Problem 07 OIMU

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Problem 07 OIMU

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Source: OIMU 2024

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#1 h Dec 21, 2024, 3:14 AM

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Show that the equacion $x^{3}+2y^{3}+3z^{3}=4$ has infinitely many solutions with $x,y,z$ rational numbers.

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#2 h Mar 30, 2025, 12:21 AM

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Let $C$ be the surface $x^3 + 2y^3 + 3z^3 = 4$ , and let $l$ be a line tangent to $C$ at $(1,0,1)$
with parametrization $(1+a\lambda,\, b\lambda,\, 1+c\lambda)$ .

The intersections of $C$ with $l$ are given by the roots of the polynomial
$P(\lambda) \;=\; (1+a\lambda)^3 \;+\; 2\,(b\lambda)^3 \;+\; 3\,(1+c\lambda)^3 \;-\; 4.$
Since $l$ and $C$ are tangent at $(1,0,1)$ , it follows that $0$ is a root of multiplicity $2$ of $P$ ,
which implies $a + 3c = 0$ . Substituting $a = -3c$ into $P$ we obtain
$P(\lambda) \;=\; \lambda^2\,\bigl[\,(2b^3 \,-\,24c^3)\,\lambda \;-\; 18c^2\,\bigr].$
Thus $P$ vanishes at
$\lambda \;=\; -\,\frac{18c^2}{\,2b^3 \;-\; 24c^3\,}.$
It follows that if $b,c$ are real numbers such that $2b^3 \;-\; 24c^3 \neq 0$ , then $C$ passes through the point
$\Bigl(\,1 \;-\; \frac{54c^3}{\,2b^3 \;-\; 24c^3\,},\;\; \frac{18bc^2}{\,2b^3 \;-\; 24c^3\,},\;\; 1 \;+\; \frac{18c^3}{\,2b^3 \;-\; 24c^3\,}\Bigr).$
Hence if $b,c$ are rational, then
$\Bigl(\,1 \;-\; \frac{54c^3}{\,2b^3 \;-\; 24c^3\,}\Bigr)^3 \;+\; 2\,\Bigl(\,\frac{18bc^2}{\,2b^3 \;-\; 24c^3\,}\Bigr)^3 \;+\; 3\,\Bigl(\,1 \;+\; \frac{18c^3}{\,2b^3 \;-\; 24c^3\,}\Bigr)^3 \;=\; 4.$
And we are done.

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#3 h Mar 30, 2025, 12:22 AM

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how in the world did u find this thread with no comments on it?

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#4 h Mar 30, 2025, 12:30 AM • 1 Y

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jkim0656 wrote:

how in the world did u find this thread with no comments on it?

Destiny put this problem in front of me.

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