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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: $(x^3 + y^3 + z^3)/(z+x)(z+y) - (z-y)/2$

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    $\begingroup$ A few numeric example would most certainly improve this question. $\endgroup$ – barak manos Dec 20 '16 at 6:10
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    $\begingroup$ 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. $\endgroup$ – barak manos Dec 20 '16 at 6:12
  • $\begingroup$ try the triple (3,4,5) this gives the prime 2. $\endgroup$ – unknownMe Dec 20 '16 at 6:14
  • $\begingroup$ the result is always an integer $\endgroup$ – unknownMe Dec 20 '16 at 6:14
  • $\begingroup$ @jevie but $y$ is odd? Maybe $(4,3,5)$? $\endgroup$ – Juniven Dec 20 '16 at 6:16
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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.

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  • $\begingroup$ btw how do you write a fraction form of the equation in latex? $\endgroup$ – unknownMe Dec 20 '16 at 7:31
  • $\begingroup$ @jevie: Using \frac{x}{y} with a dollar sign on each side. But I would generally recommend that you input LaTex into your favorite search engine and read a little. It's not too difficult. $\endgroup$ – barak manos Dec 20 '16 at 8:14

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