- #1
RamaWolf
- 95
- 2
Let's recall the
Euclidean Rule for Pythagorean Triangles:
Let (m,n) be co-prime natural numbers (m<n), then
h := n[itex]^{2}[/itex] + m[itex]^{2}[/itex]
e := 2 m n
d := n[itex]^{2}[/itex] - m[itex]^{2}[/itex] = (n - m) * (n + m)
form the hypothenuse, the even and the odd leg
of a primitive Pythagorean triangle (PPT)
If we want two primes as sides, since e is even, d and h must be prime numbers,
and that means for d, the difference (n - m) must be one, resulting in n = 1 + m
With this assignment in the formula for h, we have h = 2m * (m + 1) +1
Now let d be a prime p (> 2, since d is odd!), we with have m = [itex]\frac{p - 1}{2}[/itex],
n = 1 + m, or n = [itex]\frac{p + 1}{2}[/itex], e = [itex]\frac{p^2 - 1}{2}[/itex] and h = [itex]\frac{p^2 + 1}{2}[/itex] the following
Theorem: Let 2 < p [itex] \in \mathbb{P}[/itex] be the odd leg and [itex]\frac{p^2 - 1}{2}[/itex] be the even leg of a rectangular triangle,
then the the hypothenuse will be [itex]\frac{p^2 + 1}{2}[/itex].
Corollary: For '2 primes in PPT' we must have the hypothenuse h = [itex]\frac{p^2 + 1}{2}[/itex] be a prime
The OEIS sequence A048161 ("Primes p such that q=(p^2+1)/2 is also a prime")
starts with: 3, 5, 11, 19, 29, 59, 61, 71, 79, 101, 131, 139, 181, 199, 271, 349, 379, 409, 449
see:http://oeis.org/search?q=A048161&sort=&language=english&go=Search
Euclidean Rule for Pythagorean Triangles:
Let (m,n) be co-prime natural numbers (m<n), then
h := n[itex]^{2}[/itex] + m[itex]^{2}[/itex]
e := 2 m n
d := n[itex]^{2}[/itex] - m[itex]^{2}[/itex] = (n - m) * (n + m)
form the hypothenuse, the even and the odd leg
of a primitive Pythagorean triangle (PPT)
If we want two primes as sides, since e is even, d and h must be prime numbers,
and that means for d, the difference (n - m) must be one, resulting in n = 1 + m
With this assignment in the formula for h, we have h = 2m * (m + 1) +1
Now let d be a prime p (> 2, since d is odd!), we with have m = [itex]\frac{p - 1}{2}[/itex],
n = 1 + m, or n = [itex]\frac{p + 1}{2}[/itex], e = [itex]\frac{p^2 - 1}{2}[/itex] and h = [itex]\frac{p^2 + 1}{2}[/itex] the following
Theorem: Let 2 < p [itex] \in \mathbb{P}[/itex] be the odd leg and [itex]\frac{p^2 - 1}{2}[/itex] be the even leg of a rectangular triangle,
then the the hypothenuse will be [itex]\frac{p^2 + 1}{2}[/itex].
Corollary: For '2 primes in PPT' we must have the hypothenuse h = [itex]\frac{p^2 + 1}{2}[/itex] be a prime
The OEIS sequence A048161 ("Primes p such that q=(p^2+1)/2 is also a prime")
starts with: 3, 5, 11, 19, 29, 59, 61, 71, 79, 101, 131, 139, 181, 199, 271, 349, 379, 409, 449
see:http://oeis.org/search?q=A048161&sort=&language=english&go=Search