1998 JBMO Problems/Problem 3

Find all pairs of positive integers such that

Solution

We are given the equation: \[ x^y = y^{x - y} \] and asked to find all positive integers $(x,y)$ satisfying it.

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First Approach (Algebraic Parameterization)

Note that $x^y\ge 1$, so $y^{x-y}\ge 1$ as well. This implies $x\ge y$.

Suppose $x=a^{b+c}$, $y=a^c$ for some integer $a>0$ and integers $b,c\ge 1$ with $gcd(b,c)=1$. This ensures that $x\ge y$.

Then the original equation becomes: \[ (a^{b + c})^{a^c} = (a^c)^{a^{b + c} - a^c} \] Taking logarithms base $a$, we get: \[ (b + c) \cdot a^c = c \cdot (a^{b + c} - a^c) \] Divide both sides by $a^c$: \[ b + c = c(a^b - 1) \] Now, since $gcd(b,c)=1$, and $c$ divides the right-hand side, it must divide the left-hand side. Hence, $c=1$. Substituting back: \[ b + 1 = a^b - 1 \Rightarrow a^b = b + 2 \] We now solve for small values of $b$:

- If $b=1$: $a^1=1+2=3\Rightarrow a=3$ - If $b=2$: $a^2=2+2=4\Rightarrow a=2$ - Larger $b$ gives too big values of $a^b$, so no further solutions.

So the only valid pairs $(a,b)$ are $(3,1)$ and $(2,2)$. We now compute:

- For $(a,b)=(3,1)$, $c=1$:

 \(y = a^c = 3\), \(x = a^{b + c} = 3^2 = 9\)

- For $(a,b)=(2,2)$, $c=1$:

 \(y = 2\), \(x = 2^3 = 8\)

So the corresponding solutions are $(x,y)=(9,3)$ and $(8,2)$.

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Second Approach (Logarithmic Substitution)

Assume $x=ky$ for some integer $k>1$. Then: \[ (ky)^y = y^{ky - y} = y^{y(k - 1)} \] Taking natural logarithms: \[ y \ln(ky) = y(k - 1)\ln y \Rightarrow \ln(ky) = (k - 1)\ln y \] Expanding the left-hand side: \[ \ln k + \ln y = (k - 1)\ln y \Rightarrow \ln k = (k - 2)\ln y \Rightarrow \ln y = \frac{\ln k}{k - 2} \]

For $y$ to be an integer, the RHS must be $\ln$ of an integer.

Try small values of $k$: - $k=3$: $\ln y=\frac{\ln 3}{1}=\ln 3\Rightarrow y=3,x=9$ - $k=4$: $\ln y=\frac{\ln 4}{2}=\ln 2\Rightarrow y=2,x=8$ - $k=5$: $\ln y=\frac{\ln 5}{3}\notin \ln (\mathbb{Z})$ ⟹ discard

No other values of $k$ give integer $y$. Also, checking small $y$ values directly:

- $y=1$: Then $x^1=1^{x-1}=1\Rightarrow x=1$ ⟹ $(1,1)$ is a solution.

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Final Answer

The only positive integer solutions to the equation $x^y=y^{x-y}$ are: \[ \boxed{(x, y) = (1, 1),\ (8, 2),\ (9, 3)} \]

We are given the equation: \[ x^y = y^{x - y} \] and asked to find all positive integers $(x,y)$ that satisfy it.

\textbf{Step 1: Try small values of $y$}

We begin by checking small values of $y$:

- If $y=1$, then the equation becomes:

 \[
 x^1 = 1^{x - 1} = 1 \Rightarrow x = 1
 \]
 So \((x, y) = (1, 1)\) is a solution.

- If $y=2$, try $x=8$:

 \[
 x^y = 8^2 = 64,\quad y^{x - y} = 2^6 = 64
 \]
 So \((x, y) = (8, 2)\) is a solution.

- If $y=3$, try $x=9$:

 \[
 x^y = 9^3 = 729,\quad y^{x - y} = 3^6 = 729
 \]
 So \((x, y) = (9, 3)\) is a solution.

\textbf{Step 2: General approach using logarithms}

Assume $x=ky$ for some integer $k>1$. Then the equation becomes: \[ (ky)^y = y^{ky - y} = y^{y(k - 1)} \] Taking natural logarithms: \[ y \ln(ky) = y(k - 1)\ln y \Rightarrow \ln(ky) = (k - 1)\ln y \] Expanding the left-hand side: \[ \ln k + \ln y = (k - 1)\ln y \Rightarrow \ln k = (k - 2)\ln y \Rightarrow \ln y = \frac{\ln k}{k - 2} \] So for $y$ to be an integer, $\frac{\ln k}{k-2}$ must be the logarithm of an integer.

Try small values of $k$: - If $k=3$: $\ln y=\frac{\ln 3}{1}=\ln 3\Rightarrow y=3,x=9$ - If $k=4$: $\ln y=\frac{\ln 4}{2}=\ln 2\Rightarrow y=2,x=8$ - If $k=5$: $\ln y=\frac{\ln 5}{3}\notin \ln (\mathbb{Z})$, so no integer solution for $y$

No other values of $k$ give integer solutions for $y$.

\textbf{Final Answer:} The only positive integer solutions are: \[ \boxed{(x, y) = (1, 1),\ (8, 2),\ (9, 3)} \]

See also

1998 JBMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
All JBMO Problems and Solutions