show that $x^2+y^2=z^5+z$ Has infinitely many relatively prime integral solutions How to show that this equation:
$$x^2+y^2=z^5+z$$
Has infinitely many relatively prime integral solutions 
 A: The number $z^4+1$ is a sum of two relatively prime squares. Let $z$ be the sum of two relatively prime squares. Then the product $(z^4+1)z$ is a sum of two squares, by the Brahmagupta identity
$$(s^2+t^2)(u^2+v^2)=(su\pm tv)^2+(sv\mp tu)^2.$$ 
Now we take care of the relatively prime part. Suppose that $m$ has a representation as a sum of two squares, but no primitive representation. Then $m$ is divisible by $4$ or by some prime of the form $4k+3$. In our case, primes of the form $4k+3$ are irrelevant. And if $z$ is a sum of two relatively prime squares, then $(z^4+1)z$ cannot be divisible by $4$.  
So pick for example $z$ a power of $5$, or a prime of the form $4k+1$. 
A: It is necessary to write down the formula!
In the equation:
$X^2+Y^2=Z^5+Z$
I think this formula should be written in a more general form:
$Z=a^2+b^2$
$X=a(a^2+b^2)^2+b$
$Y=b(a^2+b^2)^2-a$
And yet another formula:
$Z=\frac{a^2+b^2}{2}$
$X=\frac{(a-b)(a^2+b^2)^2-4(a+b)}{8}$
$Y=\frac{(a+b)(a^2+b^2)^2+4(a-b)}{8}$
$a,b$ - arbitrary integers.
Solutions can be written as follows:
$Z=\frac{(a^2+b^2)^2}{2}$
$X=\frac{((a^2+b^2)^4+4)a^2+2((a^2+b^2)^4-4)ab-((a^2+b^2)^4+4)b^2}{8}$
$Y=\frac{((a^2+b^2)^4-4)a^2-2((a^2+b^2)^4+4)ab-((a^2+b^2)^4-4)b^2}{8}$
where $a,b$ - any integers asked us.
Well, a simple solution:
$Z=(a^2+b^2)^2$
$X=a^2+2(a^2+b^2)^4ab-b^2$
$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$
