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How to show that this equation:

$$x^2+y^2=z^5+z$$

Has infinitely many relatively prime integral solutions

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2 Answers 2

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:

$X^2+Y^2=Z^5+Z$

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:

$Z=(a^2+b^2)^2$

$X=a^2+2(a^2+b^2)^4ab-b^2$

$Y=(a^2+b^2)^4a^2-2ab-(a^2+b^2)^4b^2$

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