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How to show that this equation:


Has infinitely many relatively prime integral solutions

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up vote 8 down vote accepted

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 $4k+1$.

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It is necessary to write down the formula!

In the equation:


I think this formula should be written in a more general form:




And yet another formula:




$a,b$ - arbitrary integers.

Solutions can be written as follows:




where $a,b$ - any integers asked us.

Well, a simple solution:




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