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

(Created page with "==Problem== Find all pairs <math>(a,b)</math> of integers <math>a,b \ge 1</math> that satisfy the equation <math>a^{b^{2}}=b^{a}</math> ==Solution== {{solution}}")
 
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==Solution==
 
==Solution==
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{{solution}}
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Case 1: <math>(1 \le a \le b)</math>
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<math>(a^{b})^{b}=b^{a}</math>
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Looking at this expression since <math>b \ge a</math> then <math>a^{b} \le b</math>.
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Here we look at subcase <math>a>1</math> which gives <math>a^{b}>b</math> for all <math>(1 < a \le b)</math>.  This contradicts condition <math>a^{b} \le b</math>, and thus <math>a</math> can't be more than one giving the solution of <math>a=1</math> with <math>b \ge 1</math>.  So we substitute the value of <math>a=1</math> into the original equation to get <math>1^{b^2}=b^{1}</math> which solves to <math>b=1</math> and our first pair <math>(a,b)=(1,1)</math>
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Case 2: <math>(1 \le b < a)</math>
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<math>a^{b^2}=b^{a}</math>
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since <math>a>b</math>, then <math>b^{2}<a</math> and we multiply both sides of the equation by <math>b^{-2b^{2}}</math> to get:
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<math>b^{-2b^{2}}a^{b^2}=b^{-2b^{2}}b^{a}</math>
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<math>(ab^{-2})^{b^{2}}=b^{a-2b^{2}}</math>
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Since <math>b^{2}<a</math>, then <math>(ab^{-2})^{b^{2}}>1</math> and <math>b^{a-2b^{2}}>0</math>. This gives <math>a-2b^{2}>1</math>
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This implies that <math>a>2b^{2}</math> for <math>b>1</math>
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Let <math>a=kb^{2}</math> with <math>k \in \mathbb{Z} ^{+}</math>. Since <math>a>2b^{2}</math>, then <math>k \ge 3</math>
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<math>(kb^{2}b^{-2})^{b^{2}}=b^{kb^{2}-2b^{2}}</math>
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<math>k^{b^{2}}=b^{(k-2)b^{2}}</math>
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<math>k=b^{k-2}</math>, which gives <math>b \ge 2</math>
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subcase <math>k=3</math>:
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<math>3=b^{3-2}=b</math> and <math>a=kb^{2}=(3)(3)^{2}=27</math>. which provides 2nd pair <math>(a,b)=(27,3)</math>
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subcase <math>k=4</math>:
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<math>4=b^{4-2}=b^{2}</math>, thus <math>b=2</math> and <math>a=kb^{2}=(4)(2)^{2}=16</math>. which provides 3rd pair <math>(a,b)=(16,2)</math>
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subcase <math>k \ge 5</math>:
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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 approaches <math>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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{{alternate solutions}}

Revision as of 17:39, 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 approaches $1$ 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 is $(a,b)=(1,1); (27,3); (16,2)$




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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