crack RSA: NP, or NP-complete?

I've heard differing opinions/statements of fact from different professors and sources as to whether cracking RSA is "thought" to be in NP, or known to be an NP-complete problem. Can anyone shed light on this discrepancy? which is it, or do we know?

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migrated from cstheory.stackexchange.comAug 16 '11 at 5:02

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This question is not research level. Read wiki Integer factorization difficulty and complexity. –  Pratik Deoghare Aug 16 '11 at 2:20
George Bush: Great president, or greatest president? –  Aaron Roth Aug 16 '11 at 2:57
Please read the FAQ to understand the scope of cstheory. I am closing the question as off-topic and migrating to Math.SE. –  Kaveh Aug 16 '11 at 5:02

Proving that breaking any cryptosystem is $\mathsf{NP}$-complete would be a great achievement in cryptography, since $\mathsf{P} \neq \mathsf{NP}$ is probably the most "standard" hardness assumption around. For the case of RSA, security of the system is based on hardness of the integer factorization problem. This problem lies in the intersection of $\mathsf{NP}$ and $\mathsf{coNP}$, and is thus unlikely to be $\mathsf{NP}$-complete. However, it's obviously in $\mathsf{NP}$ (just guess the private key, and decrypt).