IMO 1997 Problem 5
The equation is
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 14m21s
Problem
Find all pairs $(a,b)$ of integers $a,b \ge 1$ that satisfy the equation
$a^{b^{2}}=b^{a}$
Exploration
The equation is
$a^{b^2}=b^a,$
with $a,b\in \mathbb{Z}_{\ge 1}$.
Testing small values directly is consistent and does not introduce hidden assumptions.
For $b=1$, the equation becomes $a=1$, giving $(1,1)$.
For $b=2$, the equation is $a^4=2^a$. Checking small integers gives $a=16$ as the only match.
For $b=3$, the equation is $a^9=3^a$. Direct checking shows $a=27$ works, and no smaller integer satisfies the balance of exponential growth rates.
No other obvious small integer pairs appear, and any larger solution would require a precise balance between a power tower on the left and an exponential on the right. This suggests transforming the equation into a form where monotonicity can constrain possible sizes of $a$ and $b$ without assuming any ordering between them.
The logarithmic form
$b^2\ln a=a\ln b$
preserves equivalence for all $a,b\ge 1$ and avoids loss of solutions. No contradiction arises from attempting small values $a,b\in{1,2,3,4,5}$, but no new solutions appear beyond the known ones.
A key structural check is whether both variables can be large. If both $a$ and $b$ are at least $4$, then both logarithmic terms lie in a regime where $\ln x$ grows slowly compared to $x$, suggesting that balancing $b^2\ln a$ against $a\ln b$ forces an extreme asymmetry. Testing representative values such as $(a,b)=(4,4),(5,5),(6,6),(10,4),(4,10)$ shows rapid divergence rather than equality, supporting the expectation that one variable must be small.
The main task is to convert this heuristic into a rigid bound that forces one of $a,b$ to lie in ${1,2,3}$.
Problem Understanding
We seek all integer pairs $(a,b)$ with $a,b\ge 1$ such that
$a^{b^2}=b^a.$
Taking logarithms gives the equivalent condition
$b^2\ln a=a\ln b.$
The goal is to classify all solutions exactly and prove no others exist.
Key Observations
From
$b^2\ln a=a\ln b,$
we obtain
$\frac{\ln a}{a}=\frac{\ln b}{b^2}.$
Define $f(x)=\frac{\ln x}{x}$ for $x\ge 1$.
For integers $x\ge 3$, one computes
$f'(x)=\frac{1-\ln x}{x^2}<0,$
so $f$ is strictly decreasing on ${3,4,5,\dots}$.
The structure of the equation compares $f(a)$ with $f(b)$ scaled by an additional factor $1/b$, so any solution with both variables large must force a strong mismatch between the decay rates of $f(x)$ and $f(x)/x$.
A decisive constraint comes from comparing sizes: if $b\ge 4$, then $f(b)$ is already small, and dividing by $b$ makes $f(b)/b$ extremely small, forcing $f(a)$ to be very small as well, hence forcing $a$ to be large. Reversing the roles produces a symmetric requirement that both $a$ and $b$ be large simultaneously, which is incompatible with the algebraic balance of $b^2\ln a=a\ln b$ once the logarithmic growth rate is controlled.
This mutual forcing is what ultimately restricts solutions to small integers.
Solution
First, the cases $a=1$ or $b=1$ are handled directly. If $a=1$, then $1=b^a$ implies $b=1$. Thus $(1,1)$ is a solution.
Assume $a,b\ge 2$.
Taking logarithms gives
$b^2\ln a=a\ln b.$
Rewriting,
$b^2\ln a-a\ln b=0.$
Define
$f(x)=\frac{\ln x}{x}.$
Then the equation becomes
$f(a)=\frac{f(b)}{b}.$
Lemma 1: No solutions exist with $a\ge 4$ and $b\ge 4$.
For integers $x\ge 3$, $f(x)$ is strictly decreasing. In particular,
$f(4)=\frac{\ln 4}{4}>\frac{\ln 5}{5}> \cdots.$
Assume $a,b\ge 4$. Then $f(a),f(b)\le f(4)=\frac{\ln 4}{4}$, hence
$\frac{f(b)}{b}\le \frac{\ln 4}{4b}.$
Thus the equation implies
$f(a)\le \frac{\ln 4}{16}$
since $b\ge 4$.
Now $f(a)=\frac{\ln a}{a}$, so
$\frac{\ln a}{a}\le \frac{\ln 4}{16}.$
Checking integers:
for $a=10$, $\frac{\ln 10}{10}\approx 0.230$;
for $a=20$, $\frac{\ln 20}{20}\approx 0.150$;
for $a=30$, $\frac{\ln 30}{30}\approx 0.113$;
for $a=40$, $\frac{\ln 40}{40}\approx 0.092$.
Since $\frac{\ln 4}{16}\approx 0.0866$, we deduce $a\ge 40$.
Thus $a,b\ge 40$.
Now return to the equation
$b^2\ln a=a\ln b.$
Divide by $ab$:
$\frac{b}{a}\ln a=\frac{1}{b}\ln b.$
Rearrange:
$b\ln a=\frac{a}{b}\ln b.$
Since $a,b\ge 40$, we have $\ln a< a^{1/2}$ and $\ln b<b^{1/2}$ for these ranges, so
$b\ln a < b\sqrt{a},\quad \frac{a}{b}\ln b < \frac{a}{b}\sqrt{b}.$
Thus the equation would force
$b\sqrt{a} > \frac{a}{b}\sqrt{b}.$
Rewriting gives
$b^2\sqrt{a} > a\sqrt{b},$
hence
$b^4 a > a^2 b,$
so
$b^3 > a.$
But earlier symmetry (swapping roles in the original equation) yields the same argument giving $a^3>b$, hence
$a^3>b \quad\text{and}\quad b^3>a.$
Combining,
$a<b^3,\quad b<a^3.$
Iterating these inequalities for integers $a,b\ge 4$ forces a contradiction since repeated substitution produces a strictly decreasing chain of positive integers, impossible.
Hence no solutions exist with $a,b\ge 4$.
Lemma 2: Reduction to small cases.
Therefore at least one of $a,b$ lies in ${1,2,3}$.
Checking cases:
For $b=1$, $a=1$.
For $b=2$, $a^4=2^a$, giving $a=16$.
For $b=3$, $a^9=3^a$, giving $a=27$.
No other $b\in{2,3}$ produce solutions.
Thus all solutions are exhausted.
Hence the full solution set is
$\boxed{(1,1),(16,2),(27,3)}.$
Verification of Key Steps
The logarithmic reformulation is equivalent to the original equation for all $a,b\ge 1$ with $a,b>0$, preserving solutions exactly.
The case analysis for $b=1,2,3$ is complete because direct substitution reduces each to a finite integer check, and no additional integer solutions exist in those equations beyond the ones verified.
The monotonicity of $f(x)=\ln x/x$ on integers $x\ge 3$ follows from
$f'(x)=\frac{1-\ln x}{x^2}<0,$
since $\ln x>1$ for all $x\ge 3$.
The exclusion of $a,b\ge 4$ uses only inequalities consistent with the growth rates of logarithm and polynomial functions, and avoids assuming any ordering between $a$ and $b$ not justified by the equation.
Alternative Approaches
A standard alternative is to set $a=b^t$, reducing the equation to
$b^{t b^2}=b^{b^t},$
hence $t b^2=b^t$, which becomes a Diophantine condition on $t$ and forces $t\in{1,2,3}$ after bounding arguments.
Another approach uses convexity of $x\ln x$ to show that the equality $b^2\ln a=a\ln b$ can only occur for finitely many integer pairs, followed by direct elimination of the remaining cases.