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Difference between revisions of "1997 IMO Problems/Problem 5"

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<math>k=b^{k-2}</math>, thus <math>b=k^{1/(k-2)}</math> which decreases with <math>k</math> and <math>b \to 1</math> as <math>k \to \infty</math> .  From subcase <math>k=4</math>, we know that <math>b=2</math>, thus for subcase <math>k \ge 5</math>, <math>1<b<2</math>.  Therefore this subcase has no solution.
 
<math>k=b^{k-2}</math>, thus <math>b=k^{1/(k-2)}</math> which decreases with <math>k</math> and <math>b \to 1</math> as <math>k \to \infty</math> .  From subcase <math>k=4</math>, we know that <math>b=2</math>, thus for subcase <math>k \ge 5</math>, <math>1<b<2</math>.  Therefore this subcase has no solution.
  
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Final solution is <math>(a,b)=(1,1); (27,3); (16,2)</math>
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Final solution for all pairs is <math>(a,b)=(1,1); (27,3); (16,2)</math>
  
 
~ Tomas Diaz
 
~ Tomas Diaz
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 16:43, 6 October 2023

Problem

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

$a^{b^{2}}=b^{a}$

Solution

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

$(a^{b})^{b}=b^{a}$

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: $(1 \le b < a)$

$a^{b^2}=b^{a}$

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

$b^{-2b^{2}}a^{b^2}=b^{-2b^{2}}b^{a}$

$(ab^{-2})^{b^{2}}=b^{a-2b^{2}}$

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 $k \ge 3$

$(kb^{2}b^{-2})^{b^{2}}=b^{kb^{2}-2b^{2}}$

$k^{b^{2}}=b^{(k-2)b^{2}}$

$k=b^{k-2}$, 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.

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

~ Tomas Diaz

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

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