1993 USAMO Problems/Problem 1

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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 1

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 $a^{n}=a+1$ 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 $a>b$.

Solution 2

Define $f(x)=x^n-x-1$ and $g(x)=x^{2n}-x-3a$. By Descarte's Rule of Signs, both polynomials' only positive roots are $a$ and $b$, respectively. With the Intermediate Value Theorem and the fact that $f(1)=-1$ and $f(2)=2^n-3>0$, we have $a\in(1,2)$. Thus, $-3a\in(-6,-3)$, which means that $g(1)=-3a<0$. Also, we find that $g(a)=a^{2n}-4a$. All that remains to prove is that $g(a)>0$, or $a^{2n}-4a>0$. We can then conclude that $b$ is between $1$ and $a$ from the Intermediate Value Theorem. From the first equation given, $a^{2n}=(a+1)^2=a^2+2a+1$. Subtracting $4a$ gives us $a^2-2a+1>0$, which is clearly true, as $a\neq1$. Therefore, we conclude that $1<b<a<2$.

See Also

1993 USAMO (ProblemsResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions

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