1997 IMO Problems/Problem 5


Find all pairs $(a,b)$ of integers $a,b \ge 1$ that satisfy the equation



Case 1: $(1 \le a \le b)$


Looking at this expression since $b \ge a$ then $a^{b} \le b$.

Here we look at subcase $a>1$ which gives $a^{b}>b$ for all $(1 < a \le b)$. This contradicts condition $a^{b} \le b$, and thus $a$ can't be more than one giving the solution of $a=1$ with $b \ge 1$. So we substitute the value of $a=1$ into the original equation to get $1^{b^2}=b^{1}$ which solves to $b=1$ and our first pair $(a,b)=(1,1)$

Case 2: $(1 \le b < a)$


since $a>b$, then $b^{2}<a$ and we multiply both sides of the equation by $b^{-2b^{2}}$ to get:



Since $b^{2}<a$, then $(ab^{-2})^{b^{2}}>1$ and $b^{a-2b^{2}}>0$. This gives $a-2b^{2}>1$

This implies that $a>2b^{2}$ for $b>1$

Let $a=kb^{2}$ with $k \in \mathbb{Z} ^{+}$. Since $a>2b^{2}$, then $k \ge 3$



$k=b^{k-2}$, which gives $b \ge 2$

subcase $k=3$:

$3=b^{3-2}=b$ and $a=kb^{2}=(3)(3)^{2}=27$. which provides 2nd pair $(a,b)=(27,3)$

subcase $k=4$:

$4=b^{4-2}=b^{2}$, thus $b=2$ and $a=kb^{2}=(4)(2)^{2}=16$. which provides 3rd pair $(a,b)=(16,2)$

subcase $k \ge 5$:

$k=b^{k-2}$, thus $b=k^{1/(k-2)}$ which decreases with $k$ and $b \to 1$ as $k \to \infty$ . From subcase $k=4$, we know that $b=2$, thus for subcase $k \ge 5$, $1<b<2$. Therefore this subcase has no solution because it contradicts the condition for Case 2 of $b \ge 2$.

Final solution for all pairs is $(a,b)=(1,1); (27,3); (16,2)$

~ Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1997 IMO (Problems) • Resources
Preceded by
Problem 4
1 2 3 4 5 6 Followed by
Problem 6
All IMO Problems and Solutions