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). 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 $a>b$.

See also

 1993 USAMO (Problems • Resources) Preceded byFirst question Followed byProblem 2 1 • 2 • 3 • 4 • 5 All USAMO Problems and Solutions
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