A090682 Integer part of one leg of a right triangle where the other leg and hypotenuse are twin primes.

4, 4, 6, 8, 10, 12, 15, 16, 20, 20, 23, 24, 26, 27, 28, 30, 30, 32, 33, 35, 37, 40, 41, 42, 45, 47, 48, 49, 50, 51, 56, 57, 57, 58, 59, 63, 64, 64, 65, 66, 67, 70, 71, 71, 72, 72, 75, 76, 76, 77, 80, 80, 81, 82, 82, 84, 86, 86, 87, 88, 89, 90, 91, 91, 91, 92, 92

Offset

3,1

Comments

Except for the first term, the real value of these numbers is irrational. We have for prime p>2, (p+2)^2 - p^2 = a^2 or 4*p+4 = a^2. This implies a = 4*k, and 4*p+4 = 16*k^2 or p = 4*k^2-1 = (2*k+1)*(2*k-1) = composite for k > 1. Therefore for a = 4, p = 3 is the only integer solution. Suppose 4*p+4 = a^2/b^2 for (a,b) = 1. 4*p*b^2 - 4*b^2 = a^2 = 16*k^2 or p*b^2 = 4*k^2 - b^2 = (2*k-b)*(2*k+b) => k = b*r. Then p*b^2 = (2*b*r-b)*(2*b*r+b) or p = (2*r-1)*(2*r+1) and p is prime only r = 1. So p*b^2 = (b)*(3*b) and p = 3 is the smallest and only rational solution.

Links

Harvey P. Dale, Table of n, a(n) for n = 3..1000

Mathematica

Floor[Sqrt[#[[2]]^2-#[[1]]^2]]&/@Select[Partition[Prime[Range[500]], 2, 1], #[[2]]-#[[1]]==2&] (* Harvey P. Dale, Aug 01 2025 *)

Prog

(PARI) /* Twin right triangles */ twinright2(n) = { forprime(x=3, n, y=x+2; if(isprime(y), print1(sqrtint(y^2-x^2)", ") ) ) }

Crossrefs

Sequence in context: A273308 A105790 A338895 * A011982 A282939 A175216

Adjacent sequences: A090679 A090680 A090681 * A090683 A090684 A090685

Keyword

easy,nonn

Author

Cino Hilliard, Dec 18 2003

Status

approved