[Solved]: Why is FACTOR in Co-NP?

Problem Detail: 

I'm having trouble wrapping my head around the problems PRIME, COMPOSITE, FACTOR and how they're related in terms of complexity. I understand that PRIME has been shown to be in $P$ by the AKS primality test, and I believe this works for COMPOSITE as well.

As for FACTOR,

$$FACTOR = \{(m,r) :\;\; \exists s \text{ such that} 1<s<r \text{ and } s \text{ divides } m\} $$

from what I have read it seems that it is in $NP \cap Co-NP$. I see that it is in $NP$ since a certificate would consist of a prime divisor of $m$ less than $r$. But what kind of certificate can establish that there is no such prime divisor (in polynomial time)?

Asked By : Fequish

Answered By : Yuval Filmus

A certificate for there being no non-trivial divisor of $m$ smaller than $r$ is the factorization of $m$. We can check in polynomial time that all factors are indeed prime (since primality is in P by AKS primality test), that their product is $m$, and that all of them are at least $r$.

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Question Source : http://cs.stackexchange.com/questions/52722

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