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  "body": "\n## Problem\n\nGiven a [[composite number|008BH74D7R]] $n$, written in binary, find integers $a$ and $b$ with $1 \u003c a, b \u003c n$ and $n = ab$.\nRepeating this step gives the [[prime factorization|02EKEPPJ3R]] of $n$.\nThe open question is whether there is a classical algorithm that does this in a number of bit operations bounded by a polynomial in $\\log n$.\n\n## Remarks\n\nThe prime factorization exists and is unique by the [[existence of prime factorization|020K0X7HVR]] and the [[fundamental theorem of arithmetic|084HH1NDCB]], so the problem is about the cost of finding it.\nTrial division by all integers up to $\\sqrt n$ always works, but it needs about $\\sqrt n$ steps, which is exponential in the number of digits.\nNo polynomial-time classical algorithm is known.\nThe fastest known general method for large numbers is the general number field sieve, whose heuristic running time is\n$$\\exp\\left( \\left( \\sqrt[3]{64/9} + o(1) \\right) (\\ln n)^{1/3} (\\ln \\ln n)^{2/3} \\right).$$\nThe decision version, which asks whether $n$ has a factor $d$ with $1 \u003c d \u003c k$ for a given $k$, lies in both NP and co-NP.\nFor this reason it is not expected to be NP-complete, but this is not proved.\nOn a quantum computer, Shor's algorithm factors $n$ in polynomial time.\nDeciding whether $n$ is prime is easier: the [[AKS primality test|XMHQCDWZW1]] does it in deterministic polynomial time, without finding any factor.\n\n## Explanations\n\n### L1 human\n\nMultiplying two large primes is quick, but recovering the two primes from their product seems to be very slow.\nFor example, $91 = 7 \\cdot 13$ is easy to factor by hand, but a product of two primes with three hundred digits each is beyond every known method.\nThe security of the [[RSA cryptosystem|0HM584C465]] rests on this gap.\nNobody has proved that factoring is really hard, so this remains an open problem.\n",
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