Wikipedia:Reference desk/Archives/Mathematics/2024 October 3

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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)

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)

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

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)

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)

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