Difference between revisions of "1993 USAMO Problems/Problem 1"

(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+...)
 
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 15:00, 22 March 2008

Problem

For each integer $n\ge 2$, determine, with proof, which of the two positive real numbers $a$ and $b$ satisfying \[a^n=a+1,\qquad b^{2n}=b+3a\] 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 $a$ clearly cannot equal 0 (Otherwise $a^n=0\ne0+1$). 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 $b>a$, then $b^{2n}>a^{2n}$ since $a$, $b$, and $n$ are all positive. Adding the two would mean $b^{2n}-b>a^{2n}-a$, a contradiction, so $a>b$. However, when $n$ equals 0 or 1, the first equation becomes meaningless, so we conclude that for each integer $n\ge 2$, we always have $a>b$.