[math-fun] Another prime question
Given prime p, do there necessarily exist primes q, r such that p+q+r | pqr ?
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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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
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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* Warut Roonguthai <warut822@gmail.com> [Apr 15. 2011 15:11]:
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.
This really want to be a sequence in the OEIS! (how did you determine these?)
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.
This as well ? forprime(p=3,10^4,if( (p%4==3)&&isprime((p+1)/4)&&(!isprime((p+3)/2)),print1(p,", "))); 67, 283, 547, 787, 907, 1051, 1531, 1867, 2011, 2083, 2251, 2347, 2467, 2707, 2803, 3187, 3307, 3547, 3907, 3931, 4243, 4363, 4603, 4651, 4723, 5107, 5227, 5443, 6091, 6211, 6427, 6451, 6547, 6883, 7507, 8443, 8971, 9067, 9187, 9643, 9787, 9907,
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
[...]
Recall that p+1 = (q-1)(r-1). So if p = 4n+3 is a non-hypotenuse, then for all divisors d of p+1, either q = d+1 or r = (p+1)/d+1 is composite. Here are the first 100 such primes: 67, 127, 151, 283, 307, 367, 439, 487, 547, 571, 587, 607, 643, 683, 727, 739, 751, 787, 811, 823, 907, 947, 967, 991, 1051, 1063, 1087, 1163, 1307, 1327, 1423, 1447, 1459, 1471, 1523, 1531, 1567, 1579, 1627, 1663, 1699, 1747, 1783, 1831, 1867, 1987, 1999, 2003, 2011, 2083, 2131, 2143, 2239, 2251, 2287, 2311, 2347, 2411, 2467, 2503, 2531, 2539, 2543, 2647, 2659, 2671, 2683, 2707, 2767, 2791, 2803, 2887, 3067, 3079, 3187, 3259, 3307, 3319, 3343, 3347, 3499, 3511, 3547, 3559, 3583, 3607, 3631, 3727, 3803, 3847, 3907, 3919, 3923, 3931, 3967, 4007, 4027, 4099, 4111, 4211 In my previous e-mail, the sequence there had one wrong term, and the condition I gave for a special class of non-hypotenuse primes was also wrong; it should read: If p is a prime of the form 4n+3 and (p+1)/4 is also prime but p+2 and (p+3)/2 are composite, then p cannot be a hypothenuse. If you think this stuff is interesting enough for OEIS, feel free to submit any related sequences. I'm not familiar with the new OEIS yet, and I don't have much time now. Thanks, Warut On Fri, Apr 15, 2011 at 9:20 PM, Joerg Arndt <arndt@jjj.de> wrote:
* Warut Roonguthai <warut822@gmail.com> [Apr 15. 2011 15:11]:
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.
This really want to be a sequence in the OEIS! (how did you determine these?)
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.
This as well
? forprime(p=3,10^4,if( (p%4==3)&&isprime((p+1)/4)&&(!isprime((p+3)/2)),print1(p,", "))); 67, 283, 547, 787, 907, 1051, 1531, 1867, 2011, 2083, 2251, 2347, 2467, 2707, 2803, 3187, 3307, 3547, 3907, 3931, 4243, 4363, 4603, 4651, 4723, 5107, 5227, 5443, 6091, 6211, 6427, 6451, 6547, 6883, 7507, 8443, 8971, 9067, 9187, 9643, 9787, 9907,
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
[...]
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participants (5)
-
Allan Wechsler -
David Wilson -
Joerg Arndt -
Victor Miller -
Warut Roonguthai