Is the complement of this decision problem in \$P\$?

Are there any two primes that are NOT a factor of $$M$$ that multiply up to $$M$$?

Fact: Any two primes that multiply up to $$M$$. Must be factors of $$M$$!

Thus because of the fact above an $$O(1)$$ algorithm exists. It always outputs $$NO$$

Complement

Are there any two primes that are a factor of $$M$$ that multiply up to $$M$$?

Fact: A complement of a decision problem does not always require to always return $$YES$$ or $$NO$$. It can be either one!

(eg. $$M$$ = 6 and two primes that multiply up to $$M$$ are $$3$$,$$2$$.)

Well, this I find interesting this is deciding $$Semi-Primes$$.

Question

Shouldn’t $$Semi-Primes$$ be in $$P$$, because what was shown above?