# 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

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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 $4k+1$.