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October 3

Prime Pythagorean triples

Are there infinitely many Pythagorean triples where both of a and c are prime?? (Using the rule that a is always the short leg and b is the long leg, I conjecture [please disprove me if possible] that there are no Pythagorean triples where b is prime.)

Properties:

c-b is always 1.
Except in the case of 3,4,5, and 5,12,13 b is always a multiple of 60.

Georgia guy (talk) 23:38, 3 October 2024 (UTC)[reply]

The formula for Pythagorean triples, assuming the numbers are relatively prime, is p=x²-y², q=2xy, r=x²+y², where x>y>0, x and y are relatively prime, and x and y are not both odd. Your a and b are p and q, possibly in a different order, and your c is r. Suppose b is prime. Then b cannot be q since q is even, so b is p and p > q. Further, x-y is a factor of p, and since p is prime, x-y = 1. So we get p = 2y+1, q=2y(y+1). Then p>q implies 2y²-1 < 0, which is impossible for y>0. As a further result of this proof, the only Pythagorean triples with a prime side are of the from 2y+1, 2y(y+1), 2y²+2y+1, which includes the 3, 4, 5 and 5, 12, 13 examples. It also generates at least a few more: 11, 60, 61; 19, 180, 181; 29, 420, 421. I think you're right about b being a multiple of 60; I haven't written out a proof, but I don't expect it would be that difficult. I suspect there are infinitely many triples where a and c are prime. This amounts to saying there are infinitely many primes p so that (p²+1)/2 is also prime. Statements like this are usually difficult to prove though. For example it's unknown if there are infinitely many primes p such that p+2 is prime. Unless there is a congruence or set of congruences, or a polynomial factorization which can prove it easily then the likelihood is that it's extremely difficult; there's seldom a middle ground. --RDBury (talk) 04:23, 4 October 2024 (UTC)[reply]

For al such triples with 5 < p < 100000, q ≡ 0 (mod 60). --Lambiam 07:55, 4 October 2024 (UTC)[reply]

It is easily seen that (q mod 60) = 0 iff (y mod 15) ∈ {0, 5, 9, 14}. In each of the 11 other cases, by elementary modular arithmetic, at least one of p and r is divisible by at least one of 3 and 5. This includes the two triples (3, 4, 5) and (5, 12, 13). --Lambiam 08:24, 4 October 2024 (UTC)[reply]

A somewhat more interesting issue is the divisibility properties of p, q and r when they aren't restricted to primes. That q is divisible by 4 is trivial. Either p or q is divisible by 3 (but not both) is seen by considering the 9 pairs of remainders of x and y mod 3. The x%3=y%3=0 is eliminated because x and y are assumed relatively prime. With a similar case breakdown, either p, q or r is divisible by 5. So if neither p or r are divisible by 3 or 5, then q is divisible by 60. This idea breaks down for divisibility by 7, but one can say that r is never divisible by 7, nor by any other prime s with s%4=3. --RDBury (talk) 13:31, 4 October 2024 (UTC)[reply]

For reference, the OEIS sequence of hypotenuses for such triples is OEIS:A067756. GalacticShoe (talk) 13:57, 4 October 2024 (UTC)[reply]

October 15

Is this really a prime-generating polynomial?

In this article, an inequality involving a set of Diophantine equations is given as an example. But does that even qualify as a "formula for primes"? It looks more like a primality test of sorts, ie iff the inequality holds then (k + 2) is prime, so one would still need to generate a likely prime candidate to begin with. Which brings me to the next issue. Setting all of the variables to positive integers and then k to "some prime minus two", the inequality fails regardless. So there must be some special set of rules for selecting the values for each variable? Earl of Arundel (talk) 20:48, 15 October 2024 (UTC)[reply]

The inequality only fails if one of the terms isn't zero. If all terms are zero, then the polynomial evaluates to k+2. Thus, the set of primes is precisely the set of positive values taken by this function. The arguments needed to produce the primes are not constructed, but one imagines plotting the function for all integer values of the arguments. Tito Omburo (talk) 21:59, 15 October 2024 (UTC)[reply]

Sorry, I don't quite understand. I thought none of the 26 variables could be zero? Also, the fact that "the primes are not constructed" leads me to believe that this is indeed not a "prime-generating polynomial" per se. Is that correct? Earl of Arundel (talk) 23:00, 15 October 2024 (UTC)[reply]

The terms alpha_i have a simultaneously zero only when k+2 is prime. So the sum of the squares of the alpha_i always exceeds one unless k+2 is prime. I don't know what "prime generating polynomial" means. This is certainly a polynomial whose range in the positive integers consists only of the primes. Tito Omburo (talk) 23:06, 15 October 2024 (UTC)[reply]

Perhaps it's not that obvious from the article, but the polynomial is not meant to be a practical method for generating primes. Plugging in random values for the variables will give positive, hence prime, values only a small fraction of the time. Finding values of the variables for which the polynomial is positive will be at least as difficult as just computing primes using conventional methods. The point is demonstrate that such a function is possible, but that doesn't mean you'd actually use it. It's possible to program a Turing machine to compute √2 to 10 decimals, but that doesn't mean you should go out and buy a Turing machine if you want to know the diagonal of a square. --RDBury (talk) 04:46, 16 October 2024 (UTC)[reply]

Speaking of that, some years ago I wrote a program to try to find a prime (any prime) by plugging in values for the variables. It ran for hours without finding a 26-tuple that works. Are there any 26-tuples known that yield a prime? Bubba73 ^{You talkin' to me?} 06:44, 16 October 2024 (UTC)[reply]

The section describes a prime-generating inequality. Its terms are polynomials, so this inequality is a polynomial inequality, The bracketing is not as in "[prime-generating polynomial] inequality", but as in "prime-generating [polynomial inequality]". To make the inequality actually produce primes, one has to turn the crank really hard. --Lambiam 05:13, 16 October 2024 (UTC)[reply]

October 16