Because of the odd nature of my question, I am interested in partial solutions if a complete solution is not known. Specifically, the question I am trying to answer is the following:

Is there a method to show that a number does not have precisely two prime factors if we know it has at most three?

More generally, any condition uniquely satisfied by integers with an odd number of prime factors, or with an even number of prime factors, but not numbers with both, will be extremely helpful.

Edit: As per helpful comments, I am assuming the integer is entirely square free; that is, my question refers to distinct prime factors. (If there is an answer or partial answer that does not refer to specifically distinct prime factors, however, it would still be appreciated).

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    $\begingroup$ A related problem is to compute the value of the Mobius function. You can see links at here and maybe here if you're interested. $\endgroup$ – Ben Jan 11 at 23:11
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    $\begingroup$ It would be interesting to know if you consider $p^2$ to have 1 or 2 prime factors, for this problem. $\endgroup$ – Ingix Jan 11 at 23:14
  • $\begingroup$ To determine the number of prime factors is not significantly easier than integer factorization (even if we assume we know that the number is squarefree). If you can find a non-trivial factor, you can much more efficient distinguish between two or three prime factors because in this case you only need a primality test. $\endgroup$ – Peter Jan 12 at 12:45

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