Difference between revisions of "1988 OIM Problems/Problem 5"
(Created page with "== Problem == Consider expressions in the form: <math>x+yt+zt^2</math> with <math>x</math>, <math>y</math>, and <math>z</math> rational numbers and <math>t^3=2</math>. Prove...") |
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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: 
with 
, 
, and 
rational numbers and 
.
Prove that if 
, then there exist 
, 
, and 
as rational numbers such that:
![\[(x + yt + z^2)(u + vt + wt^2) = 1\]](http://latex.artofproblemsolving.com/c/7/2/c7243d4e1ddb2b3415825a692eedec348be5243f.png)
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
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