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text: 'The proof that there are infinitely many [4mf prime numbers] is a [-46z].\n\nFirst, we note that there is indeed a prime: namely $2$.\nWe will also state a lemma: that every number greater than $1$ has a prime which divides it.\n(This is the easier half of the [5rh], and the slightly stronger statement that "every number may be written as a product of primes" is proved there.)\n\nNow we can proceed to the meat of the proof.\nSuppose that there were finitely many prime numbers $p_1, p_2, \\ldots, p_n$.\nSince we know $2$ is prime, we know $2$ appears in that list.\n\nThen consider the number $P = p_1p_2\\ldots p_n + 1$ — that is, the product of all primes plus 1.\nSince $2$ appeared in the list, we know $P \\geq 2+1 = 3$, and in particular $P > 1$.\n\nThe number $P$ can't be divided by any of the primes in our list, because it's 1 more than a multiple of them.\nBut there is a prime which divides $P$, because $P>1$; we stated this as a lemma.\nThis is immediately a contradiction: $P$ cannot be divided by any prime, even though all integers greater than $1$ can be divided by a prime.\n\n# Example\n\nThere is a common misconception that $p_1 p_2 \\dots p_n+1$ must be prime if $p_1, \\dots, p_n$ are all primes.\nThis isn't actually the case: if we let $p_1, \\dots, p_6$ be the first six primes $2,3,5,7,11,13$, then you can check by hand that $p_1 \\dots p_6 + 1 = 30031$; but $30031 = 59 \\times 509$.\n(These are all somewhat ad-hoc; there is no particular reason I knew that taking the first six primes would give me a composite number at the end.)\nHowever, we *have* discovered a new prime this way (in fact, two new primes!): namely $59$ and $509$.\n\nIn general, this is a terrible way to discover new primes.\nThe proof tells us that there must be some new prime dividing $30031$, without telling us anything about what those primes are, or even if there is more than one of them (that is, it doesn't tell us whether $30031$ is prime or composite).\nWithout knowing in advance that $30031$ is equal to $59 \\times 509$, it is in general very difficult to *discover* those two prime factors.\nIn fact, it's an open problem whether or not prime factorisation is "easy" in the specific technical sense of there being a polynomial-time algorithm to do it, though most people believe that prime factorisation is "hard" in this sense.',
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