here is my question:
(1) Is the OPN Problem undecidable?
(2) What is the main obstruction to proving that the OPN Problem is undecidable?
Possible (Philosophical) Answer to Question (2) Well, one main obstruction would of course be if the OPN Problem is, in fact, decidable. But how would one then prove that the problem is indeed decidable? How would one even start?
It is potentially undecidable. We don't know.
Note, if it is undecidable, then it is "intuitively" true. If there is an odd perfect number, we'd expect to be able to find it and write a proof that it is odd and perfect. That is our intuition about the natural numbers, and it would be an intuition not encoded in the Peano postulates.
I like to think of undecidability in this case as showing a limit of what induction can do.
Induction, intuitively, allows you to write an outline for an infinite proof. Once you have an induction proof for $P(n)$, you can write proofs for $P(10)$, $P(1000)$, or $P(100000)$ which do not use induction, but just standard deductions.
If there is no odd perfect number (in the "intuitive" natural numbers) then there would be an infinite proof of this fact - you'd just prove in each case that $\sigma(1)\neq 2,\sigma(3)\neq 6,\sigma(5)\neq 10,\dots.$ But if the question is undecidable, what this means is that the limited sort of infinite proof afforded you by induction is not enough to get the same result.
A similar statement can be made about Goldbach's conjecture - if it is undecidable, then it is intuitively true.
What this means is that induction is a fairly blunt instrument, but it is known that any such reasonable axiom scheme is still too blunt too cover all the theorems that might have intuitive infinite proofs.
Now, if there is a statement $P(n)$ of one variable such that we can write a decision process for $P(n)$ for each $n$ (such as this case or Goldbach,) then, if $\forall n: P(n)$ is undecidable, we will never be able to prove it undecidable. It turns out, if we can prove there is no counter-example, then we can prove the theorem is true, so the theorem would not be undecidable.