There are primes of the form 4n+3 that are not hypotenuses; the smallest ones are: 67, 127, 151, 227, 283, 307, 367, 439, 487, 547, 571, 587, 607, 643, 683, 727, 739, 751, 787, 811, 823, 907, 947, 967, 991, 1051. If p is a prime of the form 4n+3 and (p+1)/4 is also prime but (p+3)/2 is composite (e.g., 67, 283, 547, 787, 907, 1051), then p cannot be a hypothenuse. So, assuming the first Hardy–Littlewood conjecture, there should be infinitely many non-hypotenuse primes of the form 4n+3. Also by assuming the same conjecture, for every prime q there should be infinitely many primes r such that (q-1)(r-1)-1 is prime. In particular, every prime should be a leg. Warut On Fri, Apr 15, 2011 at 2:04 AM, Allan Wechsler <acwacw@gmail.com> wrote:
I'm just catching up with you, Victor. So all the smaller members of twin primes pairs are "hypotenuses"; other hypotenuses are all 3 mod 4. So the non-hypotenuses should include primes of the form 4n+1, where 4n+3 is composite. Note that we haven't proven that all primes of the form 4n+3 are hypotenuses, but so far I haven't found any counterexamples. (3, 2, 5), (7, 3, 5), (11, 3, 7), (19, 3, 11), (23, 3, 13), (31, 3, 17), (43, 3, 23), (47, 5, 13), (59, 3, 31) ...
It certainly is starting to look like every prime is a *leg*. The correponding conjecture is: for every prime q there is a prime r such that (q-1)(r-1)-1 is prime. The scariest case I have found so far is 47, but then to my relief I found (1033, 23. 47).
On Thu, Apr 14, 2011 at 1:29 PM, Victor Miller <victorsmiller@gmail.com>wrote:
Given an *odd* prime p, there exist primes q, r such that p+q+r = qr.
This is more profitably written as
p+1 = (q-1)(r-1)
So if q=2 then p,r must be twin primes If both q and r are odd, then p = 3 mod 4.
Victor
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