Prime numbers yield from Pythagoras triples Pythagoras theorem $$a^2+b^2=c^2$$
we got $$P_{prime}(a,b)={a^4+b^4+(a+b)^4\over a^2+b^2+(a+b)^2}$$
Where $(a,b,c)$ are Pythagoras theorem triples, this function $P_{prime}(a,b)$ always produce a prime number for the values of a and b.
Examples: $P_{prime}(3,4)=37$, $P_{prime}(5,12)=229$, $P_{prime}(68,285)=105229$ and so on...
I have checked a lot of values, it seem to be prime so far.
My question is: Does the function $P_{prime}(a,b)$ always produce prime numbers?

Let $a=2xy,$ and $b=x^2-y^2$.
Hence, we obtain:
$$\frac{(2xy)^4+(x^2-y^2)^4+(2xy+x^2-y^2)^4}{(2xy)^2+(x^2-y^2)^2+(2xy+x^2-y^2)^2}=x^4+y^4+2x^3y-2xy^3+2x^2y^2.$$
We can get this expression by the following.
$$\frac{a^4+b^4+(a+b)^4}{a^2+b^2+(a+b)^2}=\frac{2(a^4+2a^3b+3a^2b^2+2ab^3+b^4)}{2(a^2+ab+b^2)}=$$
$$=\frac{(a^2+ab+b^2)^2}{a^2+ab+b^2}=a^2+ab+b^2=x^4+y^4+2x^3y-2xy^3+2x^2y^2.$$
Now, easy to find a counterexample: $x=10$ and $y=1$, which gives $a=20$ and $b=99.$  
