1993 USAMO Problems/Problem 1

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 $a>b$ .

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1993 USAMO Problems/Problem 1

Problem

Solution