[memkit]

> While it is true that their presence in NP-Hard does not require their pressence in NP (and thus are not proved NP Complete)

You're confused here. The two conditions for a problem being NP-complete are (1) it being NP-hard and (2) it being in NP.

You suggest (2) is the issue, but usually it's harder to prove (1) rather than (2). In the context of factorization problems, the factors are simply the certificate that satisfy condition (2).