# infinitude of primes with primitive pythagorean triple variables

I just want to share what I found, I don't know if this is something useful or worth knowing:

Let $(x,y,z)$ be a primitive pythagorean triple, odd $y$, then there are infinitely many primes of the form: $(x^3 + y^3 + z^3)/(z+x)(z+y) - (z-y)/2$

• A few numeric example would most certainly improve this question. – barak manos Dec 20 '16 at 6:10
• BTW, if this expression also yields infinitely many non-primes (let alone, infinitely many non-integers), then I doubt if it is really "worth" anything. – barak manos Dec 20 '16 at 6:12
• try the triple (3,4,5) this gives the prime 2. – unknownMe Dec 20 '16 at 6:14
• the result is always an integer – unknownMe Dec 20 '16 at 6:14
• @jevie but $y$ is odd? Maybe $(4,3,5)$? – Juniven Dec 20 '16 at 6:16

Indeed, using the well-known parametrization of primitive Pythagorean triples $$x=2rs,\, y=r^2-s^2,\, z=r^2+s^2$$ where $r>s>0$ are coprime and of opposite parity, we obtain the simplification $$\frac{x^3+y^3+z^3}{(x+z) (y+z)}-\frac{z-y}{2} = (r-s)^2+s^2.$$ And every prime congruent to $1$ (mod $4$), as well as the prime $2$, can be written in this form by a theorem of Fermat.