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

Juniven Acapulco
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unknownMe
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    A few numeric example would most certainly improve this question. – barak manos Dec 20 '16 at 06:10
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    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 06:12
  • try the triple (3,4,5) this gives the prime 2. – unknownMe Dec 20 '16 at 06:14
  • the result is always an integer – unknownMe Dec 20 '16 at 06:14
  • @jevie but $y$ is odd? Maybe $(4,3,5)$? – Juniven Acapulco Dec 20 '16 at 06:16
  • The triple $(3,4,5)$ is only a single example (and a very trivial one). You must have found a few more, to come up with this conjecture. Also - are there also infinitely many cases of non-primes? If so, then just FYI, by Dirichlet theorem, every 1st degree polynomial whose coefficients are coprimes yields infinitely many primes (and obviously infinitely many non-primes)... So although your expression is not a 1st degree polynomial (and not even a single-variable polynomial), I doubt that it contributes anything. – barak manos Dec 20 '16 at 06:17
  • @barak manos yea I also think its not raeally worth anything just want to share if something could be deduced from the statement, etc. – unknownMe Dec 20 '16 at 06:20
  • Well, then I suggest that you by the least share some of the information that has led you to this conjecture. As I've already mentioned, a few numeric examples might be very useful for others to observe some sort of pattern that makes this expression yields infinitely many primes. – barak manos Dec 20 '16 at 06:22
  • BTW, your expression gives $242.5$ for $x=3,y=4,z=5$. In order for it to give a value closer to $2$, you need to put $(z+x)(z+y)$ in parenthesis. And if you do that, then it gives $2.5$. That's not even an integer. So it sounds to make like you haven't done any research whatsoever, but rather tried only a single numeric example, and probably without even bothering to make sure that the result is correct (I guess you've used integer division instead of floating-point division). – barak manos Dec 20 '16 at 06:29
  • @barak manos precisely correct. I noticed it also. – Juniven Acapulco Dec 20 '16 at 06:30
  • @juniven: And in addition, OP says "odd $y$" in the question, then gives an example with an even $y$ in the comment ($4$)... Feels like not much thought was put into all this... – barak manos Dec 20 '16 at 06:31
  • @barak manos I asked him already, see my comment – Juniven Acapulco Dec 20 '16 at 06:32
  • @juniven: I think that the expression doesn't even yield an integer value for most Pythagorean triplets. And to make it ($6$ times) worse - no matter how you permute them!!! – barak manos Dec 20 '16 at 06:35
  • I think you dont get my equation, the result is always an integer that is a sum of two squares – unknownMe Dec 20 '16 at 06:47
  • i dont know to to write the equation in a latex – unknownMe Dec 20 '16 at 06:48
  • first add the cubes of x, y and z. then divide the result by z+x and by z+y, then subtract the result by z-y/2 – unknownMe Dec 20 '16 at 06:50
  • example: (4,3,5), 4^3 + 3^3 + 5^3 = 216, z+x=9, z+y=8. divide 216 by 8x9=72, the result is 3. subtract (z-y)/2=(5-3)/2=1 will give 2 – unknownMe Dec 20 '16 at 06:53

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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.

Greg Martin
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  • btw how do you write a fraction form of the equation in latex? – unknownMe Dec 20 '16 at 07:31
  • @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. – barak manos Dec 20 '16 at 08:14