UK MATHEMATICAL OLYMPIAD
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Prove that , the equation $ x^2+y^2=z^5+z $ has infinitely many relatively prime integral solutions .
Re: UK MATHEMATICAL OLYMPIAD
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.
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.
Re: UK MATHEMATICAL OLYMPIAD
For a start, use the lemma: every divisor of a sum of square is a sum of square itself. So, $z$ is a bi-square too.MATHPRITOM wrote:Prove that , the equation $ x^2+y^2=z^5+z $ has infinitely many relatively prime integral solutions .
One one thing is neutral in the universe, that is $0$.