I find this particular proof a bit terse and unhelpful (again, ignoring the obviously incorrect "divisors of a prime part"--specifically, it requires a supporting lemma that escapes me at the moment.
So, we want to prove the following, in effect, for the proof as stated to be valid:
Given consecutive primes p1....pn starting at 2, the product x of p1 to pn plus one cannot be divisible by some combination of factors a1p1, a2p2.....anpn, where a1....an are integers.
A previous poster mentioned that this is doable via modular arithmetic, which I'm sure is accurate, but without thinking about it more in that vein this claim is not self evident. In particular, it's obvious that for sufficiently high values of ai, it's possible to have a1p1.......anpn greater than x, so what we're really trying to prove here is that THERE DOES NOT EXIST A COMBINATION OF INTEGERS a1.....an such that a1p1 *.... * anpn = (p1 * p2 *...... * pn) + 1, which strikes me as not immediately self-evident Can someone explain how best to go about this proof and why it might be considered obvious?
Edit: Wikipedia makes the argument clearer by the simple claim that clearly (x + 1) / pi must have a remainder of 1 since pi > 1 for all pi. Certainly the proof is simple in this context, although as stated in the comic its terseness is perhaps unwarranted....although maybe so, given its elegance and expressibility in haiku. Also, remembering that I actually took an Algebra class in college, I recall that p1^a1.....pn^an would have been a much more canonical way of expressing a composite as a product of primes. silly me.
Last edited by Tarquin
on Wed Aug 12, 2009 5:54 am UTC, edited 2 times in total.