Difference between revisions of "1993 USAMO Problems/Problem 1"
ZzZzZzZzZzZz (talk | contribs) (New page: ==Problem== For each integer <math>n\ge 2</math>, determine, with proof, which of the two positive real numbers <math>a</math> and <math>b</math> satisfying <cmath>a^n=a+1,\qquad b^{2n}=b+...) |
ZzZzZzZzZzZz (talk | contribs) |
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| Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
Square and rearrange the first equation and also rearrange the second. | Square and rearrange the first equation and also rearrange the second. | ||
| − | \begin{gather} | + | <math>\begin{gather} |
a^{2n}-a=a^2+a+1\\ | a^{2n}-a=a^2+a+1\\ | ||
b^{2n}-b=3a | b^{2n}-b=3a | ||
| − | \end{gather} | + | \end{gather}</math> |
It is trivial that | It is trivial that | ||
\begin{equation} | \begin{equation} | ||
Revision as of 14:00, 22 March 2008
Problem
For each integer 
, determine, with proof, which of the two positive real numbers 
and 
satisfying
![\[a^n=a+1,\qquad b^{2n}=b+3a\]](http://latex.artofproblemsolving.com/9/a/6/9a63b2ade70110e8161ea351e1b94044db3fc522.png)
is larger.
Solution
Square and rearrange the first equation and also rearrange the second.
$\begin{gather}
a^{2n}-a=a^2+a+1\\
b^{2n}-b=3a
\end{gather}$ (Error compiling LaTeX. Unknown error_msg)
It is trivial that
\begin{equation}
(a-1)^2>0
\end{equation}
since clearly cannot equal 0 (Otherwise 
). Thus
\begin{gather}
a^2+a+1>3a\\
a^{2n}-a>b^{2n}-b
\end{gather}
where we substituted in equations (1) and (2) to achieve (5). If 
, then 
since , , and 
are all positive. Adding the two would mean 
, a contradiction, so 
. However, when equals 0 or 1, the first equation becomes meaningless, so we conclude that for each integer , we always have .
