Art of Problem Solving

1988 OIM Problems/Problem 5: Difference between revisions

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Prove that if <math>x+yt+zt^2 \ne 0</math>, then there exist <math>u</math>, <math>v</math>, and <math>w</math> as rational numbers such that:
Prove that if <math>x+yt+zt^2 \ne 0</math>, then there exist <math>u</math>, <math>v</math>, and <math>w</math> as rational numbers such that:
<cmath>(x + yt + z^2)(u + vt + wt^2) = 1</cmath>
<cmath>(x + yt + z^2)(u + vt + wt^2) = 1</cmath>
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com


== Solution ==
== Solution ==
{{solution}}
{{solution}}
== See also ==
https://www.oma.org.ar/enunciados/ibe3.htm

Latest revision as of 12:28, 13 December 2023

Problem

Consider expressions in the form: $x+yt+zt^2$ with $x$, $y$, and $z$ rational numbers and $t^3=2$.

Prove that if $x+yt+zt^2 \ne 0$, then there exist $u$, $v$, and $w$ as rational numbers such that: \[(x + yt + z^2)(u + vt + wt^2) = 1\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

https://www.oma.org.ar/enunciados/ibe3.htm