# Difference between revisions of "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{align} a^{2n}-a&=a^2+a+1\\ b^{2n}-b&=3a \end{align} It is trivial that \begin{align*} (a-1)^2 > 0 \tag{3} \end{align*} since $a-1$ clearly cannot equal $0$ (Otherwise $a^n=1\neq 1+1$). Thus \begin{align*} a^2+a+1&>3a \tag{4}\\ a^{2n}-a&>b^{2n}-b \tag{5} \end{align*} where we substituted in equations (1) and (2) to achieve (5). Notice that from $a^{2n}=a+1$ we have $a>1$. Thus, if $b>a$, then $b^{2n-1}-1>a^{2n-1}-1. Since$a>1\Rightarrow a^{2n-1}-1>0 $, multiplying the two inequalities yields$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\$.