David, where do you *get* these things? For some reason I'm finding this one a little hard to think about. If we don't already have too many things named after Wilson, let's call (p, q, r) a *Wilsonian triple* whenever p, q, and r are prime and their sum divides their product. Only eight numbers divide pqr, so we are constraining p+q+r to be one of these eight. But in fact p+q+r can't be 1 or p or q or r, because it's too big, and it can't be pqr because it's too small, so what we are really saying is that p+q+r is qr or pr or pq. This means that each Wilsonian triple can be reordered as (p, q, r) such that p+q+r = qr, and it amuses me to call q,r the *legs* and p the *hypotenuse.* I'll probably regret that. Just messing around, I have found it very easy to erect a Wilsonian triple on any hypotenuse except 2. Since 2 can certainly be the leg of a triple (say, (2, 3, 5)), I will strengthen David's (cautiously phrased) conjecture just a little: Given an *odd* prime p, there exist primes q, r such that p+q+r = qr. Findings so far, hypotenuse first: (3, 2, 5), (5, 2, 7), (7, 3, 5), (11, 2, 13), (13, ah, hmm) ... Obviously, I spoke too soon; one can prove pretty easily that 13 is not the hypotenuse of any Wilsonian triple. Someone can extend the sequence of impossible hypotenuses 2, 13 ... and look in OEIS for it. But now I must break for lunch. Thanks, David! On Thu, Apr 14, 2011 at 7:00 AM, David Wilson <davidwwilson@comcast.net>wrote:
Given prime p, do there necessarily exist primes q, r such that p+q+r | pqr ?
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